Solubility product of a salt AB is 1 10^ -8 in a solution in whic… - Vidyadip

MCQ

Solubility product of a salt $AB$ is $1\times10^{-8}$ in a solution in which the concentration of $A^+$ ions is $10^{-3}\, M$. The salt will precipitate when the concentration of $B^-$ ions is kept

A
between $10^{-8}\, M$ to $10^{-7}\, M$
B
between $10^{-7}\, M$ to $10^{-8}\, M$
✓
$ > 10^{-5}\, M$
D
$ < 10^{-8}\, M$

✓

Answer

Correct option: C.

$ > 10^{-5}\, M$

c
$AB \leftrightarrow {A^ + } + {B^ - }$

${K_{sp}} = [{A^ + }][{B^ - }]$

Salt will precipitate if ionic conc. $ > {K_{sp}}$

$[{A^ + }][{B^ - }] > 1 \times {10^{ - 8}}$

$(1 \times {10^{ - 3}})[{B^ - }] > 1 \times {10^{ - 8}}$

$[{B^ - }] > \frac{{1 \times {{10}^{ - 8}}}}{{1 \times {{10}^{ - 3}}}}$ or $1 \times {10^{ - 5}}$

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