Đáp án:

 VD4: 

a) ${m_{BaS{O_4}}} = 46,6g$

b)

$\left[ {{H^ + }} \right] = 0,26M;\left[ {C{l^ - }} \right] = 0,3M;\left[ {SO_4^{2 - }} \right] = 0,08M;\left[ {N{a^ + }} \right] = 0,2M$

VD5:

a) ${m_{BaC{O_3}}} = 39,4g$

b) $\left[ {O{H^ - }} \right] = 1M;\left[ {N{a^ + }} \right] = 0,92M;\left[ {{K^ + }} \right] = 0,6M$; $  \left[ {CO_3^{2 - }} \right] = 0,26M$

Giải thích các bước giải:

 VD4: 

a)

$\begin{gathered}
  {n_{Ba{{(OH)}_2}}} = 0,2mol;{n_{NaOH}} = 0,2.0,5 = 0,1mol \hfill \\
  {n_{HCl}} = 0,5.0,3 = 0,15mol;{n_{{H_2}S{O_4}}} = 0,3.0,8 = 0,24mol \hfill \\ 
\end{gathered} $

$B{a^{2 + }} + SO_4^{2 - } \to BaS{O_4}$

⇒ $SO_4^{2 - }$ dư; $B{a^{2 + }}$ hết

$\begin{gathered}
   \Rightarrow {n_{BaS{O_4}}} = {n_{B{a^{2 + }}}} = 0,2mol \hfill \\
   \Rightarrow {m_{BaS{O_4}}} = 0,2.233 = 46,6g \hfill \\ 
\end{gathered} $

b) 

$\begin{gathered}
  {n_{O{H^ - }}} = {n_{NaOH}} + 2{n_{Ba{{(OH)}_2}}} = 0,1 + 0,2.2 = 0,5mol \hfill \\
  {n_{{H^ + }}} = {n_{HCl}} + 2{n_{{H_2}S{O_4}}} = 0,15 + 0,24.2 = 0,63mol \hfill \\ 
\end{gathered} $

${H^ + } + O{H^ - } \to {H_2}O$

⇒$  O{H^ - }$ hết; ${H^ + }$ dư

Dung dịch Y gồm:

$\begin{gathered}
  {n_{H_{}^ + (du)}} = 0,63 - 0,5 = 0,13mol \hfill \\
  {n_{SO_4^{2 - }(du)}} = 0,24 - 0,2 = 0,04mol \hfill \\
  {n_{C{l^ - }}} = {n_{HCl}} = 0,15mol \hfill \\
  {n_{N{a^ + }}} = {n_{NaOH}} = 0,1mol \hfill \\ 
\end{gathered} $

$\begin{gathered}
   \Rightarrow \left[ {{H^ + }} \right] = \dfrac{{0,13}}{{0,5}} = 0,26M;\left[ {C{l^ - }} \right] = \dfrac{{0,15}}{{0,5}} = 0,3M \hfill \\
  \left[ {SO_4^{2 - }} \right] = \dfrac{{0,04}}{{0,5}} = 0,08M;\left[ {N{a^ + }} \right] = \dfrac{{0,1}}{{0,5}} = 0,2M \hfill \\ 
\end{gathered} $

VD5: 

a) ${n_{Ba{{(OH)}_2}}} = 0,2mol;{n_{NaOH}} = 0,1mol;{n_{N{a_2}C{O_3}}} = 0,18mol;{n_{{K_2}C{O_3}}} = 0,15$

$ \Rightarrow {n_{B{a^{2 + }}}} = {n_{Ba{{(OH)}_2}}} = 0,2mol$

${n_{CO_3^{2 - }}} = {n_{{K_2}C{O_3}}} + {n_{N{a_2}C{O_3}}} = 0,18 + 0,15 = 0,33mol$

$B{a^{2 + }} + CO_3^{2 - } \to BaC{O_3}$

⇒ $B{a^{2 + }}$ hết; $CO_3^{2 - }$ dư

$\begin{gathered}
   \Rightarrow {n_{BaC{O_3}}} = {n_{B{a^{2 + }}}} = 0,2mol \hfill \\
   \Rightarrow {m_{BaC{O_3}}} = 0,2.197 = 39,4g \hfill \\ 
\end{gathered} $

b) ${n_{CO_3^{2 - }(du)}} = 0,33 - 0,2 = 0,13mol$

$\sum {{n_{N{a^ + }}}}  = {n_{NaOH}} + 2{n_{N{a_2}C{O_3}}} = 0,1 + 0,18.2 = 0,46mol$

${n_{{K^ + }}} = 2{n_{{K_2}C{O_3}}} = 0,15.2 = 0,3mol$

$\sum {{n_{O{H^ - }}}}  = 2{n_{Ba{{(OH)}_2}}} + {n_{NaOH}} = 2.0,2 + 0,1 = 0,5mol$

$ \Rightarrow \left[ {CO_3^{2 - }} \right] = \dfrac{{0,13}}{{0,5}} = 0,26M$

$\left[ {O{H^ - }} \right] = \dfrac{{0,5}}{{0,5}} = 1M;\left[ {N{a^ + }} \right] = \dfrac{{0,46}}{{0,5}} = 0,92M;\left[ {{K^ + }} \right] = \dfrac{{0,3}}{{0,5}} = 0,6M$