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 = $\left( \frac{500 \times 1.2 \times 10^{- 3}}{1000} \right) \times 150 = 0.09\text{ gm}$

Create a free account to view solution

View Solution Free
Topic: Ionic Equilibrium·Practice all Ionic Equilibrium questions

More Ionic Equilibrium Questions

What is the pH value of M H2SO4 ?...A buffer solution is prepared in which the concentration of NH3 is 0.30 M and the concentration of NH4+ is 0.20 M. If th...The $pK_a$ of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized...Find the number of P - O - P linkage in (HPO3)n where n = 3...A buffer solution contains 1 mole of (NH4)2SO4 and 1 mole of NH4OH (Kb = 10-5). The pH of solution will be:...