Question

A sample of $AgCl$ was treated with $5$ mL of $1.5M$ ${Na}_2{CO}_3$ solution to give ${Ag}_2{CO}_3$. The remaining solution contained $0.0026g$ of ${Cl}^{-}$ per litre. The solubility product of $AgCl$ is

($K_{sp}$ ${}_{\left({Ag}_2{CO}_3\right)}=8.2\times {10}^{-12}$)

• A. $K_{sp}=1.63\times {10}^{-10}$ $M^2$
• B. $K_{sp}=1.71\times {10}^{-10}$ $M^2$
• C. $K_{sp}=1.82\times {10}^{-10}$ $M^2$
• D. None of these

Answer 1

The correct option is B $K_{sp}=1.71\times {10}^{-10}$ $M^2$

$\underset{mmoladded7.5mmolleft(7.5-a)}{{Na}_2{CO}_3}+\underset{excessexcess}{2AgCl}\rightleftharpoons \underset{02a}{2NaCl}+\underset{0a}{{Ag}_2{CO}_3}$

Given, $\left[{Cl}^{-}\right]=\frac{0.0026}{35.5}=7.32\times {10}^{-5}M$

Also, conc. of ${Cl}^{-}$ formed $\frac{millimole}{volumeinmL}=\frac{2a}{5}$

$\therefore \frac{2a}{5}=\frac{0.0026}{35.5}$

$a=1.83\times {10}^{-4}millimole$

$\therefore$ milli mole of ${Na}_2{CO}_3$ left in $5mL=7.5-1.83\times {10}^{-4}=7.5$ or $\left[C{O}_3^{2-}\right]=\frac{7.5}{5}$

Now, $K_{{sp}_{{Ag}_2{CO}_2}}={\left[{Ag}^{+}\right]}^2\left[C{O}_3^{2-}\right]$

${\left[{Ag}^{+}\right]}^2=\frac{8.2\times {10}^{-12}}{7.5/5}=5.46\times {10}^{-12}$

$\left[{Ag}^{+}\right]=2.34\times {10}^{-6}$

Now,
$K_{sp}$ of $AgCl=\left[{Ag}^{+}\right]\left[{Cl}^{-}\right]$

$=2.34\times {10}^{-6}\times \frac{0.0026}{35.5}$

$K_{sp}=1.71\times {10}^{-10}$ ${mol}^2{litre}^{-2}$