Ionic EquilibriumHard

Question

The solubility product of PbI2 is 7.2 × 10−9. The maximum mass of NaI which may be added in 500 ml of 0.005 M-Pb(NO3)2 solution without any precipitation of PbI2 is (I = 127)

Options

A.0.09 g
B.1.2 × 10−3 g
C.6 × 10−4 g
D.1.08 × 10−5 g

Solution

PbI2 (s) $\rightleftharpoons$Pb2+ + 2I–

To prevent ppt, Q ≤ Ksp

or, 0.005 × [I–]2 ≤ 7.2 × 10–9

∴ [I– ]max = 1.2 × 10–3 M

Hence, maximum mass of NaI = $\left( \frac{500 \times 1.2 \times 10^{- 3}}{1000} \right) \times 150 = 0.09\text{ gm}$

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