A student followed the procedure of this experiment to determine the Ksp of zinc

Question

A student followed the procedure of this experiment to determine the Ksp of zinc(II) iodate, Zn(IO3)2. Solutions of Zn(NO3)2 of known concentrations were titrated with 0.200 M KIO3 solutions to the first appearance of a white precipitate. For each of the zinc(II) nitrate solution concentrations below, calculate the expected concentration of iodate that would be required to initiate precipitation of zinc(II) iodate. Show all calculations. (Assume that Ksp = 3.9 x 10-6 at 25oC for zinc(II) iodate). a) 0.100 M Zn(NO3)2: b) 0.0100 M Zn(NO3)2:

Explanation / Answer

Solution :-

Ksp of the Zn(IO3)2 is 3.9*10^-6

Using the ksp value lets calculate the concentration of the iodate needed to initiate the precipitation of the each Zn(NO3)2 solution

Zn(IO3)2 ----- > Zn^2+    + 2IO3^-

                              x    2x.

a) 0.100 M Zn(NO3)2

ksp = [Zn^2+] [IO3^-]^2

3.9*10^-6 = [0.100][2x]^2

3.9*10^-6/0.100 = 4x^2

3.9*10^-5 / 4 =x^2

9.75*10^-6 =x^2

taking square root of both sides we get

3.12*10^-3 = x

so the concentration of the iodate needed to initiate the precipitation is 9.92*10^-3 M

b) Calculating the concentration of the iodate when 0.0100 M Zn(NO3)2 is present.

ksp = [Zn^2+] [IO3^-]^2

3.9*10^-6 = [0.0100][2x]^2

3.9*10^-6/0.0100 = 4x^2

3.9*10^-4 / 4 =x^2

9.75*10^-5 =x^2

taking square root of both sides we get

9.87*10^-3 = x

So the concnetration of the iodate needed to initiate the precipitation is 9.87*10^-3 M

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