Chemical and Ionic Equilibrium - Result Question 114
####58. A sample of $AgCl$ was treated with $5.00 mL$ of $1.5 M$ $Na _2 CO _3$ solution to give $Ag _2 CO _3$. The remaining solution contained $0.0026 g$ of $Cl^{-}$ions per litre. Calculate the solubility product of $AgCl$. $\left[K _{sp}\left(Ag _2 CO _3\right)=8.2 \times 10^{-12}\right]$
$(1997,5 M)$
Show Answer
Solution:
- $2 AgCl(s)+CO _3^{2-} \rightleftharpoons Ag _2 CO _3(s)+2 Cl^{-}$
$$ \begin{aligned} K & =\frac{\left[Cl^{-}\right]^{2}}{\left[CO _3^{2-}\right]}=\frac{\left[Cl^{-}\right]^{2}}{\left[CO _3^{2-}\right]} \times \frac{\left[Ag^{+}\right]^{2}}{\left[Ag^{+}\right]^{2}}=\frac{\left[K _{\text {sp }}(AgCl)\right]^{2}}{K _{\text {sp }}\left(Ag _2 CO _3\right)} \ {\left[Cl^{-}\right] } & =\frac{0.0026}{35.5} M=7.3 \times 10^{-5} M \end{aligned} $$
The above concentration of $Cl^{-}$indicates that $\left[CO _3^{2-}\right]$ remains almost unchanged.
$$ \begin{aligned} & \frac{7.3 \times 10^{-5}}{1.5}=\frac{\left[K _{\text {sp }}(AgCl)\right]^{2}}{8.2 \times 10^{-12}} \ & K _{\text {sp }}(AgCl)=2 \times 10^{-8} \end{aligned} $$
