Ionic EquilibriumMedium

Question

The solubility of Pb(OH)2 in water is 6.0 × 10–6 M. The solubility of Pb(OH)2 in a buffer solution of pH = 8 is

Options

A.8.64 M
B.2.16 × 10−16 M
C.8.64 × 10−16 M
D.8.64 × 10−4 M

Solution

For saturated solution of Pb(OH)2 in water, Ksp = 4.S3 = 4 × (6 × 10–6)3 = 8.64 × 10–16

Now, for the given buffer, [OH–] = 10–6 M

∴ Solubility, S = $\left\lbrack Pb^{2 +} \right\rbrack = \frac{K_{sp}}{\left\lbrack OH^{-} \right\rbrack^{2}} = \frac{8.64 \times 10^{- 16}}{\left( 10^{- 6} \right)^{2}} = 8.64 \times 10^{- 4}M$

Create a free account to view solution

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

More Ionic Equilibrium Questions

The amino acid glycine (NH2CH2COOH) is basic because of its –NH2 group and acidic because of its –COOH group. By a proce...50 mL of 0.1 M NaOH is added to 60 mL of 0.15 M H3PO4 solution (K1, K2 and K3 for H3PO4 are 10-3, 10-8 and 10-13 respect...Which of the following compounds is not an antacid?...The curve in the figure shows the variation of pH during the course of titration of a weak acid, HA with a strong base (...A solution is 0.1 M CH3COOH and 0.1 M CH3COONa. Which of the following solution will change its pH significantly ?...