Question

Calculate the molar solubility of Ca(IO₃)₂ in

a) 0.060 M Ca(NO₃)₂

b) 0.060 M NaIO₃

Answer 1

Ca(IO₃)₂(s) <-----------> Ca²⁺(aq) + 2(IO₃)^-(aq)

K_sp = [Ca²⁺]*[IO₃^-]²

Thus, if molar solubility of Ca(IO₃)₂ is 's' M

Then, K_sp = s*(2s)² = 4s³ = 7.1*10^-7

Thus, s = 5.62*10^-3

Now,

a) when 0.06 M of Ca(NO₃)₂ is mixed, Ca²⁺ ions will be in more amount then required for equilibrium.This, is called common ion effect due to which the molar solubility of Ca(IO₃)₂ will decrease

Thus, at eqb, let [Ca²⁺] = 0.06 + s ; [IO₃^-] = 2s ; where 's' = molar solubility of Ca(IO₃)₂

Thus, K_sp = (0.06+s)*(2s)²

or, 7.1*10^-7 = 0.24s² + 4s³

or, s = 0.0017 M

b) when 0.06 M of Na(IO₃) is mixed, IO₃^- ions will be in more amount then required for equilibrium.This, is called common ion effect due to which the molar solubility of Ca(IO₃)₂ will decrease

Thus, at eqb, let [Ca²⁺] = s ; [IO₃^-] = 2s+0.06 ; where 's' = molar solubility of Ca(IO₃)₂

Thus, K_sp = (s)*(2s+0.06)²

or, 7.1*10^-7 = 0.24s² + 4s³ + 0.0036s

or, s = 0.000195 M