Question

The solubility product of PbI₂ is \( 8.0 \times 10^{-9} \). The solubility of lead iodide in 0.1 molar solution of lead nitrate is \( x \times 10^{-6} \) mol/L. The value of \( x \) is __________. (Rounded off to the nearest integer) [Given \( \sqrt{2} = 1.41 \)]

Hint:

The Common Ion Effect significantly {decreases} the solubility of a salt.

Answer 1

Correct Answer: 141

Step 1: $PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)$. $K_{sp} = [Pb^{2+}][I^-]^2$.
Step 2: In $0.1 \text{ M } Pb(NO_3)_2$, total $[Pb^{2+}] = (0.1 + s) \approx 0.1 \text{ M}$ (since $s$ is very small).
Step 3: $[I^-] = 2s$.
Step 4: $8.0 \times 10^{-9} = (0.1) \times (2s)^2 = 0.1 \times 4s^2 = 0.4s^2$.
Step 5: $s^2 = \frac{8.0 \times 10^{-9}}{0.4} = 20 \times 10^{-9} = 200 \times 10^{-10}$.
Step 6: $s = \sqrt{200} \times 10^{-5} = 10\sqrt{2} \times 10^{-5} = 1.41 \times 10^{-4} = 141 \times 10^{-6} \text{ mol/L}$. Thus, $x = 141$.