Acids, Bases and Salts - Class 10 Science - Chapter 2 - Notes, NCERT Solutions & Extra Questions
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Notes - Acids, Bases and Salts | Class 10 NCERT | Science
Understanding the concepts of acids, bases, and salts is crucial for every class 10 student. These fundamental concepts in chemistry not only have significant importance in academics but also play an integral role in our daily lives. Let's delve into the vital aspects of acids, bases, and salts with detailed explanations and practical examples.
Introduction to Acids, Bases, and Salts
Acids, bases, and salts are three broad categories of compounds that are essential in chemistry. Acids are substances that taste sour and change blue litmus paper to red, like lemon juice and vinegar. Bases, on the other hand, are bitter and change red litmus paper to blue, like baking soda and soap. Salts are ionic compounds that result from the neutralization reaction between acids and bases.
Properties of Acids and Bases
Acids and bases have distinct properties:
Acids: Sour taste, turn blue litmus red, react with metals to produce hydrogen gas.
Bases: Bitter taste, turn red litmus blue, feel slippery, and react with acids to produce salt and water.
Identifying Acids and Bases Using Indicators
To identify whether a substance is acidic or basic without tasting it, we use indicators:
Natural Indicators
Litmus: Extracted from lichen, turns red in acids and blue in bases.
Turmeric: Turns reddish-brown in bases and remains yellow in acids.
Red Cabbage Leaves and Flower Petals: Change colors in different pH environments.
Synthetic Indicators
Methyl Orange: Turns red in acidic, yellow in neutral, and orange in basic solutions.
Phenolphthalein: Colorless in acidic and neutral solutions, turns pink in basic solutions.
Reactions Involving Acids and Bases
Reaction with Metals
When acids react with metals, they produce hydrogen gas and a salt.
$$ \text{Acid} + \text{Metal} \rightarrow \text{Salt} + \text{Hydrogen gas} $$
Reaction with Metal Carbonates and Hydrogencarbonates
These reactions produce carbon dioxide, water, and a salt.
$$ \text{Carbonate/Hydrogencarbonate} + \text{Acid} \rightarrow \text{Salt} + \text{CO}_2 + \text{H}_2\text{O} $$
Neutralization Reaction
Acids and bases neutralize each other to form salt and water.
$$ \text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water} $$
Understanding the Chemical Behavior in Aqueous Solutions
Conductivity of Acidic and Basic Solutions
Acids and bases conduct electricity in aqueous solutions due to the presence of ions.
Role of Water in Acid-Base Behavior
Water allows the dissociation of acids and bases into ions, which is crucial for their properties and reactions.
The pH Scale
The pH scale ranges from 0 to 14 and measures the acidity or basicity of a solution.
pH < 7: Acidic
pH = 7: Neutral
pH > 7: Basic
The universal indicator can be used to measure pH, with color changes indicating different pH values.
Importance of pH in Everyday Life
Body pH: Essential for bodily functions, maintained between 7.0 to 7.8.
Acid Rain: pH less than 5.6 can harm aquatic life.
Soil pH: Important for plant growth.
Stomach Acid: Necessary for digestion, but excess acid can cause indigestion.
Tooth Decay: Starts when mouth pH drops below 5.5.
Applications and Importance of Salts
Salts are formed during the neutralization reaction of acids and bases and have various applications:
Sodium Hydroxide
Used in making soap and paper.
Bleaching Powder
Used for bleaching fabrics and disinfecting water.
Baking Soda
Used in cooking and as an antacid.
Washing Soda
Used in glass, soap, and paper industries, and for softening water.
Activities and Practical Experiments
Performing activities to observe acid-base reactions, testing pH, and recognizing chemical changes with indicators adds practical understanding to theoretical knowledge.
Safety and Precautionary Measures
Always handle acids and bases with care:
Add acid to water, not water to acid to prevent exothermic reactions.
Wear protective gear during experiments.
Conclusion
Acids, bases, and salts are fundamental to chemistry and everyday life. Understanding their properties, reactions, and applications helps in grasping more advanced concepts in science. These notes should provide a solid foundation for class 10 students to excel in their studies and apply chemistry knowledge effectively.
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Extra Questions - Acids, Bases and Salts | NCERT | Science | Class 10
Which has the maximum number of molecules?
A. $100 \mathrm{~mL}$ of $0.1 \mathrm{M} \mathrm{HCl}$
B. $50 \mathrm{~mL}$ of $0.5 \mathrm{M}~ \mathrm{H}_{2} \mathrm{SO}_{4}$
C. $10 \mathrm{~mL}$ of $2 \mathrm{M} \mathrm{HNO}_{3}$
D. $200 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{HClO}_{4}$
To determine which solution has the maximum number of molecules, we need to calculate the number of moles of solute in each solution since the number of moles is directly proportional to the number of molecules. The number of moles can be calculated using the formula: $$ \text{number of moles} = Molarity (M) \times Volume (L) $$ where Volume is given in liters. Let’s calculate this for each option:
Option A: $100 , \text{mL}$ of $0.1 , M , \text{HCl}$
Volume in liters = $0.100 , \text{L}$
Number of moles = $(0.1 , \text{M}) \times (0.100 , \text{L}) = 0.010 , \text{moles}$
Option B: $50 , \text{mL}$ of $0.5 , M , \text{H}_2\text{SO}_4$
Volume in liters = $0.050 , \text{L}$
Number of moles = $(0.5 , \text{M}) \times (0.050 , \text{L}) = 0.025 , \text{moles}$
Option C: $10 , \text{mL}$ of $2 , M , \text{HNO}_3$
Volume in liters = $0.010 , \text{L}$
Number of moles = $(2 , \text{M}) \times (0.010 , \text{L}) = 0.020 , \text{moles}$
Option D: $200 , \text{mL}$ of $0.2 , M , \text{HClO}_4$
Volume in liters = $0.200 , \text{L}$
Number of moles = $(0.2 , \text{M}) \times (0.200 , \text{L}) = 0.040 , \text{moles}$
Comparison of the number of moles indicates that option D ($200 , \text{mL}$ of $0.2 , M , \text{HClO}_4$) has the maximum number of molecules because it contains the largest number of moles at $0.040 , \text{moles}$. Hence, the correct answer is:
$$ \textbf{D} \quad 200 , \text{mL} , \text{of} , 0.2 , M , \text{HClO}_4 $$
Which of the following solutions in water will have the lowest vapor pressure?
A) $0.1 , \mathrm{M~ NaCl}$
B) $0.1 , \mathrm{M~ Na}_{3} \mathrm{PO}_{4}$
C) $0.1 , \mathrm{M~ BaCl}_{2}$
D) $0.1 , \mathrm{M}$ Sucrose
The correct answer is Option B: $0.1 , \mathrm{M} , \mathrm{Na}_3 \mathrm{PO}_4$.
Vapor pressure depression is primarily governed by a colligative property that depends on the number of solute particles in solution. More particles result in lower vapor pressure. We need to compare the number of ions each compound dissociates into when dissolved in water:
$\mathrm{NaCl}$: Dissociates into 2 ions: $\mathrm{Na}^+$ and $\mathrm{Cl}^-$.
$\mathrm{Na}_3 \mathrm{PO}_4$: Dissociates into 4 ions: 3 $\mathrm{Na}^+$ and 1 $\mathrm{PO}_4^{3-}$.
$\mathrm{BaCl}_2$: Dissociates into 3 ions: $\mathrm{Ba}^{2+}$ and 2 $\mathrm{Cl}^-$.
Sucrose: Does not dissociate; it remains as 1 molecule.
Here is a summary table for clarity:
Compound | Number of Ions per Formula Unit |
---|---|
$\mathrm{NaCl}$ | 2 |
$\mathrm{Na}_3\mathrm{PO}_4$ | 4 |
$\mathrm{BaCl}_2$ | 3 |
Sucrose | 1 |
$\mathrm{Na}_3\mathrm{PO}_4$ dissociates to produce the highest number of particles (ions) in the solution. Therefore, it will result in the lowest vapor pressure among the given options.
0.1 mole of HCl is added to 2 litres of buffer solution. The decrease in pH of that buffer by 0.5. Then the buffer capacity is:
A. 0.5
B. 0.2
C. 0.1
D. 2
The correct option is 0.1.
Using the formula for buffer capacity:
$$ \phi = \frac{\frac{\text{No. of moles of strong acid}}{\text{strong base added per liter of buffer}}}{\text{change in pH}} $$
We calculate:
Number of moles of HCl added: 0.1 mole
Volume of buffer solution: 2 liters
Change in pH: 0.5
Substitute these values into the formula:
$$ \phi = \frac{\frac{0.1 \text{ moles}}{2 \text{ liters}}}{0.5} $$
Simplify the expression:
$$ \phi = \frac{0.05}{0.5} = 0.1 $$
Thus, the buffer capacity is 0.1.
Phosphoric acid is prepared in a two-step process. Find the mass of $\mathrm{H}{3} \mathrm{PO}{4}$ formed when $124 \mathrm{~g}$ of $\mathrm{P}_{4}$ reacts with an excess of oxygen and then treated with sufficient water.
$$ \begin{array}{l} \mathrm{P}{4} + \mathrm{O}{2} \rightarrow \mathrm{P}{4} \mathrm{O}{10} \ \mathrm{P}{4} \mathrm{O}{10} + \mathrm{H}{2} \mathrm{O} \rightarrow \mathrm{H}{3} \mathrm{PO}_{4} \end{array}$$
(Molar mass of $P=31 \mathrm{~g/mol}$) A $150 \mathrm{~g}$ B $288 \mathrm{~g}$ C $560 \mathrm{~g}$ D $392 \mathrm{~g}$
The correct answer is D: $392 , \mathrm{g}$.
Let's determine the mass of $\mathrm{H}{3} \mathrm{PO}{4}$ formed when $124 , \mathrm{g}$ of $\mathrm{P}_{4}$ reacts with excess oxygen and then with sufficient water.
Given reactions:
$$ \begin{aligned} \quad & \mathrm{P}{4} + \mathrm{O}{2} \rightarrow \mathrm{P}{4} \mathrm{O}{10} \ \quad & \mathrm{P}{4} \mathrm{O}{10} + \mathrm{H}{2} \mathrm{O} \rightarrow \mathrm{H}{3} \mathrm{PO}_{4} \end{aligned} $$
Molar mass of P: $31 , \mathrm{g/mol}$.
First, we calculate the moles of $\mathrm{P}_{4}$: $$ \text{Moles of } \mathrm{P}{4} = \frac{\text{mass of } \mathrm{P}{4}}{\text{molar mass of } \mathrm{P}_{4}} = \frac{124 , \mathrm{g}}{31 , \mathrm{g/mol} \times 4} = 1 , \text{mol} $$
From the reaction stoichiometry: $$ 4 , \text{moles of } \mathrm{P}{4} \Rightarrow 1 , \text{mole of } \mathrm{P}{4} \mathrm{O}_{10} $$
Given $1$ mole of $\mathrm{P}{4}$, it forms $1$ mole of $\mathrm{P}{4} \mathrm{O}_{10}$.
Next, the second reaction indicates: $$ 1 , \text{mole of } \mathrm{P}{4} \mathrm{O}{10} \Rightarrow 4 , \text{moles of } \mathrm{H}{3} \mathrm{PO}{4} $$
Thus, $1$ mole of $\mathrm{P}{4} \mathrm{O}{10}$ forms $4$ moles of $\mathrm{H}{3} \mathrm{PO}{4}$.
Molar mass of $\mathrm{H}{3} \mathrm{PO}{4}$: $$ \mathrm{H}{3} \mathrm{PO}{4} = 3(1) + 31 + 4(16) = 98 , \mathrm{g/mol} $$
Therefore, the mass of $\mathrm{H}{3} \mathrm{PO}{4}$ formed: $$ \text{Mass of } \mathrm{H}{3} \mathrm{PO}{4} = \text{moles of } \mathrm{H}{3} \mathrm{PO}{4} \times \text{molar mass of } \mathrm{H}{3} \mathrm{PO}{4} = 4 , \text{moles} \times 98 , \mathrm{g/mol} = 392 , \mathrm{g} $$
Hence, the mass of phosphoric acid ($\mathrm{H}{3} \mathrm{PO}{4}$) produced is $392 , \mathrm{g}$.
Solubility products of $\mathrm{Al}(\mathrm{OH})_{3}$ and $\mathrm{Zn}(\mathrm{OH})_{2}$ are $2.7 \times 10^{-23}$ and $3.2 \times 10^{-14}$ respectively. If to a solution of $0.1 \mathrm{M}$ each of $\mathrm{Al}^{3+}$ and $\mathrm{Zn}^{2+}$ ions, $\mathrm{NH}_{4} \mathrm{OH}$ is added in increasing amounts which of the following will be precipitated first? A. $\mathrm{Zn}(\mathrm{OH})_{2}$ B. $\mathrm{Al}(\mathrm{OH})_{3}$ C. Both of them D. None of them
The correct answer is B: $\mathrm{Al}(\mathrm{OH})_{3}$.
To determine which compound will precipitate first, we need to compare the solubilities given by the solubility product constants ($ K_{sp} $) for $\mathrm{Al}(\mathrm{OH})_{3}$ and $\mathrm{Zn}(\mathrm{OH})_{2}$.
For $\mathrm{Al}(\mathrm{OH})_{3}$:
The solubility product is given by $ K_{sp} = 2.7 \times 10^{-23} $.
The dissociation of $\mathrm{Al}(\mathrm{OH}){3}$ can be represented as: $$ \mathrm{Al}(\mathrm{OH})_{3(\mathrm{s})} \rightleftharpoons \mathrm{Al}^{3+} + 3\mathrm{OH}^{-} $$
Setting up the expression for $ K_{sp} $: $$ K_{sp} = [\mathrm{Al}^{3+}][\mathrm{OH}^{-}]^3 = 2.7 \times 10^{-23} $$
Assuming the solubility $ S $ of $\mathrm{Al}(\mathrm{OH})_{3}$ in $\mathrm{M}$ is: $$ S \rightarrow [\mathrm{Al}^{3+}] = S \quad \text{and} \quad [\mathrm{OH}^{-}] = 3S $$ $$ K_{sp} = S \cdot (3S)^3 = 27S^4 $$ $$ 27S^4 = 2.7 \times 10^{-23} $$
Solving for $ S $: $$ S = 1.0 \times 10^{-6} \mathrm{M} $$
Therefore, the concentration of $\mathrm{OH}^{-}$ ions required for precipitation is: $$ [\mathrm{OH}^{-}] = 3S = 3.0 \times 10^{-6} \mathrm{M} $$
For $\mathrm{Zn}(\mathrm{OH})_{2}$:
The solubility product is given by $ K_{sp} = 3.2 \times 10^{-14} $.
The dissociation of $\mathrm{Zn}(\mathrm{OH})_{2}$ can be represented as: $$ \mathrm{Zn}(\mathrm{OH})_{2(\mathrm{s})} \rightleftharpoons \mathrm{Zn}^{2+} + 2\mathrm{OH}^{-} $$
Setting up the expression for $ K_{sp} $: $$ K_{sp} = [\mathrm{Zn}^{2+}][\mathrm{OH}^{-}]^2 = 3.2 \times 10^{-14} $$
Assuming the solubility $ S $ of $\mathrm{Zn}(\mathrm{OH})_{2}$ in $\mathrm{M}$ is: $$ S \rightarrow [\mathrm{Zn}^{2+}] = S \quad \text{and} \quad [\mathrm{OH}^{-}] = 2S $$ $$ K{sp} = S \cdot (2S)^2 = 4S^3 $$ $$ 4S^3 = 3.2 \times 10^{-14} $$
Solving for $ S $: $$ S = 2.0 \times 10^{-5} \mathrm{M} $$
Therefore, the concentration of $\mathrm{OH}^{-}$ ions required for precipitation is: $$ [\mathrm{OH}^{-}] = 2S = 4.0 \times 10^{-5} \mathrm{M} $$
Conclusion:
Since $\mathrm{Al}(\mathrm{OH})_{3}$ requires a lower concentration of $\mathrm{OH}^{-}$ ions ($3.0 \times 10^{-6} \mathrm{M}$) to start precipitating compared to $\mathrm{Zn}(\mathrm{OH})_{2}$ ($4.0 \times 10^{-5} \mathrm{M}$), $\mathrm{Al}(\mathrm{OH})_{3}$ will precipitate first when $\mathrm{NH}_4\mathrm{OH}$ is added.
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Ask Chatterbot AINCERT Solutions - Acids, Bases and Salts | NCERT | Science | Class 10
A solution turns red litmus blue, its $\mathrm{pH}$ is likely to be
(a) 1
(b) 4
(c) 5
(d) 10
A solution that turns red litmus paper blue is basic in nature. The pH scale ranges from 0 to 14, with 7 being neutral. A pH value less than 7 indicates acidity, and a pH value greater than 7 indicates basicity (alkalinity).
Among the options given:
(a) pH 1 is highly acidic.
(b) pH 4 is also acidic.
(c) pH 5 is still acidic, although closer to neutral.
(d) pH 10 is basic.
Therefore, a solution that turns red litmus blue is likely to have a basic pH. The correct answer is:
(d) 10
A solution reacts with crushed egg-shells to give a gas that turns lime-water milky. The solution contains
(a) $\mathrm{NaCl}$
(b) $\mathrm{HCl}$
(c) $\mathrm{LiCl}$
(d) $\mathrm{KCl}$
Crushed egg-shells primarily consist of calcium carbonate (( \mathrm{CaCO_3} )). When calcium carbonate reacts with an acidic solution, it produces carbon dioxide (( \mathrm{CO_2} )), which can turn lime-water milky due to the formation of calcium carbonate precipitate. The reaction is as follows:
$$ \mathrm{CaCO_3} \ (s) + 2 \mathrm{HCl} \ (aq) \rightarrow \mathrm{CaCl_2} \ (aq) + \mathrm{H_2O} \ (l) + \mathrm{CO_2} \ (g) $$
The gas produced (carbon dioxide) reacts with lime-water (( \mathrm{Ca(OH)_2} )) to form the milky precipitate of calcium carbonate:
$$ \mathrm{CO_2} \ (g) + \mathrm{Ca(OH)_2} \ (aq) \rightarrow \mathrm{CaCO_3} \ (s) + \mathrm{H_2O} \ (l) $$
From the given options, the solution that is an acid and can release ( \mathrm{H^+} ) ions to react with calcium carbonate is hydrochloric acid (( \mathrm{HCl} )).
Therefore, the solution that contains ( \mathrm{HCl} ) will have this reaction. The correct answer is (b) ( \mathrm{HCl} ).
$10 \mathrm{~mL}$ of a solution of $\mathrm{NaOH}$ is found to be completely neutralised by $8 \mathrm{~mL}$ of a given solution of $\mathrm{HCl}$. If we take $20 \mathrm{~mL}$ of the same solution of $\mathrm{NaOH}$, the amount $\mathrm{HCl}$ solution (the same solution as before) required to neutralise it will be
(a) $4 \mathrm{~mL}$
(b) $8 \mathrm{~mL}$
(c) $12 \mathrm{~mL}$
(d) $16 \mathrm{~mL}$
This problem is based on the concept of chemical stoichiometry wherein the reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) is considered. The reaction follows the equation:
$$ \mathrm{NaOH} + \mathrm{HCl} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} $$
This is a neutralization reaction where one mole of NaOH reacts with one mole of HCl to produce one mole of NaCl and one mole of water.
According to the given data, 10 mL of NaOH solution is neutralized by 8 mL of HCl solution. This indicates that the volume ratio of NaOH to HCl for complete neutralization is 10:8 or 1:0.8.
If we double the volume of NaOH solution to 20 mL, we would need to double the amount of HCl from the original 8 mL to maintain the same ratio for complete neutralization.
Hence, the answer is: $$ 8 \mathrm{~mL} \times 2 = 16 \mathrm{~mL} $$
So, the correct option is:
(d) $16 \mathrm{~mL}$
Which one of the following types of medicines is used for treating indigestion?
(a) Antibiotic
(b) Analgesic
(c) Antacid
(d) Antiseptic
The type of medicine used for treating indigestion is:
(c) Antacid
Antacids are medicines that neutralize the excess acid in the stomach, which is often the cause of indigestion.
Write word equations and then balanced equations for the reaction taking place when -
(a) dilute sulphuric acid reacts with zinc granules.
(b) dilute hydrochloric acid reacts with magnesium ribbon.
(c) dilute sulphuric acid reacts with aluminium powder.
(d) dilute hydrochloric acid reacts with iron filings.
(a) Dilute sulfuric acid reacts with zinc granules.
Word equation: Sulfuric acid + Zinc → Zinc sulfate + Hydrogen gas
Balanced chemical equation: $$ \text{H}_2\text{SO}_4(aq) + \text{Zn}(s) \rightarrow \text{ZnSO}_4(aq) + \text{H}_2(g) $$
(b) Dilute hydrochloric acid reacts with magnesium ribbon.
Word equation: Hydrochloric acid + Magnesium → Magnesium chloride + Hydrogen gas
Balanced chemical equation: $$ 2\text{HCl}(aq) + \text{Mg}(s) \rightarrow \text{MgCl}_2(aq) + \text{H}_2(g) $$
(c) Dilute sulfuric acid reacts with aluminium powder.
Word equation: Sulfuric acid + Aluminium → Aluminium sulfate + Hydrogen gas
Balanced chemical equation: $$ 3\text{H}_2\text{SO}_4(aq) + 2\text{Al}(s) \rightarrow \text{Al}_2(\text{SO}_4)_3(aq) + 3\text{H}_2(g) $$
(d) Dilute hydrochloric acid reacts with iron filings.
Word equation: Hydrochloric acid + Iron → Iron(II) chloride + Hydrogen gas
Balanced chemical equation: $$ 6\text{HCl}(aq) + 2\text{Fe}(s) \rightarrow 2\text{FeCl}_3(aq) + 3\text{H}_2(g) $$
Compounds such as alcohols and glucose also contain hydrogen but are not categorised as acids. Describe an Activity to prove it.
To demonstrate that compounds such as alcohols and glucose contain hydrogen but are not categorized as acids, you can conduct a simple activity:
Prepare two separate beakers with solutions of glucose and alcohol respectively.
Set up a simple circuit with a light bulb and a battery.
Now pour some dilute hydrochloric acid (HCl) in a separate beaker and connect this acidic solution into the circuit and switch on the current. Observe whether the light bulb glows.
Repeat the experiment by replacing the acidic solution with glucose solution and then with the alcohol solution, and observe whether the light bulb glows in these cases.
In the acidic solution, you will observe that the light bulb starts glowing, indicating that there is a flow of electric current through the solution. This is because the electric current is carried through the acidic solution by ions $ \text{H}^+ $) as cation and an anion such as $ \text{Cl}^- $ in HCl. On the other hand, with glucose and alcohol solutions, you will observe that the light bulb does not glow, indicating that these solutions do not conduct electricity. This indicates that, even though they contain hydrogen, they do not ionize in solution to form $ \text{H}^+ $ ions, and therefore are not acids .
Why does distilled water not conduct electricity, whereas rain water does?
Distilled water does not conduct electricity because it is pure water without any impurities. Pure water is a very weak conductor of electricity. For water to conduct electricity, it needs to have ions present in it, which act as charge carriers. Ions are atoms or molecules that have lost or gained electrons, resulting in a net positive or negative charge.
In contrast, rainwater conducts electricity because it is not pure; it contains dissolved salts and minerals. As rainwater falls through the atmosphere, it absorbs carbon dioxide and other gases, which can react with water to form weak acidic solutions like carbonic acid. This results in the production of ions in the water. Additionally, rain can pick up other contaminants and particles as it falls, all of which can contribute ions to the water. These ions provide charge carriers that can move through the water under the influence of an electric field, thus allowing the water to conduct electricity.
Why do acids not show acidic behaviour in the absence of water?
Acids do not show acidic behavior in the absence of water because the separation of $ H^+ $ ions from acid molecules cannot occur without the presence of water. For instance, $ H^+ $ ions in HCl are produced in the presence of water, as shown by the chemical reaction:
$$ HCl + H_2O \rightarrow H_3O^+ + Cl^- $$
Hydrogen ions $ (H^+) $ cannot exist alone; they are only produced when acids are dissolved in water, forming hydronium ions $ (H_3O^+) $. The hydronium ions, which are essentially hydrated hydrogen ions, are responsible for the acidic properties of a solution. Hence, in the absence of water, these hydronium ions do not form, and the substance does not exhibit acidic properties .
Five solutions $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ and $\mathrm{E}$ when tested with universal indicator showed $\mathrm{pH}$ as $4,1,11,7$ and 9 , respectively. Which solution is
(a) neutral?
(b) strongly alkaline?
(c) strongly acidic?
(d) weakly acidic?
(e) weakly alkaline?
Arrange the $\mathrm{pH}$ in increasing order of hydrogen-ion concentration.
The pH scale is a measure of the acidity or alkalinity of a solution. It ranges from 0 to 14, where 7 is neutral, values below 7 are acidic, and values above 7 are alkaline (or basic). The lower the pH, the higher the hydrogen ion concentration, and the higher the pH, the lower the hydrogen ion concentration.
Based on the pH values given for solutions A, B, C, D, and E, we can categorize them as follows:
(a) Neutral: A solution with a pH of 7 is neutral, which means it has neither acidic nor alkaline properties.
Solution D has a pH of 7, so it is neutral.
(b) Strongly alkaline: A solution with a high pH value that is significantly above 7 is considered strongly alkaline.
Solution C has a pH of 11, so it is strongly alkaline.
(c) Strongly acidic: A solution with a low pH value that is significantly below 7 is considered strongly acidic.
Solution B has a pH of 1, so it is strongly acidic.
(d) Weakly acidic: A solution with a pH value just below 7 is considered weakly acidic.
Solution A has a pH of 4, so it is weakly acidic.
(e) Weakly alkaline: A solution with a pH value just above 7 is considered weakly alkaline.
Solution E has a pH of 9, so it is weakly alkaline.
To arrange the pH in increasing order of hydrogen-ion concentration, we list the solutions from the highest pH (lowest hydrogen-ion concentration) to the lowest pH (highest hydrogen-ion concentration):
Solution C (pH 11)
Solution E (pH 9)
Solution D (pH 7 - neutral, has the reference concentration of hydrogen ions)
Solution A (pH 4)
Solution B (pH 1)
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid $(\mathrm{HCl})$ is added to test tube $\mathrm{A}$, while acetic acid $\left(\mathrm{CH}_{3} \mathrm{COOH}\right)$ is added to test tube B. Amount and concentration taken for both the acids are same. In which test tube will the fizzing occur more vigorously and why?
Fizzing during a reaction generally indicates the release of a gas. In this case, when magnesium reacts with an acid, hydrogen gas is released, which causes the fizzing. The reaction between magnesium ribbon and hydrochloric acid ($ \text{HCl} $) is generally more vigorous than the reaction with acetic acid ($ \text{CH}_3\text{COOH} $) due to the differences in the acids' properties.
Hydrochloric acid is a strong acid, which means it completely ionizes in water and releases a large number of hydrogen ions ($ \text{H}^+ $) that can react with magnesium. On the other hand, acetic acid is a weak acid, which means it only partially ionizes in water, producing fewer hydrogen ions available for the reaction with magnesium.
Given that the concentration and amount of the acids are the same, test tube A with hydrochloric acid will have a higher concentration of hydrogen ions compared to test tube B with acetic acid. As a result, the reaction with hydrochloric acid in test tube A will be more vigorous, leading to more robust fizzing due to the rapid production of hydrogen gas.
The reactions can be represented as follows:
For hydrochloric acid: $$ \text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \uparrow $$
For acetic acid: $$ \text{Mg} + 2\text{CH}_3\text{COOH} \rightarrow \text{(CH}_3\text{COO)}_2\text{Mg} + \text{H}_2 \uparrow $$
It is important to note that the upward arrow indicates the production of hydrogen gas escaping from the solution.
Fresh milk has a pH of 6 . How do you think the $\mathrm{pH}$ will change as it turns into curd? Explain your answer.
When fresh milk turns into curd (yogurt), the pH tends to decrease, meaning it becomes more acidic. This occurs due to the fermentation process, where lactic acid bacteria present in the milk convert lactose, the sugar in milk, into lactic acid. As the amount of lactic acid increases in the milk, the pH decreases below 6, which contributes to the sour taste of the curd. Consequently, you would expect the pH of the curd to be lower than that of fresh milk.
A milkman adds a very small amount of baking soda to fresh milk.
(a) Why does he shift the $\mathrm{pH}$ of the fresh milk from 6 to slightly alkaline?
(b) Why does this milk take a long time to set as curd?
(a) A milkman might add a very small amount of baking soda (sodium bicarbonate, NaHCO₃) to fresh milk to shift its pH from 6, which is slightly acidic, to a slightly alkaline level. This is done primarily to prevent the milk from souring quickly. Fresh milk contains lactose, which over time is broken down by bacteria into lactic acid. This process lowers the pH of the milk, leading to souring and eventual curdling. By increasing the pH slightly above neutral (pH 7), the milkman inhibits the growth of these bacteria, thereby slowing down the formation of lactic acid and prolonging the milk's shelf life.
(b) The milk takes a long time to set as curd because the process of curdling involves the conversion of lactose into lactic acid by lactic acid bacteria, which thrive in slightly acidic conditions. When milk is slightly alkaline due to the addition of baking soda, the growth of these bacteria is inhibited, and hence the production of lactic acid is slowed. Since it is the lactic acid that causes the milk proteins (casein) to coagulate and form curd, a higher pH will slow down the curdling process. As a result, it will take a longer time for the milk to set as curd under these conditions.
Plaster of Paris should be stored in a moisture-proof container. Explain why?
Plaster of Paris (chemical formula: $\text{CaSO}_4\cdot\frac{1}{2}\text{H}_2\text{O}$) should be stored in a moisture-proof container because it is a hygroscopic substance, meaning it readily absorbs water from the air. When Plaster of Paris comes into contact with moisture, it undergoes a chemical reaction, known as hydration, and converts into gypsum (chemical formula: $\text{CaSO}_4\cdot2\text{H}_2\text{O}$), which is hard.
The reaction is:
$$\text{CaSO}_4\cdot\frac{1}{2}\text{H}_2\text{O} + \frac{3}{2}\text{H}_2\text{O} \rightarrow \text{CaSO}_4\cdot2\text{H}_2\text{O}$$
This is an exothermic reaction, which means it releases heat. The formed gypsum is a hard solid that is not useful as a plaster. Therefore, in order to maintain the effectiveness and workability of Plaster of Paris, it must be kept dry until it is ready to be used. Moisture-proof containers prevent the entry of moisture and ensure that the Plaster of Paris retains its properties and is ready to be mixed with water in controlled quantities for use.
What is a neutralisation reaction? Give two examples.
A neutralisation reaction is a chemical process in which an acid and a base react with each other to form a salt and water. This reaction effectively "neutralizes" the acid and base, producing substances with neutral pH.
Here are two examples of neutralisation reactions:
When sodium hydroxide (a base) reacts with hydrochloric acid (an acid), it forms sodium chloride (a common table salt) and water: $$\text{NaOH} , (aq) + \text{HCl} , (aq) \rightarrow \text{NaCl} , (aq) + \text{H}_2\text{O} , (l)$$
The reaction between copper oxide (a metallic oxide) and dilute hydrochloric acid (an acid) leads to the formation of copper(II) chloride (a salt) and water: $$\text{Copper oxide} + \text{Hydrochloric acid} \rightarrow \text{Copper(II) chloride} + \text{Water}$$
These reactions demonstrate the general form of a neutralisation reaction: $$\text{Base} + \text{Acid} \rightarrow \text{Salt} + \text{Water}$$ .
Give two important uses of washing soda and baking soda.
Important uses of washing soda and baking soda are as follows:
Washing Soda (Sodium Carbonate) uses:
Used in the glass, soap, and paper industries.
Used for removing permanent hardness of water.
Baking Soda (Sodium Hydrogencarbonate) uses:
For making baking powder, which helps bread and cakes to rise, making them soft and spongy.
As an ingredient in antacids, neutralizing excess stomach acid and providing relief.
These uses are outlined in the provided NCERT class 10 Science book .
The blue-colored copper sulfate crystals turn white when treated with sulfuric acid. This is due to the nature of the acid.
A) dehydrating
B) oxidizing
C) reducing
D) color changing
The correct answer is: A) dehydrating
Sulfuric acid is recognized as a potent dehydrating agent. The striking blue color of copper sulfate crystals ($\mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}$) is primarily due to the water of crystallization embedded within its structure. When sulfuric acid is added to these crystals, it effectively removes the water of crystallization, converting copper sulfate to its anhydrous form, which is white.
The chemical reaction can be represented as follows:
$$ \mathrm{CuSO}_4 \cdot 5\mathrm{H}_2\mathrm{O}(\mathrm{s}) + \text{Conc.} \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{CuSO}_4(\mathrm{s}) + 5\mathrm{H}_2\mathrm{O}(\mathrm{g}) $$
This transformation leads to the color change from blue to white, confirming the dehydrating action of sulfuric acid.
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Ask Chatterbot AIExtra Questions and Answers - Acids, Bases and Salts | NCERT | Science | Class 10
Calculate the molarity of $\mathrm{KOH}$ in a solution prepared by dissolving $5.6 \mathrm{g}$ in enough water to form $500 \mathrm{mL}$ of solution.
A) $2 \mathrm{M}$
B) $0.2 \mathrm{M}$
C) $5.6 \mathrm{M}$
D) $500 \mathrm{M}$
The correct answer is Option B: $0.2 \mathrm{M}$.
Step 1: Calculate the molar mass of $\mathrm{KOH}$.
The atomic masses are as follows: K = 39 g/mol, O = 16 g/mol, H = 1 g/mol.
Therefore, the molar mass of $\mathrm{KOH}$ is: $$ 39 \text{ g/mol} + 16 \text{ g/mol} + 1 \text{ g/mol} = 56 \text{ g/mol} $$
Step 2: Determine the number of moles of $\mathrm{KOH}$.
Use the formula for number of moles, $$ \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.6 \text{ g}}{56 \text{ g/mol}} = 0.1 \text{ mol} $$
Step 3: Calculate the molarity of the solution.
Convert the solution volume from milliliters to liters: $$ 500 \text{ mL} = 500 \times 10^{-3} \text{ L} = 0.5 \text{ L} $$
Molarity is calculated as: $$ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume of solution in liters}} = \frac{0.1 \text{ mol}}{0.5 \text{ L}} = 0.2 \text{ M} $$
Thus, the molarity of the $\mathrm{KOH}$ solution is 0.2 M.
State whether true or false: There are two types of salts of $\mathrm{CO}_{2}$.
A) True
B) False
The correct answer is A) True.
There are indeed two types of salts of $\mathrm{CO}_{2}$:
Carbonates
Hydrogen Carbonates
You have been provided with three test tubes. One of them contains distilled water, and the other two have an acidic solution and a basic solution, respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?"
Start by testing each test tube: Insert the provided red litmus paper into each test tube. The test tube containing the basic solution will turn the red litmus paper blue.
Next, check for the acidic solution: After identifying the basic solution, use the now blue litmus paper and introduce it into the other two test tubes. The test tube that changes the blue litmus paper back to red contains the acidic solution.
Identify distilled water: The test tube, which has no effect on the litmus paper, regardless of its colour (whether it stays red or blue), contains distilled water.
Alternatively, the identification can be summarized as:
Distilled Water: No change in the colour of the red litmus paper.
Acidic Solution: If it's the only test tube left unidentified and does not change the colour of the paper, this indicates an acidic environment by process of elimination.
Basic Solution: Confirmed by the change of red litmus paper to blue.
Describe how you could show that phosphorous acid, $\mathrm{H}_{3} \mathrm{PO}_{3}$, is a weaker acid than hydrochloric acid.
To demonstrate that phosphorous acid ($\mathrm{H}_3\mathrm{PO}_3$) is a weaker acid compared to hydrochloric acid ($\mathrm{HCl}$), one can consider the trend in acid strength across the periodic table.
Acid strength generally increases as we move across a period in the periodic table. This is attributed to the increase in electronegativity as one goes from left to right within a period. Higher electronegativity in an element such as chlorine (than phosphorous) results in a stronger attraction to the shared electron pair in the hydrogen-atom bond.
In $\mathrm{HCl}$, chlorine's higher electronegativity compared to phosphorus in $\mathrm{H}_3\mathrm{PO}_3$, allows the bond between hydrogen and chlorine to weaken more easily, facilitating faster proton (H+) dissociation.
Therefore, due to the stronger electronegativity and capability to dissociate protons more readily, $\mathrm{HCl}$ is demonstrated to be a stronger acid than phosphorous acid.
Colourless compound X obtained from seawater used in daily meals is taken in a test tube. Concentrated H2SO4 is added to the test tube. A pungent-smelling gas Y comes out, which does not affect dry blue litmus paper but turns moist blue litmus paper red. Identify X and Y. Write the chemical reactions involved.
The colourless compound X identified from the scenario is sodium chloride (NaCl), obtained from seawater and commonly used as salt in meals. When sodium chloride reacts with concentrated sulfuric acid (H₂SO₄), a chemical reaction occurs producing hydrochloric acid gas (HCl). This gas Y is what emerges from the reaction mixture and demonstrates its acidic properties by turning moist blue litmus paper red, while having no effect on dry litmus paper. The chemical equation representing this reaction is:
$$ \text{NaCl}_{(\text{s})} + \text{H}2\text{SO}4{(\text{l})} \rightarrow \text{HCl}{(\text{g})} + \text{NaHSO}4{(\text{s})} $$
In this equation, $\text{NaCl (sodium chloride)}$ and $\text{H₂SO₄ (sulfuric acid)}$ react to produce $\text{HCl (hydrochloric acid gas)}$ and $\text{NaHSO₄ (sodium bisulfate)}$, highlighting that the HCl gas has acidic properties, evident from its effect on moist blue litmus paper.
Identify A, B, & C in the given figure (Krebs cycle).
A. Oxaloacetic acid, B. Succinic acid, C. Alpha-ketoglutaric acid
B. Succinic acid, B. Alpha-ketoglutaric acid, C. Oxaloacetic acid
C. Alpha-ketoglutaric acid, B. Oxaloacetic acid, C. Succinic acid
D. Oxaloacetic acid, B. Alpha-ketoglutaric acid, C. Succinic acid
Correct Choice: D - A: Oxaloacetic Acid, B: Alpha-ketoglutaric Acid, C: Succinic Acid
Explanation:
In the Krebs cycle, Acetyl CoA (a 2C molecule) combines with Oxaloacetic Acid (a 4C molecule) initiating the cycle by forming Citric Acid (a 6C molecule).
Citric Acid is oxidized, reducing NAD+ to NADH and losing a carbon atom to produce CO₂, resulting in the formation of Alpha-ketoglutaric Acid (a 5C molecule).
Alpha-ketoglutaric Acid further breaks down, again losing a carbon and forming Succinic Acid (a 4C molecule), while another molecule of NAD+ is reduced to NADH. Concurrently, FAD is reduced to FADH₂ as part of these reactions.
The cycle continues, eventually regenerating Oxaloacetic Acid, thus enabling continuous operation of the Krebs cycle.
These conversions confirm that the components at points A, B, and C in the given diagram are Oxaloacetic Acid, Alpha-ketoglutaric Acid, and Succinic Acid, respectively.
Which of the following substances is generally present in common salt?
A. $\mathrm{Na}$
B. $\mathrm{Cl}$
C. 1
D. $\mathrm{Br}$
The correct option is B. $\mathrm{Cl}$
Common salt is primarily composed of sodium chloride ($\mathrm{NaCl}$), where $\mathrm{Na}$ represents sodium and $\mathrm{Cl}$ represents chlorine. Both substances are critical in common salt, but $\mathrm{Cl}$ (chlorine) is specifically noted here as a major component. The inclusion of iodine in table salt is for health reasons (iodized salt), to provide necessary iodine to the diet, but it is not a primary component of common salt. Therefore, $\mathrm{Cl}$ is the appropriate choice from the options given.
How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO3? The concentrated acid is 70% HNO3.
A. 70.0 g conc. HNO3 B. 54.0 g conc. HNO3 C. 45.0 g conc. HNO3 D. 90.0 g conc. HNO3
The correct option is C) 45.0 g concentrated $\mathrm{HNO_3}$.
First, we calculate the mass of pure $\mathrm{HNO_3}$ needed using the formula for molarity: $$ \text{Molarity} = \frac{W \times 1000}{M_{w} \times \text{Sol}_{\text{(ml)}}} $$ Here:
$W$ is the mass of the solute (nitric acid) in grams,
$M_w$ is the molar mass of $\mathrm{HNO_3}$ approximately $63 , \mathrm{g/mol}$,
$\text{Sol}_{\text{(ml)}}$ is the volume of the solution in milliliters.
Given the molarity is $2.0 , \mathrm{M}$ and $\text{Sol}_{\text{(ml)}} = 250 , \mathrm{ml}$, plugging in the values we get: $$ 2 = \frac{W \times 1000}{63 \times 250} $$
Solving for $W$: $$ W = \frac{2 \times 63 \times 250}{1000} = 31.5 , \mathrm{g} $$
Now we consider the concentration of the nitric acid solution which is 70%. Thus, the actual mass of the concentrated solution needed is calculated as: $$ \text{Mass of concentrated acid} = \frac{\text{Mass of pure HNO}_3}{\text{Fraction of HNO}_3} = \frac{31.5}{0.70} = 45.0 , \mathrm{g} $$
Therefore, to prepare $250 , \mathrm{mL}$ of $2.0 , \mathrm{M} , \mathrm{HNO}_3$, 45.0 g of concentrated nitric acid is required. The answer is option C.
$9.8 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{SO}_{4}$ is present in 2 liters of a solution. The molarity of the solution is:
A) $0.1 \mathrm{M}$
B) $0.05 \mathrm{M}$
C) $0.2 \mathrm{M}$
D) $0.01 \mathrm{M}$
The correct answer is Option B) 0.05 M.
To determine the molarity of the solution, we start by calculating the number of moles of $\mathrm{H}_2\mathrm{SO}_4$. Molar mass of $\mathrm{H}_2\mathrm{SO}_4$ = $2 \times 1 + 32 + 4 \times 16 = 98$ g/mol. The amount of $\mathrm{H}_2\mathrm{SO}_4$ given is $9.8$ g.
Number of moles of $\mathrm{H}_2\mathrm{SO}_4$: $$ \text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{9.8 , \text{g}}{98 , \text{g/mol}} = 0.1 , \text{mol} $$
We know that molarity ($ M $) is defined as the number of moles of solute per liter of solution. Since the volume of the solution is $2$ liters, the molarity is: $$ M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} = \frac{0.1 , \text{mol}}{2 , \text{L}} = 0.05 , \text{M} $$
Thus, the molarity of the solution is 0.05 M.
When water is added to magnesium oxide, magnesium hydroxide is produced, which is also present in lime water. So, is it correct that the magnesium hydroxide present in lime water and the magnesium hydroxide formed from magnesium oxide and water are one and the same?
When magnesium oxide (MgO) is dissolved in water, a chemical reaction occurs, producing magnesium hydroxide (Mg(OH)₂). The reaction can be described by the chemical equation:
$$ \text{MgO} + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2 $$
This implies a chemical change where a new substance, magnesium hydroxide, is formed.
Now, turning to lime water, it's crucial to clarify that lime water traditionally contains calcium hydroxide, Ca(OH)₂, not magnesium hydroxide. Thus, although both are hydroxides, they are chemically distinct compounds. Therefore, the magnesium hydroxide present when magnesium oxide reacts with water is not the same as the compound in lime water.
Acid + Base $\rightarrow \mathrm{X} + \mathrm{Y}$ What are X and Y?
A $\mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}$
B Salt + $\mathrm{H}_{2}$
C Salt + $\mathrm{H}_{2} \mathrm{O}$ (D) $\mathrm{CO}_{2} + \mathrm{H}_{2}$
The correct answer is Option C: Salt + $\mathrm{H}_2\mathrm{O}$.
The reaction between an acid and a base typically results in the formation of a salt and water. This type of reaction is known as a neutralization reaction. Here's how it generally looks: $$ \text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water} $$ Thus, when an acid reacts with a base, the products formed are a salt and water.
How is sodium stored?
Sodium is an extremely reactive metal, which makes its storage quite critical. To prevent it from reacting with oxygen and moisture in the air, sodium is typically stored in kerosene. Exposure to moisture would lead to a chemical reaction where sodium forms sodium hydroxide ($ \text{NaOH} $), accompanied by a release of substantial heat. This reaction is strongly exothermic, meaning that a lot of heat is generated in the process.
The acid dissociation constant of $\mathrm{H}_{2} \mathrm{S}$ and $\mathrm{HS}^{-}$ are $10^{-9}$ and $10^{-13}$ respectively. The $\mathrm{pH}$ of a $0.1$ M aqueous solution of $\mathrm{H}_{2} \mathrm{S}$ will be:
A) 2
B) 3
C) 4
D) 5
The correct answer is D) 5.
To find the $\mathrm{pH}$ of a $\mathbf{0.1}$ M solution of $\mathrm{H}_{2} \mathrm{S}$:
We use the acid dissociation constants:
$\mathrm{K}_{\mathrm{a1}}$ for $\mathrm{H}_{2}\mathrm{S}$ is $\mathbf{10^{-9}}$
$\mathrm{K}_{\mathrm{a2}}$ for $\mathrm{HS}^{-}$ is $\mathbf{10^{-13}}$
Application of the formula for the calculation of $\mathrm{pH}$ considering the first dissociation of $\mathrm{H}{2}\mathrm{S}$: $$ \mathrm{pH} = \frac{1}{2}\left[\mathrm{pK}{\mathrm{a1}} - \log [\mathrm{initial~concentration}]\right] $$
$\mathrm{pK}_{\mathrm{a1}} = -\log \mathrm{K}_{\mathrm{a1}} = -\log(10^{-9}) = 9$
$\log [\mathrm{initial~concentration}] = \log [0.1] = -1$
Putting values into the formula: $$ \mathrm{pH} = \frac{1}{2}\left[9 + 1\right] = \frac{1}{2}[10] = 5 $$
Thus, the $\mathrm{pH}$ of the solution is 5.
$\mathrm{P}_{4} \mathrm{O}_{6}$ is hydrolysed to
(A) $\left(\mathrm{HPO}_{3}\right){n}$
(B) $\mathrm{H}_{3} \mathrm{PO}_{4}$
(C) $\mathrm{H}_{3} \mathrm{PO}_{3}$
(D) $\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}$
The correct option is (C).
Phosphorus trioxide ($\mathrm{P}_4\mathrm{O}_6$) undergoes hydrolysis as shown in the equation below:
$$ \mathrm{P}_4\mathrm{O}_6 + 6\mathrm{H}_2\mathrm{O} \rightarrow 4\mathrm{H}_3\mathrm{PO}_3 $$
Through this reaction, each molecule of $\mathrm{P}_4\mathrm{O}_6$ reacts with water to produce four molecules of phosphorous acid ($\mathrm{H}_3\mathrm{PO}_3$). Thus, the product of hydrolyzing $\mathrm{P}_4\mathrm{O}_6$ is $\mathrm{H}_3\mathrm{PO}_3$.
Calculate the molality of a 1-liter solution of 98% $\mathrm{H}_2\mathrm{SO}_4$ (weight/volume). The density of the solution is $1.88 , \mathrm{g/mL}$.
To determine the molality of the given $98% , (w/v)$ sulfuric acid ($\mathrm{H}_2\mathrm{SO}_4$) solution with a density of $1.88 , \mathrm{g/mL}$, we shall follow these calculations:
Concentration Specification: $98% , (w/v)$ means that there are $98 , \mathrm{g}$ of $\mathrm{H}_2\mathrm{SO}_4$ in every $100 , \mathrm{mL}$ of the solution.
Calculate the total mass of the solution for $1000 , \mathrm{mL}$ (which is 1 liter): $$ \text{Mass of the solution} = 1000 , \mathrm{mL} \times 1.88 , \mathrm{g/mL} = 1880 , \mathrm{g} $$
Calculate the mass of $\mathrm{H}_2\mathrm{SO}_4$ in $1000 , \mathrm{mL}$: $$ \text{Mass of } \mathrm{H}_2\mathrm{SO}_4 = 980 , \mathrm{g} \quad \text{(proportional to the concentration in 1000 mL)} $$
Deduce the mass of the solvent (water in this case): $$ \text{Mass of solvent} = \text{Total mass} - \text{Mass of } \mathrm{H}_2\mathrm{SO}_4 = 1880 , \mathrm{g} - 980 , \mathrm{g} = 900 , \mathrm{g} = 0.90 , \mathrm{kg} $$
Calculate the number of moles of $\mathrm{H}_2\mathrm{SO}_4$:
Molar mass of $\mathrm{H}_2\mathrm{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 , \mathrm{g/mol}$ $$ \text{Number of moles} = \frac{980 , \mathrm{g}}{98 , \mathrm{g/mol}} = 10 , \text{moles} $$
Calculate the molality (moles of solute per kilogram of solvent): $$ \text{Molality} = \frac{10 , \text{moles}}{0.90 , \mathrm{kg}} \approx 11.11 , \mathrm{m} $$
Thus, the molality of the sulfuric acid solution is 11.11 m.
Acid rain is considered to be one of the major causes for discoloration of the Taj Mahal. Say True or False.
A) True
B) False
The correct answer is A) True.
Taj Mahal is constructed from marble, which contains calcium. When sulfur and nitrogen pollutants enter the atmosphere, they transform the rainwater into acid rain, particularly forming sulphurous and nitrous acids. This acidic rain reacts with the marble, leading to the discoloration of the Taj Mahal.
The historical recorded use of litmus dyes was by the Spanish alchemist around AD.
A) Arnaldus de Villa Nova, 300 BC
B) Raymond Lulli, 1300 BC
C) Arnaldus de Villa Nova, 1300 AD
D) Raymond Lulli, 1300 AD
Correct Answer: C) Arnaldus de Villa Nova, 1300 AD
Explanation: The documented historical use of litmus dyes traces back to the Spanish alchemist Arnaldus de Villa Nova. This development occurred around the year 1300 AD. Thus, Arnaldus de Villa Nova was integral to early experiments involving these dyes during that period.
Nitric acid turns blue litmus red.
A) True
B) False
The correct answer is A) True.
Nitric acid, like all acids, turns blue litmus paper red. This characteristic reaction is utilized in the litmus test, which is a basic method to determine whether a substance is acidic.
Is toothpaste acidic or alkaline?
Toothpaste is typically alkaline in nature, with an average pH of about 8. This slightly alkaline property arises because the various components in toothpaste are designed to neutralize the acids in the mouth. These acids are produced by bacteria that break down food particles. By brushing our teeth with alkaline toothpaste, the acidic environment created after eating can be neutralized effectively.
What is the accurate representation of a Hydrated Proton in aqueous solution?
(A) $\mathrm{H}_{9}\mathrm{O}_{5}^{+}$
(B) $\mathrm{H}_{9}\mathrm{O}_{4}^{-}$
(C) $\mathrm{H}^{+}$
(D) $\mathrm{H}_{3}\mathrm{O}^{+}$
The correct option is (D): $$ \mathrm{H}_{3} \mathrm{O}^{+} $$
In aqueous solution, the hydrated proton is most commonly represented as $\mathrm{H}_{3} \mathrm{O}^{+}$. This ion is also known as the hydronium ion, which is a more accurate representation compared to a bare proton ($\mathrm{H}^{+}$), reflecting its state in an aqueous environment. While there are more complex forms like $\mathrm{H}_{9} \mathrm{O}_{4}^{+}$, $\mathrm{H}_{3} \mathrm{O}^{+}$ is the standard notation for a hydrated proton in basic aqueous chemistry. Therefore, the correct answer is (D).
Sulphuric acid is a dense, oily liquid.
A) True
B) False
The statement is True. Sulphuric acid is indeed a dense, oily liquid.
The correct relation between hydrolysis constant $K_{a}$ and degree of hydrolysis (h) for the following equilibrium is:
$$ \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{H}_{2}\mathrm{O} \stackrel{\text{hydrolysis}}{\rightleftharpoons} \mathrm{CH}_{3}\mathrm{COOH} + \mathrm{OH}^{-} $$
A) $h = \sqrt{\frac{K_{w} \cdot C}{K_{a}}}$
B) $h = \sqrt{\frac{K_{w}}{K_{a} \cdot C}}$
C) $h = \sqrt{\frac{K_{a} \cdot C}{K_{w}}}$
D) $h = \sqrt{\frac{K_{a}}{K_{w} \cdot C}}$
To determine the relationship between the hydrolysis constant $ K_a $ and the degree of hydrolysis (h) for the given equilibrium:
$$ \mathrm{CH}_{3}\mathrm{COO}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{CH}_{3}\mathrm{COOH} + \mathrm{OH}^{-} $$
We start by analyzing the equilibrium of the reaction, where $ C $ is the initial concentration of $ \mathrm{CH}_{3}\mathrm{COO}^{-} $ and ( h ) is the degree of hydrolysis. At equilibrium, the concentrations are:
$ \mathrm{CH}_{3}\mathrm{COO}^{-} $: $ C(1-h) $
$ \mathrm{CH}_{3}\mathrm{COOH} $ and $ \mathrm{OH}^{-} $: Each has concentration $ Ch $
The hydrolysis constant $ K_h $ is defined by:
$$ K_{h} = \frac{[\mathrm{CH}{3}\mathrm{COOH}][\mathrm{OH}^{-}]}{[\mathrm{CH}{3}\mathrm{COO}^{-}]} = \frac{(Ch)^2}{C(1-h)} $$
For the weak acid $ \mathrm{CH}{3}\mathrm{COOH} $ dissociating, we have:
$$ K_a = \frac{[\mathrm{CH}{3}\mathrm{COO}^{-}][H^{+}]}{[\mathrm{CH}_{3}\mathrm{COOH}]} $$
And for water, the autoionization is given by:
$$ K_w = [H^{+}][OH^{-}] $$
From the equation of $ K_h $, and using $ K_h = \frac{K_w}{K_a} $, the equation becomes:
$$ \frac{K_w}{K_a} = \frac{(Ch)^2}{C(1-h)} $$
Assuming $ h $ is small compared to 1, $ 1-h \approx 1 $ simplifies the expression to:
$$ \frac{K_w}{K_a} = (Ch)^2 $$
Solving for $ h $, we derive:
$$ h = \sqrt{\frac{K_w}{K_a \cdot C}} $$
Therefore, the correct answer is Option B:
$$ h = \sqrt{\frac{K_w}{K_a \cdot C}} $$
To $20 \mathrm{~mL}$ of $1.5 \mathrm{M} \mathrm{HCl}$ solution, $30 \mathrm{~mL}$ of water is added, and to this resultant solution again $75 \mathrm{~mL}$ of $0.4 \mathrm{M} \mathrm{HCl}$ is added. The molarity of the resultant solution is:
A $0.40 \mathrm{M}$
B $\quad 2.10 \mathrm{M}$
C $0.56 \mathrm{M}$
D $\quad 5.5 \mathrm{M}$
The correct answer is C $0.56 \mathrm{M}$.
In the initial step, when $20 , \mathrm{mL}$ of $1.5 \mathrm{M} , \mathrm{HCl}$ is diluted by adding $30 , \mathrm{mL}$ of water, the resulting solution's volume becomes $50 , \mathrm{mL}$. To compute the new molarity ($M_2$), we use the equation: $$ M_1 V_1 = M_2 V_2 $$ Where $M_1 = 1.5 , \mathrm{M}$, $V_1 = 20 , \mathrm{mL}$, and $V_2 = 50 , \mathrm{mL}$. Thus, $$ M_2 = \frac{1.5 \times 20}{50} = 0.6 , \mathrm{M} $$
In the subsequent step, $75 , \mathrm{mL}$ of $0.4 \mathrm{M} , \mathrm{HCl}$ is added to the $50 , \mathrm{mL}$ of $0.6 \mathrm{M} , \mathrm{HCl}$ previously prepared. The total volume is now $125 , \mathrm{mL}$. To find the final molarity ($M_3$) of the combined solution, we use: $$ M_3 = \frac{M_{\text{init}} V_{\text{init}} + M_{\text{add}} V_{\text{add}}}{V_{3}} $$ Where:
$M_{\text{init}} = 0.6 , \mathrm{M}$
$V_{\text{init}} = 50 , \mathrm{mL}$
$M_{\text{add}} = 0.4 , \mathrm{M}$
$V_{\text{add}} = 75 , \mathrm{mL}$
$V_3 = 125 , \mathrm{mL}$
Plugging in the values: $$ M_3 = \frac{(0.6 \times 50) + (0.4 \times 75)}{125} = 0.56 , \mathrm{M} $$
Therefore, the final molarity of the resultant solution is $0.56 \mathrm{M}$.
In Hall's process, the ore is leached by heating it with:
A) sodium carbonate
B) sodium bicarbonate
C) sodium hydroxide
The correct answer is A) sodium carbonate. In Hall's process, the powdered ore is leached by heating it with sodium carbonate, which transforms it into sodium aluminate. The chemical reaction involved is: $$ \mathrm{Al}_2\mathrm{O}_3 \cdot \mathrm{H}_2\mathrm{O}(\mathrm{s}) + \mathrm{Na}_2\mathrm{CO}_3(\mathrm{aq}) \rightarrow 2 \mathrm{NaAlO}_2(\mathrm{aq}) + \mathrm{CO}_2(\mathrm{g}) + \mathrm{H}_2\mathrm{O}(\mathrm{l}) $$ This transformation is crucial for the extraction process in Hall's method.
The potassium oxide forms potassium hydroxide when dissolved in water.
A) True
B) False
The correct option is A) True
Potassium oxide reacts with water to form potassium hydroxide. This reaction can be illustrated by the equation:
$$ \text{Potassium oxide} + \text{Water} \rightarrow \text{Potassium hydroxide} $$
Potassium hydroxide is a base, known for its property to turn red litmus paper blue.
Baking soda is which type of salt?
A. Basic salt
B. Acidic salt
Explanation:
Baking soda is classified as a basic salt.
The chemical formula for baking soda is $\mathrm{NaHCO}_{3}$. This compound is derived from a weak acid ($\mathrm{H}_{2}\mathrm{CO}_{3}$, carbonic acid) and a strong base ($\mathrm{NaOH}$, sodium hydroxide). Due to the weak nature of carbonic acid, it does not fully dissociate.
Consequently, the solution of baking soda tends to be basic. This results from the hydrolysis process, where additional $\mathrm{OH}^{-}$ ions are produced.
Therefore, the correct answer is option A (Basic salt).
The pair in which phosphorous atoms have a formal oxidation state of +3:
A) Pyrophosphorous and hypophosphoric acids
B) Orthophosphorus and pyrophosphorus acids
C) Pyrophosphorous and hypophosphoric acids
D) Orthophosphorus and hypophosphoric acids
The correct answer is Option B which includes Orthophosphorus and Pyrophosphorus acids.
To determine the oxidation states of phosphorus in these acids, consider the following calculations:
For Orthophosphorus acid $\left(\mathrm{H}_{3} \mathrm{PO}_{3}\right)$:
Each hydrogen contributes +1, and the oxidation number of oxygen is -2. Balancing the total oxidation state to be zero in the compound:
$$ 3(+1) + x + 3(-2) = 0 $$
Solving for $x$ (oxidation state of P): $$ 3 + x - 6 = 0 \implies x = 3 $$ Therefore, phosphorus has an oxidation state of +3 in Orthophosphorus acid.For Pyrophosphorous acid $\left(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{5}\right)$:
Here, each hydrogen again contributes +1, and the oxidation number for oxygen is -2. Considering there are two phosphorus atoms: $$ 4(+1) + 2x + 5(-2) = 0 $$ Solving for $x$: $$ 4 + 2x - 10 = 0 \implies 2x = 6 \implies x = 3 $$ Hence, each phosphorus atom also has an oxidation state of +3 in Pyrophosphorous acid.
Thus, the phosphorus atoms in both Orthophosphorus acid and Pyrophosphorous acid have oxidation states of +3, making Option B the correct choice.
Mercury does not react with sulphuric acid. This happens because:
A Mercury is a liquid.
B Sulphuric acid is a strong acid.
C Mercury is more reactive than hydrogen.
D Mercury is less reactive than hydrogen.
The correct answer is D: Mercury is less reactive than hydrogen.
Mercury is a metal that is less reactive than hydrogen, which means it does not have the ability to displace hydrogen in a chemical reaction with sulphuric acid. This lack of reactivity is why mercury does not react with sulphuric acid.
Phenolphthalein shows pink color with which one of the following substances?
A) Vinegar
B) Orange juice
C) Grape juice
D) Milk of magnesia
The correct option is D) Milk of magnesia
Phenolphthalein changes color in the presence of basic substances. Milk of magnesia is considered basic because it contains magnesium hydroxide ($\text{Mg(OH)}_2$), which is a base. When in contact with milk of magnesia, phenolphthalein turns pink, indicating its basic nature.
Ptyalin of saliva acts in:
A. Slightly acidic medium
B. Slightly alkaline medium
C. Highly alkaline medium
D. All types of media
The correct answer is:
A. Slightly acidic medium
Ptyalin in saliva functions optimally in a neutral or faintly acidic medium with an optimal pH of around 6.5.
The solution of borax in water is:
A) Neutral
B) Basic
C) Slightly acidic
D) Strongly acidic
The correct answer is B) Basic.
When borax (sodium borate) dissolves in water, the equation for the reaction is:
$$ \text{Na}_2\text{B}_4\text{O}_7 + 7\text{H}_2\text{O} \rightarrow 2\text{NaOH} + 4\text{H}_3\text{BO}_3 $$
Here, $\text{NaOH}$ (sodium hydroxide) is formed, which is a strong base, and $\text{H}_3\text{BO}_3$ (boric acid or orthoboric acid), which is a weak acid. Considering that the base formed is stronger than the acid, the overall solution of borax is basic in nature.
Which of the following is largest?
(A) $\mathrm{Cl}^{-}$
(B) $\mathrm{S}^{2-}$
(C) $\mathrm{Na}^{+}$
(D) $\mathrm{F}^{-}$
The correct answer is (B) $ \mathrm{S}^{2-} $.
Among the options given, $ \mathrm{S}^{2-} $ is the largest in size. Although both $ \mathrm{S}^{2-} $ and $ \mathrm{Cl}^{-} $ are isoelectronic (they have the same number of electrons), the nuclear charge (number of protons in the nucleus) of $ \mathrm{Cl}^{-} $ is greater than that of $ \mathrm{S}^{2-} $.
A higher nuclear charge attracts the electrons more strongly, reducing the size of the ion. Therefore, with a lower nuclear charge, $ \mathrm{S}^{2-} $ has a larger size compared to $ \mathrm{Cl}^{-} $, making it the largest among the given options.
If $\mathrm{K}_{\mathrm{a}} = 1.8 \times 10^{-5}$, calculate the degree of hydrolysis of a $0.1 \mathrm{M}$ solution of sodium acetate at $25^{\circ} \mathrm{C}$:
(A) $7.45 \times 10^{-5}$ B) $7.45 \times 10^{-5}$ (C) $7.45 \times 10^{-1}$ (D) $5.5 \times 10^{-2}$
Sodium acetate is a salt derived from a strong base (NaOH) and a weak acid (acetic acid, $\mathrm{CH}_3\mathrm{COOH}$). When sodium acetate dissociates in water, it hydrolyzes according to the following equilibrium: $$ \mathrm{CH}_3\mathrm{COO}^- (\mathrm{aq}) + \mathrm{H}_2\mathrm{O} (\mathrm{l}) \rightleftharpoons \mathrm{CH}_3\mathrm{COOH} (\mathrm{aq}) + \mathrm{OH}^- (\mathrm{aq}) $$ The concentration of sodium acetate ($C$) is given as $0.1 , \mathrm{M}$. Letting $h$ denote the degree of hydrolysis, the concentrations of $\mathrm{CH}_3\mathrm{COOH}$ and $\mathrm{OH}^-$ at equilibrium can be expressed as $Ch$.
The hydrolysis constant ($K_h$) for this reaction is: $$ K_h = \frac{[\mathrm{CH}_3\mathrm{COOH}][\mathrm{OH}^-]}{[\mathrm{CH}_3\mathrm{COO}^-]} = \frac{(Ch)^2}{C(1-h)} \approx \frac{(Ch)^2}{C} $$ since $h$ is typically very small and thus $1-h \approx 1$.
From the ionization of water ($K_w$) and the acid dissociation constant of acetic acid ($K_a$), the relationship between $K_h$, $K_w$, and $K_a$ is: $$ K_h = \frac{K_w}{K_a} $$ Thus, substituting the values and solving for $h$: $$ h = \sqrt{\frac{K_w}{K_a \times C}} = \sqrt{\frac{10^{-14}}{1.8 \times 10^{-5} \times 0.1}} $$ Calculating the above, $$ h = 7.45 \times 10^{-5} $$
Therefore, the degree of hydrolysis of the $0.1 \mathrm{M}$ sodium acetate solution at $25^\circ C$ is $7.45 \times 10^{-5}$, and the correct answer is (A) $7.45 \times 10^{-5}$.
$\mathrm{H}^{+}$ ion comes from the ionization of bases.
A) True
B) False
The correct answer is $\mathbf{B}$, False.
An acid typically ionizes in a chemical reaction by releasing $\mathrm{H}^{+}$ ions into the solution, as illustrated by the general reaction: $$ \mathrm{HA} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}^{+} + \mathrm{A}^{-} $$ For example, hydrochloric acid ionizes as follows: $$ \mathrm{HCl} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}^{+}_{(\mathrm{aq})} + \mathrm{Cl}^{-}_{(\mathrm{aq})} $$
In contrast, a base ionizes by accepting $\mathrm{H}^{+}$ ions from water, producing $\mathrm{OH}^{-}$ (hydroxide ions). The general reaction for a base ionizing in water is: $$ \mathrm{BOH} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{B}^{+} + (\mathrm{OH})^{-} $$ An example using sodium hydroxide: $$ \mathrm{NaOH} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{Na}^{+}_{(\mathrm{aq})} + \mathrm{OH}^{-}_{(\mathrm{aq})} $$
Therefore, $\mathrm{H}^{+}$ ions originate from the ionization of acids, not bases.
In an acidic buffer, the ratio [acid]/[salt] is 0.5. If we double the concentration of the acid, what is the value of $10\Delta \mathrm{pH}$? [$\log 2 = 0.3$]
For an acidic buffer, the pH is given by the formula: $$ \mathrm{pH} = \mathrm{pK}_\mathrm{a} + \log \frac{[\text{Salt}]}{[\text{Acid}]} $$
Initially, the ratio of acid to salt is $0.5$, which means: $$ \frac{[\text{Acid}]}{[\text{Salt}]} = 0.5 $$ or, rewriting this, $$ \frac{[\text{Salt}]}{[\text{Acid}]} = 2 $$ Thus, the initial pH of the buffer is: $$ \mathrm{pH} = \mathrm{pK}_\mathrm{a} + \log 2 $$
If the concentration of the acid is doubled, the new ratio of salt to acid becomes: $$ \frac{[\text{Salt}]}{2[\text{Acid}]} = 1 $$ which implies: $$ \frac{[\text{Salt}]}{[\text{Acid}]} = 1 $$ Under this condition, the new pH will be: $$ \mathrm{pH} = \mathrm{pK}\mathrm{a} + \log 1 = \mathrm{pK}\mathrm{a} $$
The change in pH, denoted as $\Delta \mathrm{pH}$, is then: $$ \Delta \mathrm{pH} = \mathrm{pK}{\mathrm{a}} - (\mathrm{pK}{\mathrm{a}} + \log 2) = -\log 2 = -0.3 $$ Therefore, $10 \Delta \mathrm{pH}$ calculates to: $$ 10 \Delta \mathrm{pH} = 10 \times (-0.3) = -3 $$
However, if considering absolute change, $10 \Delta \mathrm{pH}$ equals $3$ (ignoring the negative sign which represents direction of change). Thus, $10 \Delta \mathrm{pH} = 3$.
The molecule which can disproportionate is:
A $\mathrm{HClO}_{4}$
B $\mathrm{H}_{3} \mathrm{PO}_{4}$
C $\mathrm{H}_{2} \mathrm{SO}_{4}$
D $\mathrm{H}_{3} \mathrm{PO}_{3}$
The correct option is D $\mathrm{H}_{3} \mathrm{PO}_{3}$.
This molecule can undergo disproportionation because phosphorus (P) in this compound is in the +3 oxidation state. It can undergo both oxidation (increase in oxidation state) and reduction (decrease in oxidation state). The reaction is given by:
$$ 4 \mathrm{H}{3} \mathrm{PO}{3} \stackrel{\Delta}{\Rightarrow} 3 \mathrm{H}{3} \mathrm{PO}{4}+\mathrm{PH}_{3} $$
In this reaction, phosphorus is both oxidized to $\mathrm{H}_{3} \mathrm{PO}_{4}$ (where P is in +5 oxidation state) and reduced to $\mathrm{PH}_{3}$ (where P is in -3 oxidation state).
Other options A, B, and C are incorrect because in $\mathrm{HClO}_{4}$, $\mathrm{H}_{3} \mathrm{PO}_{4}$, and $\mathrm{H}_{2} \mathrm{SO}_{4}$, the elements Cl, P, and S are all in their maximum oxidation states (+7, +5, and +6, respectively), preventing them from undergoing further oxidation.
When water droplets mixed with acids fall down as rain, it is called
A) volcano
B) acid rain
C) sulphur rain
D) acid water
The accurate answer is B) acid rain.
Acid rain refers to precipitation that contains elevated levels of acids due to atmospheric pollution. These contaminants, primarily oxides of sulfur and nitrogen, react with water vapor in the atmosphere forming acids. As a result, when this mixture precipitates, it falls as acid rain, which can have harmful effects on the environment and living organisms.
Calculate the molality of 58.5 g of NaCl (Molar mass of NaCl = 58.5 g/mol) dissolved in 500 g of water.
A) 1 m B) 2 m C) 0.5 m D) 1.5 m
The correct option is B) 2 m
Molality (*m*) is calculated using the formula:
$$ m = \frac{n_{\text{solute}}}{w_{\text{solvent}}(g)} \times 1000 $$
where $n_{\text{solute}}$ is the number of moles of the solute, and $w_{\text{solvent}}$ is the mass of the solvent in grams.
We can also express this relationship as:
$$ m = \frac{W_{\text{solute}} \times 1000}{M_{\text{solute}} \times W_{\text{solvent}}} $$
Here, $W_{\text{solute}}$ is the mass of the solute, and $M_{\text{solute}}$ is the molar mass of the solute.
Substituting the known quantities:
$$ m = \frac{58.5 \times 1000}{58.5 \times 500} $$
Simplifying this gives:
$$ m = 2 , \text{m} $$
Therefore, the molality of the solution is 2 molal (2 m).
What is Crystallisation and Crystalline Salt?
Crystallization is the process (either natural or artificial) in which solid crystals form by precipitating from a solution or melt, and in rare instances, deposit directly from a gas. The process involves mass transfer of a solute from the liquid solution to a solid crystalline phase. This is commonly carried out in a device known as a crystallizer and is fundamental in chemical engineering. Crystallization is a type of precipitation that depends on variations in solubility conditions of the solute in the solvent, differing from precipitation caused by chemical reactions.
Crystalline salt refers to salts that form crystals, typically shaped like cubes. Examples include sodium chloride (common table salt) or calcium fluoride. These types of salts showcase a regular crystalline structure that is easily recognizable.
Taj Mahal is badly affected by acid rain. This is because:
A. Taj Mahal is made of marble.
B. Acid rain attacks the steel structure of Taj Mahal.
C. Acid rain has a pH above 5.6.
D. The paint on Taj Mahal is soluble in acids.
The correct answer is A. Taj Mahal is made of marble. The primary material of the Taj Mahal is marble, which is primarily composed of calcium carbonate (CaCO3). When marble comes into contact with sulfuric acid from acid rain, it undergoes a chemical reaction that can lead to pitting, discoloration, and loss of luster. Thus, acid rain negatively impacts the structure mainly because of its reaction with the marble.
Why are acids good conductors of electricity?
Acids are excellent conductors of electricity when they are in an aqueous solution. The key to this property lies in the behavior of acids upon dissolving in water. Acids, upon dissolution, release hydrogen ions ($\text{H}^+$) into the solution along with a significant amount of anions.
The presence of these charged particles, or ions, in the solution allows electrical current to flow through it. Essentially, the ionic nature of the solution facilitates the movement of charges, making it conducive to conducting electricity. Moreover, the strength of the acid influences its conductivity. A stronger acid dissociates more completely, releasing more $\text{H}^+$ ions and anions, thereby enhancing the conductivity of the solution.
You are provided with a $500 \mathrm{~mL}$ of hard water, containing $0.005 \mathrm{~mole}$ of $\mathrm{CaCl}_{2}$, and two sulfuric acid samples of $0.001 \mathrm{M}$ and $0.02 \mathrm{M}$ concentrations. The solubility product of calcium sulfate in water at $25^{\circ} \mathrm{C}$ is $2.4 \times 10^{-5}$. Choose the most appropriate answer:
A. Both samples of $\mathrm{H}_{2} \mathrm{SO}_{4}$ will cause precipitation.
B. $0.02 ~\mathrm{M} ~\mathrm{H}_{2} \mathrm{SO}_{4}$ will cause precipitation, but $0.001 \mathrm{M}$ sample doesn't.
C. $0.001 ~\mathrm{M}~ \mathrm{H}_{2} \mathrm{SO}_{4}$ will cause precipitation, but $0.02 \mathrm{M}$ sample doesn't.
D. Neither will cause precipitation.
The correct answer is Option B: $0.02~ \mathrm{M}~ \mathrm{H}_{2} \mathrm{SO}_{4}$ will cause precipitation, but $0.001 \mathrm{M}$ sample doesn't.
To solve this problem, let's assume we are mixing $500 \mathrm{~mL}$ of each sulfuric acid solution with the hard water. As there's $0.005$ mole of $\mathrm{CaCl}_{2}$ in $500 \mathrm{~mL}$ of water, the concentration of $\mathrm{Ca}^{2+}$ ions is: $$ \left[\mathrm{Ca}^{2+}\right] = \frac{0.005 \text{ mole}}{0.5 \text{ L}} = 0.01 \mathrm{~M} $$ When the solution is mixed, assuming the volumes are additive, the new volume is $1 \mathrm{~L}$. Hence, the concentrations of ions will adjust accordingly.
For $0.001 \mathrm{M}$ $\mathrm{H}_{2} \mathrm{SO}_{4}$:
Sulfate ion concentration, $\left[\mathrm{SO}_{4}^{2-}\right] = \frac{0.001 \mathrm{~M}}{2} = 0.0005 \mathrm{~M}$ (due to dilution after mixing).
The ionic product of $\mathrm{CaSO}{4}$ is then: $$ \left[\mathrm{Ca}^{2+}\right] \times \left[\mathrm{SO}_{4}^{2-}\right] = 0.005 \mathrm{~M} \times 0.0005 \mathrm{~M} = 2.5 \times 10^{-6} $$ This is less than the solubility product ($\mathrm{K}_{sp} = 2.4 \times 10^{-5}$), hence, precipitation will not occur.
For $0.02 \mathrm{M}$ $\mathrm{H}{2} \mathrm{SO}{4}$:
Sulfate ion concentration, $\left[\mathrm{SO}_{4}^{2-}\right] = \frac{0.02 \mathrm{~M}}{2} = 0.01 \mathrm{~M}$.
The ionic product of $\mathrm{CaSO}{4}$ is: $$ \left[\mathrm{Ca}^{2+}\right] \times \left[\mathrm{SO}_{4}^{2-}\right] = 0.005 \mathrm{~M} \times 0.01 \mathrm{~M} = 5 \times 10^{-5} $$ This value is greater than the solubility product ($\mathrm{K}_{sp} = 2.4 \times 10^{-5}$), indicating that precipitation will occur.
Thus, Option B is correct: $0.02 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ will cause precipitation, but $0.001 \mathrm{M}$ will not.
Statement 1: Metal oxides are basic in nature. Statement 2: Metal oxides neutralize the effect of acids and release water, similar to the reaction of a base with an acid.
A) Statement 1 and Statement 2 are correct, but Statement 2 is not the correct explanation for Statement 1.
B) Statement 1 is correct, but Statement 2 is incorrect.
C) Statement 1 is incorrect, but Statement 2 is correct.
D) Both statements are correct, and Statement 2 is the correct reason for Statement 1.
The correct option is D - Both statements are correct, and Statement 2 provides the correct explanation for Statement 1.
Metal oxides such as $ \mathrm{MgO} $ demonstrate basic properties because they react with acids to yield salt and water. This behavior is characteristic of a base, supporting the assertion that metal oxides are basic in nature. For instance, when magnesium oxide reacts with hydrochloric acid, the chemical equation is:
$$ \mathrm{MgO} + 2\mathrm{HCl} \rightarrow \mathrm{MgCl}_2 + \mathrm{H}_2\mathrm{O} $$
This reaction is akin to how bases operate, forming water and a salt when reacting with acids. Thus, Statement 1 is accurate as it declares metal oxides are basic, and Statement 2 is also correct, providing a valid explanation for why metal oxides are considered basic, as they neutralize acids similar to bases.
Name the ion responsible for unmasking of active sites for myosin for cross-bridge activity during muscle contraction.
A. Potassium
B. Calcium
C. Magnesium
D. Sodium
The correct answer is B. Calcium
Calcium ions (Ca²⁺) are crucial as they bind to a unit of troponin. This interaction leads to the unmasking of active sites on actin that are initially covered by tropomyosin, thereby allowing myosin to bind to actin. This binding is essential for initiating the cross-bridge cycling, a fundamental process in muscle contraction.
The chemical formula of baking soda is $\mathrm{NaHCO}_{3}$.
A) True
B) False
The correct answer is A) True.
Baking soda is chemically known as sodium hydrogen carbonate, which is represented by the chemical formula: $$ \mathrm{NaHCO}_3 $$
Which of the following is a tetrabasic acid?
A) Orthophosphorus acid
B) Orthophosphoric acid
C) Metaphosphoric acid
D) Pyrophosphoric acid
The correct answer is D) Pyrophosphoric acid.
Pyrophosphoric acid has the chemical formula $\mathrm{H}_{4}\mathrm{P}_{2}\mathrm{O}_{7}$.
This acid is classified as tetrabasic because it contains four ionizable hydrogen atoms per molecule, which means it can donate four protons ($\mathrm{H}^+$). Thus, Pyrophosphoric acid is capable of fully ionizing in four steps, aligning with the definition of a tetrabasic acid.
The order in which the following oxides are arranged according to decreasing basic nature is
(A) $\mathrm{Na}_{2} \mathrm{O}, \mathrm{MgO}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{CuO}$
(B) $\mathrm{MgO}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{CuO}, \mathrm{Na}_{2} \mathrm{O}$
(C) $\mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{MgO}, \mathrm{CuO}, \mathrm{Na}_{2} \mathrm{O}$
(D) $\mathrm{CuO}, \mathrm{Na}_{2} \mathrm{O}, \mathrm{MgO}, \mathrm{Al}_{2} \mathrm{O}_{3}$
The correct option is (A):
$$ \mathrm{Na}_2\mathrm{O}, \mathrm{MgO}, \mathrm{Al}_2\mathrm{O}_3, \mathrm{CuO} $$
This sequence lists oxides in decreasing basic nature. The basis for this arrangement is linked to the metallic character of the parent metals of these oxides. A higher metallic character implies a greater tendency to donate electrons. Consequently, the oxides of metals with lower ionization energies (IE) tend to be more basic. Among the given elements, sodium ($\mathrm{Na}$) displays the lowest ionization energy and thus, $\mathrm{Na}_2\mathrm{O}$ is the most basic, followed by $\mathrm{MgO}$, $\mathrm{Al}_2\mathrm{O}_3$, and finally $\mathrm{CuO}$.
Cleansing action of soap is due to the formation of micelles.
A. True
B. False
The correct answer is A. True.
Soaps are primarily sodium or potassium salts of long chain fatty acids. A soap molecule has a structure that includes a hydrophilic (water-attracting) ionic part and a hydrophobic (water-repelling) long hydrocarbon chain.
When soap is added to water, the hydrophobic tail of the soap molecule avoids water and attaches to oil or dirt, while the hydrophilic head remains attracted to water. This scenario leads to the formation of structures known as micelles. In a micelle, the hydrophobic tails of several soap molecules are oriented towards the inside away from water, while the hydrophilic heads point outward, forming a spherical structure that is soluble in water.
Micelles encapsulate the oily dirt particles with their hydrophobic tails and allow them to be washed away with water, thanks to the hydrophilic heads that interact with the water. This property is crucial to the cleansing action of soaps, confirming that the statement about micelles is indeed true.
Which of the following is wrong?
A) $\mathrm{NH}_3 < \mathrm{PH}_3 < \mathrm{AsH}_3 <$ Acidic character B) $\mathrm{Li} < \mathrm{Be} < \mathrm{B} < \mathrm{C} <$ IE1 C) $\mathrm{Al}_2\mathrm{O}_3 < \mathrm{MgO} < \mathrm{Na}_2\mathrm{O} < \mathrm{K}_2\mathrm{O} <$ Basic character D) $\mathrm{Li}^+ < \mathrm{Na}^+ < \mathrm{K}^+ < \mathrm{Cs}^+ <$ Ionic radius
The incorrect statement among the given options is B). The statement "$\mathrm{Li} < \mathrm{Be} < \mathrm{B} < \mathrm{C} <$ IE1" suggests that the first ionization energy (IE1) increases in this order. However, beryllium ($\mathrm{Be}$) has a higher first ionization energy than boron ($\mathrm{B}$). The reason for this is that beryllium has a completely filled $2s$ orbital, which is more stable compared to the half-filled $2p$ orbital in boron. Thus, the correct order should reflect beryllium having a higher ionization energy than boron, not lower.
In terms of ionization energy: $$ \text{IE1}(\mathrm{Li}) < \text{IE1}(\mathrm{B}) < \text{IE1}(\mathrm{Be}) < \text{IE1}(\mathrm{C}) $$
Oxides of Metals are:
A. Acidic
B. Basic
C. Amphoteric oxides
D. Both basic and amphoteric oxides.
The correct answer is D. Both basic and amphoteric oxides.
Metal oxides typically exhibit basic characteristics; however, there are exceptions, such as aluminum oxide ($ Al_2O_3 $) and zinc oxide ($ ZnO $), which display amphoteric behavior. Amphoteric oxides can react with both acids and bases.
$1.6 \mathrm{~g}$ of calcium and $2.6 \mathrm{~g}$ of zinc, when treated with an acid in excess separately, produced the same amount of hydrogen gas. Find the equivalent weight of calcium, given that zinc has an equivalent weight of $32.6 \mathrm{~g}$.
A) $17 \mathrm{~g}$
B) $40 \mathrm{~g}$
C) $27 \mathrm{~g}$
D) $20 \mathrm{~g}$
The correct answer is Option D: $20 \mathrm{~g}$.
Let's denote the equivalent weight of calcium as $E_{\text{Ca}}$. According to the problem, the same amount of hydrogen gas is produced from both calcium and zinc when reacted with excess acid. This implies that the equivalents of hydrogen produced are the same from both the metals.
The equivalents of a metal in a reaction can be calculated using: $$ \text{Equivalents} = \frac{\text{Weight of the metal}}{\text{Equivalent weight of the metal}} $$
Thus, the equation for the equivalents from calcium and zinc will be equal: $$ \frac{1.6 \text{ g}}{E_{\text{Ca}}} = \frac{2.6 \text{ g}}{32.6 \text{ g}} $$
Here, $32.6 \text{ g}$ is the given equivalent weight of zinc.
By solving for $E_{\text{Ca}}$, we have: $$ E_{\text{Ca}} = \frac{1.6 \text{ g} \times 32.6 \text{ g}}{2.6 \text{ g}} = \frac{52.16}{2.6} \approx 20.06 \text{ g} $$
Rounded to the nearest whole number as per the options given, the equivalent weight of calcium is approximately $20 \text{ g}$. Hence, Option D is correct.
The nature of the solution obtained by mixing $250 , \mathrm{ml} , 0.2 , \mathrm{M} , \mathrm{H}_{2}\mathrm{SO}_{4}$ and $250 , \mathrm{ml}$ of $0.1 , \mathrm{M} , \mathrm{NaOH}$ is:
A) Acidic
B) Basic
C) Neutral
D) Amphoteric
The nature of the solution obtained from mixing 250 ml of $0.2 , \mathrm{M} , \mathrm{H}_{2}\mathrm{SO}_{4}$ (sulfuric acid) and 250 ml of $0.1 , \mathrm{M} , \mathrm{NaOH}$ (sodium hydroxide) can be determined by calculating the equivalents of each solution and comparing them.
Step 1: Calculate the equivalents of sulfuric acid (Acid)
The equivalents of acid ($N_A \cdot V_A$) are calculated using the formula:
$$ N_A \cdot V_A = \text{Normality of acid} \times \text{Volume of acid in liters} $$
Since sulfuric acid ($\mathrm{H}_2\mathrm{SO}_4$) is a diprotic acid ( (n = 2) valence factor), the normality is twice the molarity. Therefore, $ N_A \cdot V_A = 0.2 , \mathrm{M} \times 2 \times \frac{250}{1000} = 0.1 , \text{equivalents} $
Step 2: Calculate the equivalents of sodium hydroxide (Base)
The equivalents of base ($N_B \cdot V_B$) are calculated similarly:
$$ N_B \cdot V_B = \text{Normality of base} \times \text{Volume of base in liters} $$
Sodium hydroxide ($\mathrm{NaOH}$) is a monoprotic base ( (n = 1) valence factor), hence its normality equals its molarity. $$ N_B \cdot V_B = 0.1 , \mathrm{M} \times 1 \times \frac{250}{1000} = 0.025 , \text{equivalents} $$
Conclusion: Nature of the Solution
Comparing the equivalents of $ \mathrm{H}_2\mathrm{SO}_4 $ ($0.1 , \text{equivalents}$) and $\mathrm{NaOH}$ ($0.025 , \text{equivalents}$), you can see that sulfuric acid has more equivalents than sodium hydroxide. This implies that not all the acid has been neutralized by the base.
Hence, the remaining unneutralized acid makes the solution acidic. Therefore, the correct option is:
A) Acidic
For the following Assertion and Reason, the correct option is:
Assertion: The pH of water increases with an increase in temperature.
Reason: The dissociation of water into $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ is an exothermic reaction.
A. Both assertion and reason are false.
B. Assertion is not true, but reason is true.
C. Both assertion and reason are true and the reason is the correct explanation for the assertion.
D. Both assertion and reason are true, but the reason is not the correct explanation for the assertion.
The correct option is A: Both assertion and reason are false.
The dissociation of water into $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$, represented by the reaction $$ \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{H}^{+} + \mathrm{OH}^{-}, $$ is an endothermic process, contrary to the claim in the reason statement that it is exothermic.
When the temperature rises, the equilibrium constant for water dissociation, $K_w$, increases, which leads to a decrease in $pK_w$. Consequently, since the pH of water is given by $$ \text{pH} = \frac{1}{2} pK_w, $$ the pH of water actually decreases with an increase in temperature, making the assertion false as well.
Which of the following substances is/are acidic in nature?
A) Lemon B) Distilled water C) Vinegar D) Baking soda
The substances that are acidic in nature from the given list are:
A) Lemon
C) Vinegar
Lemon contains citric acid, which is why it is acidic. Similarly, vinegar is primarily composed of acetic acid, giving it its acidic property. In contrast, distilled water (B) is neutral, while baking soda (D) is basic due to its bicarbonate content.
"Some crops are suitable for growing in soil with acidic nature. Which salt can be added in order to increase the acidic nature of the soil?
A) Sodium hydroxide
B) Milk of Magnesia
C) Slaked lime
D) Ammonium sulphate"
The correct answer is D) Ammonium sulphate.
Ammonium sulphate is effective in lowering the pH of the soil, making it more acidic. This occurs because when ammonium sulphate is added to the soil, the ammonium ion ($ \text{NH}_4^+ $) is converted into nitrate ($ \text{NO}_3^- $) by soil bacteria through a process called nitrification. This process releases hydrogen ions ($ \text{H}^+ $), which contribute to the increase in soil acidity.
What is the colour change observed in dil. HCl solution when phenolphthalein is added to it?
Colourless
Pink
Yellow
Green
The correct option is:
- Colourless
Phenolphthalein is commonly used as an indicator in acid-base titrations. In acidic solutions, phenolphthalein turns colorless. Conversely, in basic solutions, it turns pink.
When phenolphthalein is added to dilute hydrochloric acid (dil. HCl), the solution remains colorless. This is due to the acidic nature of the hydrochloric acid.
Which gas is evolved when acids react with metal carbonates?
A. $\mathrm{CO}_{2}$
B. $\mathrm{H}_{2}$
C. $\mathrm{NH}_{3}$
D. $\mathrm{O}_{2}$
The correct answer is A
$$
\mathrm{CO}_2
$$
When metal carbonates react with acids, they release carbon dioxide ($\mathrm{CO_2}$). Here is an example reaction to illustrate this:
$$ \mathrm{CaCO_3} + 2\ \mathrm{HCl} \rightarrow \mathrm{CaCl_2} + \mathrm{H_2O} + \mathrm{CO_2} $$
Estimate the value x to balance the given equation.
$$ \mathrm{Mg}_{3} \mathrm{N}_{2}(\mathrm{aq}) + \mathrm{xH}_{2} \mathrm{O}(\mathrm{I}) \rightarrow 3 \mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{aq}) + 2 \mathrm{NH}_{3}(\mathrm{~g}) $$
A. 2
B. 4
C. 6
D. 8
The correct option is C) 6
To balance a chemical equation, the number of atoms for each element must be the same on both sides of the reaction.
Given Equation:
$$ \mathrm{Mg}_{3}\mathrm{N}_{2}(\mathrm{aq}) + \mathrm{xH}_{2}\mathrm{O}(\mathrm{I}) \rightarrow 3\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{aq}) + 2\mathrm{NH}_{3}(\mathrm{~g}) $$
Let's analyze the atom count for each element:
Element | Reactants | Atoms | Products | Atoms |
---|---|---|---|---|
Mg | $\mathrm{Mg}_{3}\mathrm{N}_{2}$ | 3 (3 Mg) | $3\mathrm{Mg}(\mathrm{OH})_{2}$ | 3 (3 Mg) |
N | $\mathrm{Mg}_{3}\mathrm{N}_{2}$ | 2 (2 N) | $2\mathrm{NH}_{3}$ | 2 (2 N) |
H | $\mathrm{H}_{2}\mathrm{O}$ ($x$ atoms) | $2x$ (2 $x$ H) | $3\mathrm{Mg}(\mathrm{OH})_{2}$, $2\mathrm{NH}_{3}$ | 12 (6 H from $\mathrm{Mg}(\mathrm{OH})_{2}$ + 6 H from $\mathrm{NH}_{3}$) |
O | $\mathrm{H}_{2}\mathrm{O}$ ($x$ atoms) | $x$ (1 $x$ O) | $3\mathrm{Mg}(\mathrm{OH})_{2}$ | 6 (2 $3$ O from $\mathrm{Mg}(\mathrm{OH})_{2}$) |
Both Mg and N atoms are already balanced. To balance the hydrogen (H) and oxygen (O) atoms:
Hydrogen (H): The product side has a total of 12 hydrogen atoms (6 from $3 \mathrm{Mg}(\mathrm{OH})_{2}$ and 6 from 2 $\mathrm{NH}_{3}$). Thus: [ 2x = 12 \implies x = 6 ]
Oxygen (O): The product side has a total of 6 oxygen atoms, which must be equal to $x$: $ x = 6 $
Therefore, to balance the equation, the value of $x$ should be 6.
So, the balanced equation is: $$ \mathrm{Mg}_{3}\mathrm{N}_{2}(\mathrm{aq}) + 6\mathrm{H}_{2}\mathrm{O}(\mathrm{I}) \rightarrow 3\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{aq}) + 2\mathrm{NH}_{3}(\mathrm{g}) $$
What is the name given to the salts of: (a) sulphurous acid, (b) sulphuric acid?
The names of the salts are:
Sulphurous Acid:
Hydrogen Sulphites (also known as Bisulphites)
Sulphites
Sulphuric Acid:
Sulphates
Bisulphates
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