Question

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.

1. 
   

2. Calculate volume of KCN required to reach first and second equivalence points

   and sketch by hand the expected titration curve.

3. Calculate pCN- at 8.00 mL KCN added.

Answer 1

Given,

            volume of mixture = 10 ml = 0.01 L

            molarity of Ag⁺ = 0.1 M

            molarity of Hg₂²⁺ = 0.045 M

            molarity of KCN = 0.08 M

From standard data: Ksp of (Hg₂(CN)₂) = 5·10^-40

                                   Ksp of (AgCN) = 2.2·10^-16

Hg₂²⁺ + 2CN^- -------->Hg₂(CN)₂ --------------------(1)

Ag⁺ + CN^- -------->AgCN ------------------(2)

Hg₂(CN)₂ will precipitated first, as it has low Ksp.

From eq (1) : 1 mol Hg₂²⁺ = 2 mol CN^- ---------(3)

number of moles of Hg₂²⁺ = Molarity * volume

                                     = 0.045 mol/L * 0.01 L

                                     = 0.00045 moles Hg₂²⁺

moles of CN^- = Molarity * volume

                   = 0.08 mol/L * Volume

From eq (3): 2*(0.00045 moles Hg₂²⁺ ) = 0.08 mol/L * Volume

                       Volume of CN^- = 0.01125 L = 11.25mL (First equivalence point)

Similarly;     

From eq (1) : 1 mol Ag⁺ = 1 mol CN^- ---------(3)

number of moles of Ag⁺= Molarity * volume

                                   = 0.1 mol/L * 0.01 L

                                   = 0.001 moles Ag⁺

moles of CN^- = Molarity * volume

                   = 0.08 mol/L * Volume

From eq (3): (0.001 moles Ag⁺ ) = 0.08 mol/L * Volume

                       Volume of CN^- = 0.0125 L = 12.5mL (Second equivalence point)

pCN^- = - log [CN^-]            

Hg₂^{2+ =} {0.00045 moles - (0.008L * 0.08mol/L) 0.5 } / 0.01 + 0.008     

        = 7.22 * 10^-3 M

from Ksp

CN- = 2.646 * 10^-19

pCN^- = - log [CN^-]         

pCN^- = - log 2.646 * 10^-19        

pCN^- = 18.58