Question

Calculate the molar solubility of Ca(IO3)2 in each solution below. The Ksp of calcium iodate is 7.1*10^-7.

a) 0.070 M Ca(NO3)2

b).070 M NaIO3

Answer 1

Solubility product of a salt is equal to product of concentration of the dissolved ion raised to the power ofthe stoichiometric coefficients of the ions Given that, Concentration of Ca(NO 0.07M Kp of Ca(IO)27.1x10 Consider the reaction, Ca( 103 )2 (s) Ca2. (ag)+10,(ag) The solubility constant product for Ca(IO)2 is Calcium nitrate is a strong electrolyte and due to the common ion effect, it will suppress the dissociation of lead calcium iodate Suppose the molar solubility of Ca(I0)2 in Ca(NO3)2 is 'S Then, Ca-(S+0.07)M Substitute the above concentrations in equation (1) 7.1x10-[Ca2- 1[IO 7.1x10 (S+0.07) x(2S) S-1.57x103 M Hence, the molar solubility of Ca(IO a 0.07 M Ca(NO,), solution is 1.5710 M

Given that, Concentration of NalO0.07 M Kg of Ca (I0)7.1x10 Consider the reaction, Ca(10,), (s)Ca(a)+210, (aą) The solubility constant product for NalO, is 7.1x10 Ca+ IO Sodium iodate is a strong electrolyte and due to the common ion effect, it will suppress the dissociation of calcium iodate Suppose the molar solubility of Ca(O)2 in NalO is 'S Then, Ca21-SM IO (2S+0.07)M Substitute the above concentrations in equation (1) 71-10-[Ca ]1o] 7.1x10 Sx(2S+0.07) S-1.43x104 M Hence, the molar solubility of Ca(IO,), in a 0.07 M NalO, solution is 1.43 10