Question

Copper(l) ions in aqueous solution react with NH, (aq) according to Cu* (aq) + 2 NH, (aq) – Cu(NH4)2 (aq) K = 6.3 x 10' Calculate the solubility (ing.L-) of CuBr(s) (Kyp = 6,3 x 10") in 0.46 M NH, (aq). solubility of CuBr(s): eston source URG - General Chemistry Publisher University Science to

Answer 1

Solubility equilibrium of CuBr

CuBr(s) <------> Cu⁺(aq) + Br^-(aq)

Ksp = [Cu⁺][Br^-] = 6.3×10^-9

Formation equilibrium of Cu(NH3)2⁺ is

Cu⁺(aq) + 2NH3(aq) <-------> Cu(NH3)2⁺(aq)

K_f = [Cu(NH3)^{​​​​​​}2⁺]/([Cu⁺][NH3]²) = 6.3×10¹⁰

Adding two equation

CuBr(s) + 2NH3(aq) <-------> Cu(NH3)2⁺(aq) + Br^-(aq)

K = [Cu(NH3)2⁺ ] [Br^-] / [NH3]²

K = K_sp × K_f

K = 6.3 ×10^-9 × 6.3 ×10¹⁰ = 396.9

Initial concentration

[NH3] = 0.46

[Cu(NH3)2⁺] = 0

[Br^-] = 0

Change in concentration

[NH3] = -2x

[Cu(NH3)2⁺] = +x

[Br^-] = + x

equilibrium concentration

[NH3] = 0.46 - 2x

[Cu(NH3)2⁺] = x

[Br^-] = x

so,

x²/ (0.46 - 2x)² = 396.9

x/(0.46 - 2x) = 19.92

x = 9.1632 - 39.84x

40.84x = 9.1632

x = 0.2244

Therefore ,

molar solubility of CuBr in 0.46M NH3 = 0.2244mol/L

mass of 0.2244moles of CuBr = 0.2244 mol × 143.45g/mol = 32.19g

Therefore

molar solubility of CuBr in 0.46 M NH3 = 32.19g/L