Answer
(a) At $PH=10$,
$[OH^-]=1\times10^{-4}$ and $[Mg^{2+}]=1.8\times10^{-3}$
(b) At $PH=5$,
$[OH^-]=1\times10^{-9}$ and $[Mg^{2+}]=1.8\times10^{7}$
Comment:
In case of (a), the result is plausible, unlike being impossible in case of (b) because the solution is not one of $Mg(OH)_2$ but $MgCl_2{_{(aq.)}}$; $[Mg^{+2}]$, in this case, depends on the solubility of $MgCl_2$.
Work Step by Step
(a) At $PH=10$,
$[OH^-]=1\times10^{-4}$ and $[Mg^{2+}]=1.8\times10^{-3}$
(b) At $PH=5$,
$[OH^-]=1\times10^{-9}$ and $[Mg^{2+}]=1.8\times10^{7}$
Comment:
In case of (a), the result is plausible, unlike being impossible in case of (b) because the solution is not one of $Mg(OH)_2$ but $MgCl_2{_{(aq.)}}$; $[Mg^{+2}]$, in this case, depends on the solubility of $MgCl_2$.
