Question

A sample of AgCl is treated with 5mL of 2M Na2CO3 solution to produce Ag2CO3. The remaining solution contained 0.003gm of Cl per litre. Calculate solubility product of AgCl.(Ksp of Ag2CO3=8.2×10^-12)​

Answer 1

Answer: The solubility product of AgCl is $10.73 \times 10^{-9}$.

Explanation:

The reaction equation is as follows.

$Ag_{2}CO_{3} \rightleftharpoons 2Ag^{+} + CO^{2-}_{3}$

Let us assume the concentration of $2Ag^{+}$ is 2S and concentration of $CO^{2-}_{3}$ is S. Hence, the expression for $K_{sp}$ of this reaction is as follows.

$K_{sp} = [Ag^{+}]^{2}[CO^{2-}_{3}]\\8.2 \times 10^{-12} = (2S)^{2}(S)\\8.2 \times 10^{-2} = 4S^{3}\\S = 1.27 \times 10^{-4}$

This means that $[Ag^{+}]$ is $1.27 \times 10^{-4}$. Now, the concentration of $Cl^{-}$ is calculated as follows.

$[Cl^{-}] = \frac{mass}{molar mass}\\= \frac{0.003 g}{35.5 g/mol}\\= 8.45 \times 10^{-5} M$

Hence,  $K_{sp}$ for AgCl is calculated as follows.

$K_{sp} = [Ag^{+}] \times [Cl^{-}]\\= 1.27 \times 10^{-4} \times 8.45 \times 10^{-5}\\= 10.73 \times 10^{-9}$

Thus, we can conclude that solubility product of AgCl is $10.73 \times 10^{-9}$.