General instructions You must show fulldetailed calculations

General instructions: You must show full/detailed calculations to receive full marks for each problem. Please show your FINAL answer to correct number of significant figures. During intermediate steps, keep all digits in your calculator to avoid large rounding errors in the final answer. In other words, do not round off your answers during intermediate calculation steps. Use Ksp values from textbook – Harris 8th or 9th edition 1. A mixture having a volume of 10.00 mL and containing 0.1000 M Ag* and 0.04500 M Hg22* was titrated with 0.0800 M KCN to precipitate Hg2(CN)2 and AgCN. (b) (5 marks) At the first equivalence point in the above titration, will the second precipitate already start to form? Explain why or why not and show your calculation to derive the answer. (i) (2 marks) Can you propose a better titrant to perform this titration? Justify your reasoning briefly.

Solution

Data required :

Ksp(AgCN)=1.6*10^-14

Ksp(Hg2(CN)2)=5*10^-40

[Ag+]o=initial concentration=0.1mol/L

[Hg22+]o=0.045M

As the solubility product (product of ion concentration) gives the measure of the solubility of ionic salts,so it can be inferred that Hg2(CN)2 is very less soluble than AgCN in aqueous solvent.So Hg2(CN)2 will start precipitating .

AgCN(s) <---->Ag⁺(aq) +CN-(aq)

Hg2(CN)2<--->Hg₂²⁺ (aq)+2CN-(aq)

Ksp(Hg2(CN)2)=5*10^-40=[Hg₂²⁺] [CN-(aq)]^2

given: [CN]=0.08M

[Hg22+]eq=5*10^-40/(0.08)^2=7.8125e-38M (very very less)

Similarly,[Ag+][CN-]=Ksp(AgCN)=1.6*10^-14

or, [Ag+]eq=(1.6*10^-14)/[CN-]=(1.6*10^-14)/(0.08)=2e-13M

This calculation shows that AgCN starts precipitating when [Hg22+] concentration reaches 7.8125e-38M ( at the first equivalence point,)

ii) ksp(Ag2S)=1*10^-50

ksp(HgS)=1*10^-54

The ksp difference must be high for the salt and there is no better titrant as such to form soluble salts for these two ions with a larger Ksp gap.