Question

The K. of Cul is 1.1x10 and the Krfor the [Cu(CN), complex ion is 1.0x10 - Write the equilibrium reaction associated with the given K value b) Write the equilibrium reaction associated with the given Kr value. Find the final concentration of cyanide ion in a solution containing excess Cul and 0.60 M NaCN

Answer 1

a)

CuI(s) <-------> Cu⁺(aq) + I^-(aq)

K_sp = [Cu⁺][I^-] = 1.1×10^-12

b)

Cu⁺(aq) + 2CN^-(aq) <-------> Cu(CN)₂^-(aq)

K_f = [Cu(CN)₂^-] /( [Cu⁺] [CN^-]²) = 1.0 ×10^{​​​​​​24}

c)

Adding two equilbrium

CuI(s) + 2CN^-(aq) <-------> Cu(CN)₂^- (aq) + I^-(aq)

K = [Cu(CN)2^-][I^-]/[CN^-]²

K = K_sp × K_f = 1.1×10^-12 × 1.0 ×10²⁴ = 1.1 × 10¹²

Initial concentration

[CN^-] = 0.60

[Cu(CN)2^-] = 0

[I^-] = 0

change in concentration

[CN^-] = - 2x

[Cu(CN)2^-] = + x

[I^-] = + x

Equilibrium concentration

[CN^-] = 0.60 - 2x

[Cu(CN)2^-] = x

[I^-] = x

so ,

x²/ ( 0.60 - 2x)² = 1.1 ×10¹²

x/(0.60 - 2x) = 1048809

x = 629285 - 2097618x

x 2097619 = 629285

x = 0.30

[CN^-] = 0.60 - (2× 0.30) = 0

Therefore

Final concentration of CN^- almost nil