Question

30.0mL of HNO3 is titrated with 0.150M NaOH and it required 25.0mL of NaOH to reach the

equivalence point. Calculate the molar solubility of silver cyanide (AgCN) in the titration solution:

a) Before any NaOH has been added

b) After 10.0mL of NaOH is added

c) At the equivalence point

d) After 30mL of NaOH is added

ksp AgCN = 1.6x10^-14

Answer 1

a) The molat concentration of HNO3 = 0.15 M × 25 mL/30 mL = 0.125 M

AgCN(s) + HNO3(aq) AgNO3(aq) + HCN(aq)

Ksp of AgCN = 1.6×10^-14

K value for HNO3 dissociation = 24

Therefore, the resultant K value for the above equilibrium

= 1.6×10^-14×24 = 3.84×10^-13

As the equilibrium constant increases, the solubility does also increase.

b) As NaOH is added, the solubility decreases due to the partial neutralization of HNO3 by added NaOH.

c) At the equivalence point, there is the formaion of KNO3, which further increases the solubility of AgCN, as shown below.

AgCN(s) + KNO3(aq) AgNO3(aq) + KCN(aq); the resultant K > Ksp

d) As the NaOH exceeds equivalence point, there is KCN as well as NaOH to further increase the solubility of AgCN, as shown below.

AgCN(s) + NaOH(aq) AgOH(s) + NaCN(aq) (Ksp of AgOH is more than that of AgCN).