Question

The solubility product of $A{g}_2{C}_2{O}_4$ at ${25}^{\circ }C$ is $1.29\times {10}^{-11}mo{l}^3.li{t}^{-3}$. A solution of $K_2{C}_2{O}_4$ containing 0.1520 moles in 500 ml. Water is shaken at ${25}^{\circ }C$ with excess $A{g}_{2}C{O}_3$ till the following equilibrium is reached : $A{g}_{2}C{O}_3+K_2{C}_2{O}_4\to A{g}_2{C}_2{O}_4+K_{2}C{O}_3$. At equilibrium, the solution contains 0.0358 moles of $K_{2}C{O}_3$. Assuming the degree of dissociation of $K_2{C}_2{O}_4$ and $K_{2}C{O}_3$ to be equal,calculate the solubility product of $A{g}_{2}C{O}_3$.

• A. $1.194$ x ${10}^{-10}mo{l}^2.li{t}^{-2}$
• B. $3.974$ x ${10}^{-10}mo{l}^2.li{t}^{-2}$
• C. $3.974$ x ${10}^{-12}mo{l}^3.li{t}^{-3}$
• D. $1.164$ x ${10}^{-10}mo{l}^3.li{t}^{-3}$

Answer 1

Answer: (C)

$3.974$ x ${10}^{-12}mo{l}^3.li{t}^{-3}$

Initially, the concentration of oxalate ion was 0.1520 moles.

and at equilibrium the amount of carbonate ion formed=$0.1520-0.0358=0.1162$

The concentration of potassium oxalate remaining after the reaction is $\frac{0.1162}{0.5}=0.2324M$.
The concentration of potassium carbonate (or carbonate ions) at equilibrium $=\frac{0.0358}{0.5}=0.0716M$.
For silver oxalate, $K_{sp}=\left[A{g}^{+}{]}^2\right[C_{2}O{4}^{2-}]$.
Hence, $1.29\times {10}^{-11}=[A{g}^{+}{]}^2\times 0.2324$.
$[A{g}^{+}{]}^2=\frac{1.29\times {10}^{-11}}{0.2324}$.
The expression for the solubility product of silver carbonate is $K_{sp}=\left[A{g}^{+}{]}^2\right[C{O}_3^{2-}]=\frac{1.29\times {10}^{-11}}{0.2324}\times 0.0716=3.974\times {10}^{-12}mo{l}^3.li{t}^{-3}$.