Question

What is the approximate concentration of free Cu²⁺ ion at equilibrium when 1.87×10^-2 mol copper(II) nitrate is added to 1.00 L of solution that is 1.320 M in NH₃. For [Cu(NH₃)₄]²⁺, K_f = 2.1×10¹³.

[Cu²⁺] = ____M

Answer 1

You have to realize with problems like these that the [Cu²⁺] at equilibrium is minuscule (extremely small); it has to be to result in a Kf of 2.1 x 10¹³. This allows us to make an assumptions that is completely justified and avoids nasty quadratic equations.

[Cu²⁺]   + 4NH₃   [Cu(NH₃)₄]²⁺

So,

Kf = { [Cu(NH₃)_4]²⁺ } / { [Cu²⁺] [NH₃]⁴ }

2.1 x 10¹³ = { [Cu(NH₃)₄]²⁺ } / { [Cu²⁺] [NH₃]⁴ }

moles of Cu(NO₃)₂ = 2 mole and this is present in 1.00 L

So, [Cu(NO₃)₂] = 2 mole / 1 L = 2 M

At equilibrium let [Cu²⁺] = x M

[Cu(NH₃)₄²⁺] = (2 – x) M = 2 M

[NH₃] at equilibrium = (1.32 – 4x) M = 1.32 M

(since x is very small as compared to Kf value, so 1.32 – 4x can be written as 1.32)

So,

2.1 x 10¹³ = [Cu(NH₃)₄²⁺] / [Cu²⁺][NH₃]⁴

2.1 x 10¹³ = (2 ) / { (x) (1.32)⁴}

2.1 x 10¹³ = (2) / (3.04x)

(6.38 x 10¹³)x = 2

x = 2 / (6.38 x 10¹³)

x = (3.13 x 10^-14)