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 * 10^(-9)`.

Step-by-step 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 * 10^(-12) = (2S)^(2)(S)\\8.2 * 10^(-2) = 4S^(3)\\S = 1.27 * 10^(-4)`

This means that
`[Ag^(+)]` is
`1.27 * 10^(-4)`. Now, the concentration of
`Cl^(-)` is calculated as follows.

`[Cl^(-)] = (mass)/(molar mass)\\= (0.003 g)/(35.5 g/mol)\\= 8.45 * 10^(-5) M`

Hence,
`K_(sp)` for AgCl is calculated as follows.

`K_(sp) = [Ag^(+)] * [Cl^(-)]\\= 1.27 * 10^(-4) * 8.45 * 10^(-5)\\= 10.73 * 10^(-9)`

Thus, we can conclude that solubility product of AgCl is
`10.73 * 10^(-9)`.