Ionic EquilibriumHard

Question

The solubility of AgCN in a buffer solution of pH = 3.0 is (Ksp of AgCN = 1.2 × 10−16; Ka of HCN = 4.8 × 10−10)

Options

A.1.58 × 10−5 M
B.2.0 × 10−5 M
C.1.58 × 10−4 M
D.2.5 × 10−9 M

Solution

AgCN (s) $\rightleftharpoons$ Ag+ + CN–; Ksp = 1.2 × 10–16

$CN^{-} + H^{+} \rightleftharpoons HCN;K_{eq} = \frac{1}{K_{a}} = \frac{1}{4.8 \times 10^{- 10}}$

Adding both reactions,

AgCN (s) + H+ $\rightleftharpoons$ Ag+ + HCN; $K_{eq} = \frac{K_{sp}}{K_{a}} = \frac{1}{4 \times 10^{6}}$

10–3 SM SM

Now, $\frac{1}{4 \times 10^{6}} = \frac{S.S}{10^{- 3}} \Rightarrow 1.58 \times 10^{- 5}\text{ M}$

Create a free account to view solution

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

More Ionic Equilibrium Questions

For a sparingly soluble salt ApBq, the relationship of its solubility product (LS) with its solubility (S) is...A weak acid (HA) after treatment with 12 mL of 0.1 M strong base (BOH) has a pH of 5. At the end point, the volume of sa...Ksp of AgCl is 1.96 × 10-10. 100 mL of saturated AgCl solution is titrated with 1 × 10-5 M NH4 SCN. Volume of ...Which of the following is/are correct regarding buffer solution ?...Ka for the acid HA is 1 × 10-6. The value of K for the reaction A- + H3O+ ⇋ HA + H2O is...