Answer to Question #270732 in General Chemistry for toby

Question

The equilibrium constant for the formation of Cu(CN)42- is 2.0 x 1030. Calculate the value of pCu2+, i.e. -log[Cu2+], if 2.48 g of CuCl2 is dissolved in 1.000 L of a 0.937 M NaCN solution. The addition of CuCl2 does not affect the volume (the final volume is always 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-] = 2×10³⁰

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^-     [Cu^2+ = S, CN^- =S]

K_ = S ×4S⁴ = 4S⁵

K×[Cu(CN)₄^2-] = 4S⁵

S =[(2×10³⁰) × (2.05 × 10^-5)/4]^1/5 = 100495.0737

p[Cu²⁺] or -log[Cu²⁺] = -log(100495.0737) = -5.002 (Answer)