Cho 200 ml dung dịch Ba(OH)2 0,1M vào 300 ml dung dịch NaHCO3 0,1M

* Hướng dẫn giải

Đáp án D

$$\begin{gathered} + \underset{0,02 \mathrm{mol}}{\underbrace{\mathrm{Ba}{\left(\mathrm{OH}\right)}_2}}+\underset{0,03 \mathrm{mol}}{\underbrace{{\mathrm{NaHCO}}_3}}\stackrel{}{\to }\underset{0,02 \mathrm{mol}}{\underbrace{{\mathrm{BaCO}}_3\downarrow }}+\underset{\text{dd} \mathrm{X}}{\underbrace{\left\{{\mathrm{OH}}^{-}: 0,01 \mathrm{mol} \\ {{\mathrm{CO}}_3}^{2-}: 0,01 \mathrm{mol} \\ {\mathrm{Na}}^{+}: 0,03 \mathrm{mol}\right\}}} \\ + \mathrm{Cho} \mathrm{từ} \mathrm{từ} \mathrm{HCl} \mathrm{vào} \mathrm{X} \mathrm{đến} \mathrm{khi} \mathrm{bắt} \mathrm{đầu} \mathrm{thoát} \mathrm{khí} \mathrm{ra} 2 \mathrm{phản} \mathrm{ứng}: \\ {\mathrm{H}}^{+}+{\mathrm{OH}}^{-}\stackrel{}{\to }{\mathrm{H}}_2\mathrm{O} \\ {\mathrm{H}}^{+}+{{\mathrm{CO}}_3}^{2-}\stackrel{}{\to }{{\mathrm{HCO}}_3}^{-} \\ \Rightarrow {\mathrm{n}}_{{\mathrm{H}}^{+}}=0,02\Rightarrow {\mathrm{V}}_{\mathrm{HCl}}=\frac{0,02.1000}{0,25}=\boxed{80 \mathrm{ml}} \end{gathered}$$