Đáp án:
\(\begin{array}{l}
a)\\
{m_{BaS{O_4}}} = 1,864g\\
b)\\
pH = 11,6\\
c)\\
{\rm{[}}B{a^{2 + }}{\rm{]}} = 0,002M\\
{\rm{[}}O{H^ - }{\rm{]}} = 0,004M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{Ba{{(OH)}_2}}} = 0,1 \times 0,09 = 0,009\,mol\\
{n_{{H_2}S{O_4}}} = 0,4 \times 0,02 = 0,008\,mol\\
Ba{(OH)_2} + {H_2}S{O_4} \to BaS{O_4} + 2{H_2}O\\
\dfrac{{0,009}}{1} > \dfrac{{0,008}}{1} \Rightarrow \text{ $Ba(OH)_2$ dư}\\
{n_{BaS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,008\,mol\\
{m_{BaS{O_4}}} = 0,008 \times 233 = 1,864g\\
b)\\
{n_{Ba{{(OH)}_2}}} \text{ dư}= 0,009 - 0,008 = 0,001\,mol\\
{n_{O{H^ - }}} \text{ dư}= 2{n_{Ba{{(OH)}_2}}} \text{ dư} = 0,001 \times 2 = 0,002\,mol\\
{V_{{\rm{dd}}spu}} = 0,1 + 0,4 = 0,5l\\
{\rm{[}}O{H^ - }{\rm{]}} \text{ dư}= \dfrac{{0,002}}{{0,5}} = 0,004M\\
pH = 14 - {\rm{[}} - \log (0,004){\rm{]}} = 11,6\\
c)\\
{n_{B{a^{2 + }}}} \text{ dư}= {n_{Ba{{(OH)}_2}}} \text{ dư}= 0,001\,mol\\
{\rm{[}}B{a^{2 + }}{\rm{]}} \text{ dư}= \dfrac{{0,001}}{{0,5}} = 0,002M\\
{\rm{[}}O{H^ - }{\rm{]}} \text{ dư}= \dfrac{{0,002}}{{0,5}} = 0,004M
\end{array}\)
