Question

K_sp for PbCl₂ is 10^-5. Determine if a percipitate will form when following solutions are mixed:

100mL of 0.002 M Pb(NO₂₎₂ + 100mL 0.02 M KCl

Find molar solubility of BaF₂ (K_sp=10^-7) in 0.001 M NaF

Answer 1

Q1) A precipitate will form when two solutions are mixed if the ionic product excceds solubility product of the salt.

     PbCl_2(s) ----> Pb⁺² +2Cl^-

Ksp and I.P. both have the expression [pb⁺²][Cl^-]²

thus [Pb⁺²] in the mixture = (100x 0.002)/200 = 0.001M

and [Cl] = (100 x 0.02)/200 = 0.01M

and ionic product = 0.001 x (0.01)² = 10^-7 which is less than Ksp. Hence not precipitation occurs

Q2) BaF_2(s) ----->     Ba ⁺² +     2F^-

                    s                      0             0       initial concentrations

                   -                      s               2s     equilibrium concentrations

                                         p              2p+ 0.001 in presence of NaF

Due to common ion efffect p<<<s and also << 0.001 , thus 2p +0.001 is taken as 0.001

now Ksp = [Ba⁺²][F^-]² = px (0.001)² = 10^-7

Hence p = 0.1 M