Answers
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
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