Em tham khảo nha :
\(\begin{array}{l}
a)\\
Ca{(OH)_2} + 2HN{O_3} \to Ca{(N{O_3})_2} + {H_2}O\\
{n_{Ca{{(OH)}_2}}} = 0,1 \times 1 = 0,1mol\\
{n_{HN{O_3}}} = 0,3 \times 0,5 = 0,15mol\\
\dfrac{{0,1}}{1} > \dfrac{{0,15}}{2} \Rightarrow Ca{(OH)_2}\text{ dư}\\
{n_{Ca{{(OH)}_2}d}} = 0,1 - \dfrac{{0,15}}{2} = 0,025mol\\
{n_{Ca{{(N{O_3})}_2}}} = \dfrac{{{n_{HN{O_3}}}}}{2} = 0,075mol\\
{C_{{M_{Ca{{(OH)}_2}d}}}} = \dfrac{{0,025}}{{0,1 + 0,3}} = 0,0625M\\
{C_{{M_{Ca{{(N{O_3})}_2}}}}} = \dfrac{{0,075}}{{0,1 + 0,3}} = 0,1875M\\
b)\\
\text{Quỳ tím chuyển thành màu xanh vì có Bazo dư}\\
{\rm{[}}O{H^ - }{\rm{]}} = 2{C_{{M_{Ca{{(OH)}_2}d}}}} = 0,125M\\
pOH = - \log (0,125) = 0,9\\
pH = 14 - 0,9 = 13,1\\
c)\\
NaOH + HN{O_3} \to NaN{O_3} + {H_2}O\\
{n_{NaOH}} = {n_{HN{O_3}}} = 0,15mol\\
{m_{NaOH}} = 0,15 \times 40 = 6g\\
{m_{ddNaOH}} = \dfrac{{6 \times 100}}{{20}} = 30g\\
{V_{NaOH}} = \dfrac{{30}}{{1,2}} = 25ml
\end{array}\)
