At 25 ,^oC ,, , K_ sp for PbBr_2 is equal to 8 10^ -5 . If the sa… - Vidyadip

MCQ

At $25\,^oC\,,\, K_{sp}$ for $PbBr_2$ is equal to $8 \times 10^{-5}$ . If the salt is $80\%$ dissociated, what is the solubility of $PbBr_2$ in $mol/litre$ ?

✓
${\left[ {\frac{{{{10}^{ - 4}}}}{{1.6 \times 1.6}}} \right]^{1/3}}$
B
${\left[ {\frac{{{{10}^{ - 5}}}}{{1.6 \times 1.6}}} \right]^{1/3}}$
C
${\left[ {\frac{{{{10}^{ - 4}}}}{{0.8 \times 0.8}}} \right]^{1/3}}$
D
${\left[ {\frac{{{{10}^{ - 5}}}}{{1.6 \times 1.6}}} \right]^{1/2}}$

✓

Answer

Correct option: A.

${\left[ {\frac{{{{10}^{ - 4}}}}{{1.6 \times 1.6}}} \right]^{1/3}}$

a
$PbB{r_2} \rightleftharpoons \mathop {P{b^{ + 2}}(aq)}\limits_{0.8\,S} + \mathop {2B{r^ - }(aq)}\limits_{2 \times 0.8\,S} $

$8 \times 10^{-5}=\mathrm{K}_{\mathrm{sp}}=(0.8\, \mathrm{S})(1.6 \,\mathrm{S})^{2}$

$S=\left(\frac{10^{-4}}{1.6 \times 1.6}\right)^{1 / 3}$

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