Question

The solubility of $A{g}_{2}C{O}_3$ in water is $1.26\times {10}^{-4}mole/litre$. What is its solubility in $0.02MN{a}_{2}C{O}_3$ solution?
Assume no hydrolysis of $C{O}_3^{2-}$ ion. (Take $\sqrt[3]{32}=1.26)$.

• A. $5\times {10}^{-6}M$
• B. $\sqrt{50}\times {10}^{-6}M$
• C. ${10}^{-5}M$
• D. $2\times {10}^{-5}M$

Answer 1

Answer: (C)

${10}^{-5}M$

In water,

$A{g}_{2}C{O}_3\rightleftharpoons 2A{g}^{+}+C{O}_3^{2-}$

$2s$ $s$

$K_{sp}=\left[A{g}^{+}{]}^2\right[C{O}_3^{2-}]=[2s{]}^2\left[s\right]=4s^3$

as $s=1.26\times {10}^{-4}$ mole $L^{-1}$

$K_{sp}=4\times [1.26\times {10}^{-4}{]}^3=8\times {10}^{-12}$

In $N{a}_{2}C{O}_3$ solution,

$A{g}_{2}C{O}_3\rightleftharpoons 2A{g}^{+}+C{O}_3^{2-}$

After adding $2s^{\prime }$ $s^{\prime }$

$0.02MN{a}_{2}C{O}_3$ $2s^{\prime }$ $[s^{\prime }+0.02]$

$K_{sp}=\left[A{g}^{+}{]}^2\right[C{O}_3^{2-}]$

$\Rightarrow 8\times {10}^{-12}=\left[2s^{\prime }{]}^2\right[s^{\prime }+0.02]$ as $s^{\prime }<<0.02$; so, $s^{\prime }+0.02\simeq 0.02$

$\Rightarrow 8\times {10}^{-12}=\left[2s^{\prime }{]}^2\right[0.02]$

$\Rightarrow [s^{\prime }{]}^2=1\times {10}^{-10}$

$\Rightarrow s^{\prime }={10}^{-5}moles/L$

Thus, solubility of $A{g}_{2}C{O}_3$ in $N{a}_{2}C{O}_3$ solution will be ${10}^{-5}moles/L$ .