The solubility product of is . The maximum mass of NaI which may be added in 500 ml of solution with — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

For no precipitation, the ionic product must equal $K_{sp}$: $[\text{Pb}^{2+}][\text{I}^-]^2 = 7.2 \times 10^{-9}$. Given $[\text{Pb}^{2+}] = 0.005\text{ M}$, $[\text{I}^-]^2 = \frac{7.2 \times 10^{-9}}{0.005} = 1.44 \times 10^{-6} \implies [\text{I}^-] = 1.2 \times 10^{-3}\text{ M}$. Moles of NaI required in $500\text{ ml}$ (0.5 L) $= 0.5 \times 1.2 \times 10^{-3} = 6.0 \times 10^{-4}\text{ moles}$. Molar mass of NaI $= 23 + 127 = 150\text{ g/mol}$. Mass $= 6.0 \times 10^{-4} \times 150 = 0.09\text{ g}$.