Question

Volumes and molarities for titration of CdC₂O₄ solutions.Solubility product of CdC2O4 and formation constant of [Cd(NH3)₄⁺²]

Part C Data

Volume of NH₃ added 8.35 ml

Solution is has 20 ml of H2SO4

Total Volume of Solution after Titration 128.35 ml

Molarity of Ammonia 5 M

Questions

Calculate the Total Moles ofC₂O₄

Calculate the Molarity of C₂O₄

Total moles of Cd2+ (mol)

Moles of [Cd(NH3)4 2+] (mol)

Molarity of [Cd(NH3)4 2+] (M)

Moles of NH3 added by titration (mol)

Moles of NH3 that did not react with Cd2+ (mol)

Molarity of NH3 that did not react with Cd2+ (M)

Kf for [Cd(NH3)4 2+]

Answer 1

Solubility product (Ksp) of CdC2O4 = 1.4*10^-8

Total moles of C2O4 = (1.4*10^-8)^1/2 mol/L * (128.35/1000) L = 1.52*10^-5 mol

Molarity of C2O4 = (1.4*10^-8)^1/2 mol/L = 1.183*10^-4 M

Total moles of Cd²⁺ = 1.52*10^-5 mol

Moles of [Cd(NH3)4]²⁺ = 1.52*10^-5 mol

Molarity of [Cd(NH3)4]²⁺ = 1.183*10^-4 M

Moles of NH3 added by titration = (8.35/1000) L * 5 mol/L = 0.04175 mol

Moles of NH3 that did not react with Cd²⁺ = 0.04175 mol - 1.52*10^-5 mol = 0.0417348 mol

Molarity of NH3 that did not react with Cd²⁺ = 0.0417348 mol/(128.35/1000) L = 0.325 M

Now, Kf = [Cd(NH3)4]²⁺/[NH₃]⁴ = 1.183*10^-4/(0.325)⁴ = 0.0106