The solubility product of lead bromide is 8 10^ -5 . If the salt … - Vidyadip

MCQ

The solubility product of lead bromide is $8\times10^{-5}$. If the salt is $80\%$ dissociated in saturated solution, find the solubility of the salt

A
$1.7\times10^{-4}\, M$
B
$2.3\times10^{-6}\, M$
C
$1.8\times10^{-4}\, M$
✓
$3.4\times10^{-2}\, M$

✓

Answer

Correct option: D.

$3.4\times10^{-2}\, M$

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

$\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Br}^{-}\right]^{2}$

$\Rightarrow 8 \times 10^{-5}=(0.8\, \mathrm{S}) \times(2 \times 0.8\, \mathrm{S})^{2}=2.048 \,\mathrm{S}^{3}$

$\Rightarrow \mathrm{S}=3.4 \times 10^{-2}\, \mathrm{M}$

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