Đáp án:
pH=13
a=4,4325g
b=4,925g
Giải thích các bước giải:
\(\begin{array}{l}
{n_{Ba{{(OH)}_2}}} = 0,045mol\\
{n_{HCl}} = 0,04mol\\
\to {n_{O{H^ - }}} = 2{n_{Ba{{(OH)}_2}}} = 0,09mol\\
\to {n_{{H^ + }}} = {n_{HCl}} = 0,04mol\\
{H^ + } + O{H^ - } \to {H_2}O\\
{n_{{H^ + }}} < {n_{O{H^ - }}}
\end{array}\)
Suy ra có \(O{H^ - }\) dư
\(\begin{array}{l}
\to {n_{O{H^ - }}}dư= 0,09 - 0,04 = 0,05mol\\
\to C{M_{O{H^ - }}}dư= \dfrac{{0,05}}{{0,5}} = 0,1M\\
\to pOH = - \log [C{M_{O{H^ - }}}dư{\rm{]}} = 1\\
\to pH = 14 - 1 = 13
\end{array}\)
Dung dịch X chứa: \(B{a^{2 + }},C{l^ - },O{H^ - }dư\)
Phần 1:
\(\begin{array}{l}
O{H^ - } + HC{O_3}^ - \to C{O_3}^{2 - } + {H_2}O\\
B{a^{2 + }} + C{O_3}^{2 - } \to BaC{O_3}\\
{n_{C{O_3}^{2 - }}} = \dfrac{{{n_{O{H^ - }}}dư}}{2} = 0,025mol\\
{n_{B{a^{2 + }}}} = \dfrac{{{n_{Ba{{(OH)}_2}}}}}{2} = 0,0225mol\\
\to {n_{B{a^{2 + }}}} < {n_{C{O_3}^{2 - }}}\\
\to {n_{BaC{O_3}}} = {n_{B{a^{2 + }}}} = 0,0225mol\\
\to {m_{BaC{O_3}}} = 4,4325g=a
\end{array}\)
Phần 2:
\(\begin{array}{l}
O{H^ - } + HC{O_3}^ - \to C{O_3}^{2 - } + {H_2}O\\
B{a^{2 + }} + C{O_3}^{2 - } \to BaC{O_3}\\
{n_{C{O_3}^{2 - }}} = \dfrac{{{n_{O{H^ - }}}dư}}{2} = 0,025mol\\
\to {n_{B{a^{2 + }}}} > {n_{C{O_3}^{2 - }}}\\
\to {n_{BaC{O_3}}} = {n_{C{O_3}^{2 - }}} = 0,025mol\\
\to {m_{BaC{O_3}}} = 4,925g=b
\end{array}\)
