Đáp án:
pH=13
m kết tủa =17,475 gam
Giải thích các bước giải:
\[\begin{gathered}
{n_{HCl}} = 0,75.0,5 = 0,375{\text{ mol; }}{{\text{n}}_{{H_2}S{O_4}}} = 0,25.0,5 = 0,125{\text{ mol}} \hfill \\
\to {{\text{n}}_{{H^ + }}} = {n_{HCl}} + 2{n_{{H_2}S{O_4}}} = 0,375 + 0,125.2 = 0,625{\text{ mol}} \hfill \\
{{\text{n}}_{NaOH}} = 0,75.0,8 = 0,6{\text{ mol; }}{{\text{n}}_{Ba{{(OH)}_2}}} = 0,1.0,75 = 0,075{\text{ mol}} \hfill \\
\to {{\text{n}}_{O{H^ - }}} = {n_{NaOH}} + 2{n_{Ba{{(OH)}_2}}} = 0,6 + 0,075.2 = 0,75{\text{ mol}} \hfill \\
{H^ + } + O{H^ - }\xrightarrow{{}}{H_2}O \hfill \\
\to {n_{O{H^ - }{\text{ du}}}} = 0,75 - 0,625 = 0,125{\text{ mol}} \hfill \\
{{\text{V}}_{dd}} = 500 + 750 = 1250ml = 1,25{\text{lit}} \hfill \\
\to {{\text{[}}OH]^ - }{\text{ = }}\frac{{0,125}}{{1,25}} = 0,1M \to pOH = 1 \to pH = 14 - 1 = 13 \hfill \\
\end{gathered} \]
\[\begin{gathered}
B{a^{2 + }} + S{O_4}^{2 - }\xrightarrow{{}}BaS{O_4} \hfill \\
{n_{B{a^{2 + }}}} = 0,075{\text{ mol; }}{{\text{n}}_{S{O_4}^{2 - }}} = 0,125{\text{ mol}} \to {{\text{n}}_{BaS{O_4}}} = 0,075{\text{ mol}} \hfill \\
\to {{\text{m}}_{BaS{O_4}}} = 0,075.233 = 17,475{\text{ gam}} \hfill \\
\end{gathered} \]
