Đáp án:

\(\begin{array}{l}
6)\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = 8,02\,ml\\
{C_\% }N{a_2}S{O_4} = 6,85\% \\
7)\\
{C_\% }NaOH = 0,2632\% \\
{C_\% }N{a_2}S{O_4} = 1,87\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
6)\\
{n_{NaOH}} = {C_M} \times V = 0,05 \times 1,2 = 0,06\,mol\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,06}}{2} = 0,03\,mol\\
{m_{{H_2}S{O_4}}} = n \times M = 0,03 \times 98 = 2,94g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{{m_{ct}} \times 100}}{{{C_\% }}} = \dfrac{{2,94 \times 100}}{{30}} = 9,8g\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{m}{D} = \dfrac{{9,8}}{{1,222}} = 8,02\,ml\\
{m_{{\rm{dd}}spu}} = 50 \times 1,047 + 9,8 = 62,15g\\
{n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,03mol\\
{C_\% }N{a_2}S{O_4} = \dfrac{{{m_{ct}}}}{{{m_{{\rm{dd}}}}}} \times 100\%  = \dfrac{{0,03 \times 142}}{{62,15}} \times 100\%  = 6,85\% \\
7)\\
{m_{NaOH}} = \dfrac{{{m_{{\rm{dd}}}} \times {C_\% }}}{{100}} = \dfrac{{25 \times 4}}{{100}} = 1g\\
{n_{NaOH}} = \dfrac{m}{M} = \dfrac{1}{{40}} = 0,025\,mol\\
{V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{m}{D} = \dfrac{{51}}{{1,02}} = 50\,ml\\
{n_{{H_2}S{O_4}}} = {C_M} \times V = 0,05 \times 0,2 = 0,01\,mol\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O\\
\dfrac{{0,025}}{2} > \dfrac{{0,01}}{1} \Rightarrow \text{ $NaOH$ dư}\\
{m_{{\rm{dd}}spu}} = 25 + 51 = 76g\\
{n_{N{a_2}S{O_4}}} = {n_{{H_2}S{O_4}}} = 0,01\,mol\\
{n_{NaOH}}\text{ dư} = 0,025 - 0,01 \times 2 = 0,005\,mol\\
{C_\% }NaOH\text{ dư} = \dfrac{{0,005 \times 40}}{{76}} \times 100\%  = 0,2632\% \\
{C_\% }N{a_2}S{O_4} = \dfrac{{0,01 \times 142}}{{76}} \times 100\%  = 1,87\% 
\end{array}\)