Question

Calculate the molar solubility of barium iodate in a solution that is also 0.1 M in KIO3

Answer 1

KIO3(aq) ---------------> K^+ (aq)    + IO3^- (aq)

0.1M                                                   0.1M

Ba(IO3)2 (s) ---------------> Ba^2+ (aq) +2IO3^- (aq)

                                           s                  2s+0.1

    Ksp    = [Ba^2+][IO3^-]^2

6*10^-10   = s*(2s+0.1)^2                                  [ 2s + 0.1   = 0.1 , 2s<<<<<0.1 ]

6*10^-10   = s*(0.1)^2

6*10^-10   = s*0.01

s                 = 6*10^-10/0.01   = 6*10^-8 M

The molar solubility of Ba(IO3)2   = 6*10^-8 M