What is the solubility of M(OH)2 in a 0

Both the problems are examples common ion effect.

Let x be the molar solubility of M(OH)₂. The dissociation of M(OH)₂ can be written as

M(OH)₂ (s) <======> M²⁺ (aq) + 2 OH^- (aq)

x                                      x                   2x

Again, the ionization of M(NO₃)₂ can be written as

M(NO₃)₂ (aq) ---------> M²⁺ (aq) + 2 NO₃^- (aq)

0.202 M                        0.202 M

We do not bother about the concentration of NO₃^- since it doesn’t play any role in the precipitation reaction. We next set up the ICE chart as

M(OH)₂ (aq) <=====> M²⁺ (aq) + 2 OH^- (aq)

initial                         x                                     0.202               0

change                    - x                                      + x                + 2x

equilibrium               -                                   (0.202 + x)         2x

We need to account for the M²⁺ furnished by the M(NO₃)₂. The solubility product is

K_sp = (0.202 + x)(2x)²

Next, we make an approximation. Since the value of K_sp = 6.65*10^-18 is quite low, we will expect x<<0.202 and hence, (0.202 + x) ≈ 0.202. Therefore,

K_sp = (0.202).(2x)²

===> 6.65*10^-18 = 0.202*4x²

===> x² = 8.230*10^-18

===> x = 2.868*10^-9

The molar solubility of M(OH)₂ in 0.202 M M(NO₃)₂ is 2.868*10^-9 M ≈ 2.9*10^-9 M (ans).

We follow the same logic as before. The concentration of SCN^- from KSCN is 0.600 M [1:1 dissociation as per the equation KSCN (aq) ----------> K⁺ (aq) + SCN^- (aq)].

The ionization of lead thiocyanate, Pb(SCN)₂ follows the equation

Pb(SCN)₂ (s) <=====> Pb²⁺ (aq) + 2 SCN^- (aq)

x                                      x                 2x

K_sp = (x).(2x + 0.600)²

Now, we shall again assume (2x + 0.600) ≈ 0.600; therefore,

K_sp = x.(0.600)²

===> 2.0*10^-5 = x.(0.36)

===> x = 5.555*10^-5

The molar solubility of Pb(SCN)₂ in 0.600 M KSCN is 5.555*10^-5 M ≈ 5.5*10^-5 M (ans).