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For the reaction ${BaF}_{2} (s) + 2 H^{+} (aq) ⇌ {Ba}^{2 +} (aq) + 2 HF (aq), K_{c} = 2.56 M$ the solubility of ${BaF}_{2}$
in a buffer solution of pH= 2.0 is

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a

0.03 M

b

0.08 M

c

0.06 M

d

0.04 M

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detailed solution

Correct option is D

Lets be the solubility of' ${BaF}_{2}$ , in a buffer solution of pH= 2 (i.e. $[H^{+}] = 10^{- 2} M)$ . Then ${BaF}_{2} (s) + 2 H^{+} (aq) \to \underset{s}{{Ba}^{2 +} (aq}) + \underset{2 s}{2 HF} (aq)$

$\begin{array}{l} K_{c} = \frac{[{Ba}^{2 +}] [HF]^{2}}{{[H^{+}]}^{2}} = \frac{(s) (2 s)^{2}}{{(10^{- 2} M)}^{2}}; 4 s^{3} = {(10^{- 2} M)}^{2} K_{c} = {(10^{- 2} M)}^{2} (2.56 M) \\ s = {[\frac{(10^{- 4} M^{2}) (2.56 M)}{4}]}^{1 / 3} = {(64 \times 10^{- 6} M)}^{1 / 3} = 4 \times 10^{- 2} M \end{array}$

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