Ionic EquilibriumHard

Question

The formation constant of Cu(NH3)42+ is 1.25 × 1012. What will be the equilibrium concentration of Cu2+ if 0.0125 moles of Cu is oxidized and put into 1.0 L of 0.25 M–NH3 solution?

Options

A.2.5 × 10−11 M
B.2.5 × 10−13 M
C.4 × 10−12 M
D.6.25 × 10−12 M

Solution

$Cu^{2 +} + 4NH_{3} \rightleftharpoons Cu\left( NH_{3} \right)_{4}^{2 +};K_{eq} = 1.25 \times 10^{12}$

0.0125 M 0.25 M 0

100% 0 0.25 – 0.05 = 0.20 M 0.0125 M

Equ. xM 0.2 + 4 x 0.0125 – x

≈ 0.2 M ≈ 0.0125 M

Now, $1.25 \times 10^{12} = \frac{0.0125}{x \times (0.2)^{4}} \Rightarrow x = 6.25 \times 10^{- 12}$

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