Question

A 125.0 −mL sample of a solution that is 2.9×10−3 M in AgNO₃ is mixed with a 225.0 −mL sample of a solution that is 0.12 M in NaCN. After the solution reaches equilibrium, what concentration of Ag⁺(aq) remains?

Answer 1

after mixing volume of solution = 125+225 = 350 ml

using dilution law we find final molarity of AgNO3 , M1V1 = M2V2

2.9 x 10^-3 x 125 = M2 x 350 ,     M2 = 0.001036 M , [Ag+] = 0.001036 M

for NaCN , M2 = ( 0.12 x 225 /350) = 0.077 , [CN-] = 0.077M

Ag+ (aq) + CN- (aq) --> AgCN (s) , Ksp = 1.6 x 10^-14

[CN-] left after reactiong with 0.001036 [Ag+]   = 0.077-0.001036 = 0.076 M

Ksp = [Ag+] [CN-]

1.6 x 10^-14 = [Ag+] ( 0.076)

[Ag+] = 2.1 x 10^-13 M