Question

The equilibrium constant for the formation of Cu(CN)4^2- is 2.0 x 10^-30. Calculate the value of pCu2+, or -log[Cu2+], if we were to dissolve 2.76 g of CuCl2 in 1.000 L of a solution 0.848 M in NaCN. The addition of CuCl2 does not alter the volume (the final volume is still 1.000 L).

Answer 1

CuCl₂ +4NaCN = Na₂[Cu(CN)₄] + 2NaCl

Cu(CN)₄^2- = Cu²⁺ + 4CN^-

K = [Cu²⁺][ CN^-]⁴ / [Cu(CN)₄^2-] = 2×10^-30

Molecular weight of CuCl₂ is 134.45 g/mol

Number of mol CuCl₂ is = 2.76/134.45 mol =0.0205 mol

Number of mol NaCN is = 0.848 × 1000 = 848 mol  

 [Cu(CN)₄^2-] = 0.0205 mol and C_M = 0.0205/1000 = 2.05 × 10^-5

Cu(CN)₄^2- = Cu²⁺ + 4CN^-     [Cu²⁺ = S, CN^- =S]

K = S ×4S⁴ = 4S⁵

K×[Cu(CN)₄^2-] = 4S⁵

S =[(2×10^-30) × (2.05 × 10^-5)/4]^1/5 = 1.00049× 10^-7

p[Cu²⁺] or -log[Cu²⁺] = -log(1.00049× 10^-7) = 6.999 (Answer)