What minimum mass (in g) of ${\left(N{H}_4\right)}_{2}S{O}_4$  should be added to the above mixture cause the $Mg{\left(OH\right)}_2$   formed to re-dissolve?  

$(K_{SP}forMg{\left(OH\right)}_2=6.0\times {10}^{-12}and{K}_{b}forN{H}_3=2\times {10}^{-5})$

$$\begin{gathered} MgC{l}_2+2N{H}_3+2H_{2}O\underset{ }{\overset{ }{\to }}Mg{\left(OH\right)}_2\left(s\right)\downarrow +2N{H}_{4}Cl \\ Mg{\left(OH\right)}_2\left(s\right)\rightleftharpoons M{g}^{2+}+2O{H}^{-} \end{gathered}$$

${\left(N{H}_4\right)}_{2}S{O}_4$  is acidic so it will dissolve $Mg{\left(OH\right)}_2$  solid which is formed in mix. When  all   $Mg{\left(OH\right)}_2$ will re-dissolve the conc. of  $M{g}^{2+}$  ions in the final solution will become  equal to its initial conc.
$\left[N{H}_3\right]$  in mix initially= $\frac{0.70\times 0.2}{1.0}=0.14M$
$\left[M{g}^{2+}\right]$  in mix initially= $\frac{0.30\times 0.2}{1.0}=0.06M$
When  $\left[M{g}^{2+}\right]=0.06$   $\left[O{H}^{-}\right]=\sqrt{\frac{K_{SP}}{\left[M{g}^{2+}\right]}}=\sqrt{\frac{6\times {10}^{-12}}{0.66}}={10}^{-5}M$
 
For  $N{H}_3,K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_3\right]}\Rightarrow \left[N{H}_4^{+}\right]=\frac{K_b\left[N{H}_3\right]}{\left[O{H}^{-}\right]}$

$$\begin{gathered} \left[N{H}_4^{+}\right]=\frac{2\times {10}^{-4}\times 0.14}{{10}^{-5}}=0.28M \\ Moleof{\left[N{H}_4\right]}_{2}S{O}_{4}needed=\frac{0.28}{2}=0.14 \\ Massof{\left[N{H}_4\right]}_{2}S{O}_4=132\times 0.14 =18.48g \end{gathered}$$