Question

Calculate the molar solubility of Ba(IO3)2 in a solution that is also 0.0600 M in Ba(NO3)2. (Ksp (Ba(IO3)2) = 1.57 x 10-9 )

Answer 1

Ba(NO3)2 here is Strong electrolyte

It will dissociate completely to give [Ba2+] = 0.06 M

At equilibrium:

Ba(IO3)2 <----> Ba2+ + 2 IO3-

   0.06 +s 2s

Ksp = [Ba2+][IO3-]^2

1.57*10^-9=(0.06 + s)*(2s)^2

Since Ksp is small, s can be ignored as compared to 6*10^-2

Above expression thus becomes:

1.57*10^-9=(0.06)*(2s)^2

1.57*10^-9= 0.06 * 4(s)^2

s = 8.09*10^-5 M

Answer: 8.09*10^-5 M