Question

Question 7.69 Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate Ksp = 7.4 × 10^(–8)).

Class XI Equilibrium Page 230

Answer 1

2NaIO3 + Cu(ClO3)2 ----> 2NaClO3 + Cu(IO3)2

equal volume of 0.002M solutions of NaIO3 and Cu(IO3)2 are mixed together. So, molar concentration of each species will be halved.
so, conc. Of each = 0.002M/2 = 0.001 M
e.g., [Cu2+]mix = [IO3-]mix = 0.001 M

now, Cu(IO3)2 <=> Cu2+ +2IO3-
solubility product {Ksp} = [Cu2+][IO3]²
for precipitation, ionic product > Ksp
e.g., [Cu2+][IO3]² > Ksp
But here,
0.001 × 0.001 = 10^-9 < 7.4 × 10^-8
it is less than Ksp. hence, there won't be precipitation.

Answer 2

When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., 0.001 M.Then,   Now, the solubility equilibrium for copper iodate can be written as:  Ionic product of copper iodate:  Since the ionic product (1 × 10–9 ) is less than Ksp (7.4 × 10–8 ), precipitation will not occcur