Here's my reasoning:
The equilibrium expression can be written:
#\mathbf(Fe(OH)_(2(s))rightleftharpoonsFe^(2+)(aq)+2OH^(-)(aq))#
For which:
#\mathbf(K_(sp)=[Fe_((aq))^(2+)][OH_((aq))^(-)]^2)#
Now if #\mathbf([Fe_((aq))^(2+)]=x#
Then it follows that:
#\mathbf([OH_((aq))^-]=2x#
And
#\mathbf(K_(sp)=x(2x)^2=4x^3#
Initially the #\mathbf(pH=8#
Since
#\mathbf(pH+pOH=14#
Then
#\mathbf(pOH=14-8=6#
#:.\mathbf([OH_((aq))^-]=10^(-6)" ""mol/l"#
If the #\mathbf(pH# is raised to #9# then:
#\mathbf(pOH=14-9=5#
#:.\mathbf([OH_((aq))^-]=10^(-5)" ""mol/l"#
This shows that the concentration of #\mathbf(OH^-# ions has increased by a factor of #\mathbf10#.
So the equilibrium moles of #\mathbf(OH^-# has gone from
#\mathbf(2x# to #\mathbf(2x)##\mathbf(xx10=20x#
The value of #\mathbf(K_(sp)# must remain constant as this only depends on the temperature.
If you disturb a system at equilibrium by altering the concentrations of any of the species, then the system will adjust to return the same value of #\mathbf(K# which is in accordance with Le Chatelier's Principle.
To get the same value of #\mathbf(K_(sp)# you can see that the equilibrium moles of #\mathbf(Fe^(2+)# must decrease by a factor of #\mathbf(100#:
#\mathbf(K_(sp)=x/(100)xx(20x)^2=x/cancel(100)xxcancel(400)x^2=4x^3#
So the equilibrium moles of #\mathbf(Fe^(2+)# has gone from #\mathbf(x# to #\mathbf(x/100# to conserve the value of #\mathbf(K_(sp)#.
#\mathbf(x/"Vol"# represents the solubility of the iron(II) hydroxide so this also reduces by a factor of #\mathbf(100# giving #\mathbf((4))# to be the correct response.
This is an example of "The Common Ion Effect", which, as I have hoped to demonstrate, is a consequence of the principle of Le Chatelier.
We have raised the concentration of the #\mathbf(OH^(-)# ions so the system has responded by opposing this change and shifting to the left.
This has the consequence of reducing the solubility of the iron(II) hydroxide.
