The solubility (in mol $$\text{L}^{-1}$$) of AgCl ($$K_{sp} = 1.0 \times 10^{-10}$$) in a 0.1 M KCl solution will be

For the salt $$AgCl(s)$$ the dissolution equilibrium in water is

$$AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)$$

The solubility-product expression is

$$K_{sp} = [Ag^{+}][Cl^{-}]$$

Given $$K_{sp} = 1.0 \times 10^{-10}$$.

Let $$s$$ be the molar solubility of $$AgCl$$ in the 0.1 M $$KCl$$ solution.

• When $$AgCl$$ dissolves, it adds $$s$$ mol L$$^{-1}$$ of $$Ag^{+}$$ ions.
• The chloride ion concentration is the sum of that from $$KCl$$ (0.1 M) and from the dissolved $$AgCl$$ (another $$s$$). Thus

$$[Cl^{-}] = 0.1 + s$$

Substituting in the $$K_{sp}$$ expression:

$$1.0 \times 10^{-10} = s\,(0.1 + s)$$

Because the added solubility $$s$$ will be extremely small compared with 0.1 M, we approximate $$0.1 + s \approx 0.1$$. Hence

$$1.0 \times 10^{-10} \approx s \times 0.1$$

$$\Rightarrow \; s = \frac{1.0 \times 10^{-10}}{0.1} = 1.0 \times 10^{-9}\; \text{mol L}^{-1}$$

Therefore, the solubility of $$AgCl$$ in a 0.1 M $$KCl$$ solution is $$1.0 \times 10^{-9}\; \text{mol L}^{-1}$$.

Option A which is: $$1.0 \times 10^{-9}$$