Em tham khảo nha :
\(\begin{array}{l}
{n_{Ba{{(OH)}_2}}} = 0,2 \times 2 = 0,4mol\\
{n_{NaOH}} = 0,5 \times 3 = 1,5mol\\
{n_{B{a^{2 + }}}} = {n_{Ba{{(OH)}_2}}} = 0,4mol\\
[B{a^{2 + }}] = \dfrac{{0,4}}{{2 + 3}} = 0,08M\\
{n_{N{a^ + }}} = {n_{NaOH}} = 1,5mol\\
[N{a^ + }] = \dfrac{{1,5}}{{2 + 3}} = 0,3M\\
{n_{O{H^ - }}} = 2{n_{Ba{{(OH)}_2}}} + {n_{NaOH}} = 2,3mol\\
[O{H^ - }] = \dfrac{{2,3}}{5} = 0,46M
\end{array}\)