Question

In an experiment, I mix 10ml 0.1M Cd(NO3)2 and 10ml 0.1M Na2C2O4, then I titrate the mixture with 5M NH3. The total volume of solution after titration is9.10ml. The Ksp value of CdC2O4 is 3.22x10^-9

I need to do the following: I have no idea how to do them, please help. Thanks.

Total moles of (C2O4) 2-

total moles of (Cd) 2+

Moles of NH3 that did not react with (Cd) 2+

Molarity of NH3 that did not react with (Cd) 2+

Answer 1

From the question given, I seriously feel that there are few information given incorrect. If 10ml of Cd(NO3)2 and 10ml Na2C2O4 are mixed and titration is done, how is the final volume 9.1 ml. Still, I have given the best possible answer according to your question below. Please go through it.

Equal moles of Cd(NO3)2 and Na2C2O4 in the mixture = 0.1 M*10 ml*0.001= 0.001

Now, for titration to be complete, total moles of NH3 to be used = 0.001

The first double displacement reaction involved is:

Further, the titration reaction involved is:
NH3 + H2O --> NH4OH   (For amonia in aqueous form)

Now,
Moles of CdC2O4 formed after the above double displacement reaction= 0.001

Now, for the dissociation of CdC2O4 solution, Ksp value is 3.22x10^-9.
CdC2O4 -> Cd^2+ + C2O4^2-
(0.001-x)         x              x

Ksp= 3.22x10^-9 = -> [Cd^2+][C2O4^2-]/[CdC2O4] = x*x/(0.001-x)
Solving for x, x= 0.00000179= [Cd^2+]=[C2O4^2-]

Total moles of (C2O4) 2- = 0.00000179
total moles of (Cd) 2+ = 0.00000179

Now,
Thus, Moles of NH3 that did not react with (Cd) 2+ = Moles of NH3 used - moles of Cd2+

                                                                                           = 0.001-0.00000179= 0.00099821

Molarity of NH3 that did not react with (Cd) 2+ = Moles of NH3 remaining/ Total Volume (in ltr)

                                                                                  = 0.00099821/ 0.0091 = 0.10969 M