Answer: The solubility product of AgCl is [tex]10.73 \times 10^{-9}[/tex].
Explanation:
The reaction equation is as follows.
[tex]Ag_{2}CO_{3} \rightleftharpoons 2Ag^{+} + CO^{2-}_{3}[/tex]
Let us assume the concentration of [tex]2Ag^{+}[/tex] is 2S and concentration of [tex]CO^{2-}_{3}[/tex] is S. Hence, the expression for [tex]K_{sp}[/tex] of this reaction is as follows.
[tex]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}[/tex]
This means that [tex][Ag^{+}][/tex] is [tex]1.27 \times 10^{-4}[/tex]. Now, the concentration of [tex]Cl^{-}[/tex] is calculated as follows.
[tex][Cl^{-}] = \frac{mass}{molar mass}\\= \frac{0.003 g}{35.5 g/mol}\\= 8.45 \times 10^{-5} M[/tex]
Hence, [tex]K_{sp}[/tex] for AgCl is calculated as follows.
[tex]K_{sp} = [Ag^{+}] \times [Cl^{-}]\\= 1.27 \times 10^{-4} \times 8.45 \times 10^{-5}\\= 10.73 \times 10^{-9}[/tex]
Thus, we can conclude that solubility product of AgCl is [tex]10.73 \times 10^{-9}[/tex].
