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