Question

In a solution, the concentration of hydroxide ions is $$ 8 \times 10^{-11} \mathrm{M} $$. What is the pH of the solution?

To find the pH, we first need to find the pOH using the concentration of hydroxide ions (( \mathrm{OH}^- )):

$$ \text{pOH} = -\log[ \mathrm{OH}^-] $$

Substitute the given concentration into the formula:

$$ \text{pOH} = -\log(8 \times 10^{-11}) $$

We can express this as:

$$ \text{pOH} = -\log(8) - \log(10^{-11}) $$

Using the properties of logarithms, we get:

$$ \text{pOH} = -\log(8) + 11 $$

We know that ( \log(8) \approx 0.903 ):

$$ \text{pOH} = -0.903 + 11 $$

Therefore, the pOH is:

$$ \text{pOH} \approx 10.097 $$

To find the pH, we use the relationship between pH and pOH:

$$ \text{pH} + \text{pOH} = 14 $$

Substitute the value of pOH into this equation:

$$ \text{pH} + 10.097 = 14 $$

Solving for pH:

$$ \text{pH} = 14 - 10.097 $$

$$ \text{pH} \approx 3.903 $$

Therefore, the correct option is:

C. 3.9

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Answer

To determine the $\text{pH}$ of an alcohol solution with a hydroxide ion concentration of $8 \times 10^{-11} , \text{M}$, follow these steps:

  1. Find the pOH using the formula: $$ \text{pOH} = -\log [\text{OH}^-] $$

    Given that $[\text{OH}^-] = 8 \times 10^{-11} , \text{M}$, $$ \text{pOH} = -\log (8 \times 10^{-11}) $$ Approximating, we get: $$ \text{pOH} \approx 10.1 $$

  2. Convert pOH to pH: Using the relation $\text{pH} + \text{pOH} = 14$, $$ \text{pH} = 14 - \text{pOH} $$

    Substituting the computed pOH value, $$ \text{pH} = 14 - 10.1 = 3.9 $$

Hence, the $\text{pH}$ of the alcohol solution is $\textbf{3.9}$. Thus, the correct option is:

C. 3.9


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