Ionic EquilibriumHard

Question

For the reaction [Ag(CN)2]–­ $\rightleftharpoons$ Ag+ + 2CN–, the equilibrium constant, at 25°C, is 4.0 × 10–19. Calculate the silver ion concentration in a solution which was originally 0.10 molar in KCN and 0.03 molar in AgNO3.

Options

A.0
B.0.03 M
C.3 × 10−19 M
D.7.5 × 10−18 M

Solution

$Ag^{+} + 2CN^{-} \rightleftharpoons Ag(CN)_{2}^{-};K = \frac{1}{4 \times 10^{- 19}} = 2.5 \times 10^{18}$

0.03 M 0.1 M 0

100% 0 (0.1 – 0.06)M 0.03 M

= 0.04 M

Eqn. xM (0.04 + 2x)M (0.03 – x)M

≈ 0.04 M ≈ 0.03 M

Now, $2.5 \times 10^{18} = \frac{0.03}{x \times (0.04)^{2}} \Rightarrow \left\lbrack Ag^{+} \right\rbrack = x = 7.5 \times 10^{- 18}\text{ M}$

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