Question

The solubility product of Ag₂C₂O₄ at 25⁰C is 1.29×10^-11 mol³ L^-3. Solution ofK₂C₂O₄ containing 0.1520 mole in 500 mL of water is shaken with excess of Ag₂CO₃ till the following equilibrium is reached.
Ag₂CO₃+K₂C₂O₄⇌Ag₂CO_­4 + K₂CO₃
At equilibrium the solution contains 0.0358 mole of K₂CO_3- Assuming the degree of dissociation of K₂CO₃ to be equal, the solubility product of Ag₂CO₃ will be

• A. 1.96×10^-12
• B. 8.48×10^-11
• C. 1.29×10^-11
• D. 3.97×10^-12

Answer 1

The correct option is D

3.97×10^-12

$A{g}_{2}C{O}_3+K_2{C}_2{O}_4\rightleftharpoons A{g}_2{C}_2{O}_4+K_{2}C{O}_{30}$
At $t=0,0.1520$ mol $0$
At equilibrium
$0.1520-0.0358$ $0.0358$ mol
Concentration
$2\times 0.1162$ $2\times 0.0358$ mol
$=0.2324$ M $=0.0716$ M
$K_{SP}\left(A{g}_{2}C{O}_3\right)=\left[A{g}^{+}{]}^2\right[C_2{O}_4^{2-}]$
Or $\left[A{g}^{+}\right]={\left[\frac{K_{SP}\left(A{g}_2{C}_2{O}_4\right)}{\left[C{O}_3^{2-}\right]}\right]}^{\frac{1}{2}}\cdots \left(i\right)$
And, $K_{SP}\left(A{g}_{2}C{O}_3\right)=\left[A{g}^{+}{]}^2\right[C{O}_3^{2-}]$
or $\left[A{g}^{+}\right]={\left[\frac{K_{SP}\left(A{g}_{2}C{O}_3\right)}{\left[C{O}_3^{2-}\right]}\right]}^{\frac{1}{2}}\cdots \left(ii\right)$
from eqs. $\left(i\right)$ and $\left(ii\right)$,
${\left[\frac{K_{SP}\left(A{g}_2{C}_2{O}_4\right)}{\left[C{O}_3^{2-}\right]}\right]}^{\frac{1}{2}}={\left[\frac{K_{SP}\left(A{g}_{2}C{O}_3\right)}{\left[C{O}_3^{2-}\right]}\right]}^{\frac{1}{2}}$