Here's the English translation of the given text: PbF2 (molar mass = 245 g/mol) has a solubility of 0.46 grams/liter. What is the solubility product constant value? A. 1.1 x 10^-10 B. 2.6 x 10^-8 C. 1.1 x 10^-7 D. 6.8 x 10^-9
Question
Here's the English translation of the given text:
$\mathrm{PbF}_{2}$ (molar mass = 245 g/mol) has a solubility of 0.46 grams/liter. What is the solubility product constant value?
- A. $1.1 \times 10^{-10}$
- B. $2.6 \times 10^{-8}$
- C. $1.1 \times 10^{-7}$
- D. $6.8 \times 10^{-9}$
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Answer
:
Given: Molar mass of $\mathrm{PbF}_{2}$ = 245 g/mol
Step 1: Calculate Molar Solubility
[ \text{Solubility} = 0.46 , \text{g/L} ] [ \text{Molar Solubility} = \frac{0.46}{245} , \text{mol/L} ] [ \text{Molar Solubility} = 0.00188 , \text{mol/L} ]
Step 2: Write Dissociation Equation and Express in Terms of Molar Solubility
[ \mathrm{PbF}_{2} \leftrightarrow \mathrm{Pb}^{2+} + 2\mathrm{F}^{-} ]
Let the molar solubility be $ s $: [ [\mathrm{Pb}^{2+}] = s = 0.00188 , \text{mol/L} ] [ [\mathrm{F}^{-}] = 2s = 2 \times 0.00188 = 0.00376 , \text{mol/L} ]
Step 3: Use Solubility Values to Calculate $ K_{sp} $
[ K_{sp} = [\mathrm{Pb}^{2+}] \times [\mathrm{F}^{-}]^2 = s \times (2s)^2 ] [ K_{sp} = s \times 4s^2 = 4s^3 ] [ K_{sp} = 4 \times (0.00188)^3 ] [ K_{sp} = 4 \times (6.648 \times 10^{-9}) ] [ K_{sp} = 2.6 \times 10^{-8} ]
Therefore, the value of the solubility product is:
[ \boxed{2.6 \times 10^{-8}} ]
Thus, the correct answer is B.
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