Question

Calculate the molar solubility of La(IO)s in 0.050 M NalOs by taking into account the activity coefficients of the ions involved. Utilize for the activity coefficients: Where Iz is the ionic strength: CiZi With appropriate summation over all ions, ci being the molar concentration of the i'th ion and zi being its charge. Explain why you either ignore or take into account the additional ions from dissolution of La(I03)3. If you make any approximations, check whether they are justified.

Answer 1

The dissociation of La(IO₃)₃ can be written as

La(IO₃)₃(s) $\rightleftharpoons$ La³⁺(aq) + 3IO₃?^-(aq); Ksp = 7.50 × 10^-12

Here, Ksp = [La³⁺]*[IO₃?^-]³ = 7.50*10^-12

Here, [IO₃^-] = [NaIO₃] = 0.05 M

Therefore, [La³⁺]*[0.05]³ = 7.50 × 10^-12

i.e. [La³⁺] = 7.50*10^-12/25*10^-6 = 3*10^-7 M

Therefore, the molar solubility of La(IO₃)₃ in 0.05 M NaIO₃ is 3*10^-7 M