Question

A 130.0 −mL sample of a solution that is 2.6×10−3 M in AgNO3 is mixed with a 220.0 −mL sample of a solution that is 0.13 M in NaCN. After the solution reaches equilibrium, what concentration of Ag+(aq) remains? Answer with 2 sig. figures.

Answer 1

AgNO(s) <---> Ag+ (aq) + NO3-(aq)

Ag+ (aq) + CN-(aq) ---> AgCN(s)

Ag+ formed forms precipitate with CN- ,

Ag+ moles = AGNO3 moles = M x V = ( 2.6x10^-3) x ( 130/1000) = 0.000338

CN- moles = ( 0.13 x 220/1000) = 0.0286

Thus all Ag+ moles converted to AgCN , we use Ksp of AgCN to get [Ag+]

Now AgCN (s) <---> Ag+ (aq) + CN- (aq)

Ksp = [Ag+] [CN-] = S^2 = 1.6 x 10^ -16                    ( Ksp taken from web)

S = [Ag+] = sqrt ( 1.6x10^-16) = 1.3 x 10^ -8 M