Question

A solution is 0.10 M Ba(NO3)2 and also 0.10 M AgNO3. Solid KIO3 is added slowly. (a) Calculate the concentration of iodate ion needed to saturate the solution with Ba(IO3)2(s) and with AgIO3(s). (b) Which compound precipitates first? Why?

Answer 1

Given, [Ba(NO₃)₂] = 0.10 M , [AgNO₃] = 0.10

Calculate the molarity of iodate

The Ksp values for Ba(IO₃)₂ and AgIO₃ are :

Ksp for AgIO₃ = 3.0 x 10^-8 and Ksp for Ba(IO₃)₂ = 1.57x 10^-9

First calculate the iodate concentration needed to saturate the Ba(IO₃)₂ as shown below:

Ksp = [Ba²⁺] [IO₃^-]²

1.57x 10^-9 = 0.10 * (2x)²

4x² = 1.57 x 10^-8

so, x = 6.26 x 10^-5 M

Thus, [IO₃^-] =2x = 2 *6.26 x 10^-5 M

                       = 1.25 x 10^-4

so, 1.25 x 10^-4 M iodate concentration needed to saturate the Ba(IO₃)₂

Now for AgIO₃

Ksp = [Ag⁺] [IO₃^-]

3.0 x 10^-8 = 0.10 M * x

x = 3.0 x 10^-7

3.0 x 10^-7 M iodate concentration needed to saturate the AgIO₃.

b) The Ksp value for Ba(IO₃)₂ is lower than AgIO₃, so Ba(IO₃)₂ will be first precipitate out.