Question

For which of the following concentrations , the sparingly soluble salt $CaC{O}_3$ will not precipitate at ${25}^{\circ }C$
Given,
$K_{sp}\left(CaC{O}_3\right)=3.8\times {10}^{-9}$

• A. $\left[C{a}^{2+}\right]=\left[C{O}_3^{2-}\right]={10}^{-4}M$
• B. $\left[C{a}^{2+}\right]=\left[C{O}_3^{2-}\right]={10}^{-5}M$
• C. $\left[C{a}^{2+}\right]=\left[C{O}_3^{2-}\right]={10}^{-6}M$
• D. $\left[C{a}^{2+}\right]=\left[C{O}_3^{2-}\right]={10}^{-7}M$

Answer 1

Answer: (B), (C), (D)

$\left[C{a}^{2+}\right]=\left[C{O}_3^{2-}\right]={10}^{-7}M$

Given, $CaC{O}_3(K_{sp}=3.8\times {10}^{-9})$
For precipitation , to occur the ionic product should exceed solubility product
$Q_{sp}>K_{sp}$
The equilibrium between undissociated solid and its ions can be represented as:
$CaC{O}_3\left(s\right)\rightleftharpoons C{a}^{2+}\left(aq\right)+C{O}_3^{2-}\left(aq\right)$
$Q_{sp}=\left[C{a}^{2+}\right]\left[C{O}_3^{2-}\right]$

Option A:
$Q_{sp}={10}^{-4}\times {10}^{-4}{Q}_{sp}={10}^{-8}$
Since, $Q_{sp}>K_{sp}$, precipitation occurs.
Option B:
$Q_{sp}={10}^{-5}\times {10}^{-5}{Q}_{sp}={10}^{-10}$
Since, $Q_{sp}<K_{sp}$, precipitation does not occur.
Option C:
$Q_{sp}={10}^{-6}\times {10}^{-6}{Q}_{sp}={10}^{-12}$
Since, $Q_{sp}<K_{sp}$, precipitation does not occur.
Option D:
$Q_{sp}={10}^{-7}\times {10}^{-7}{Q}_{sp}={10}^{-14}$
Since, $Q_{sp}<K_{sp}$, precipitation does not occur.
Option B,C,D are correct.