Đáp án:

a) ${C_{M{{(N{H_4})}_2}S{O_4}}} = 0,05M;\\{C_{M({K_2}S{O_4})}} = 0,06M$

b) $\left[ {{K^ + }} \right] = 0,05M;\\\left[ {B{a^{2 + }}} \right] = 0,0375M;\\\left[ {O{H^ - }} \right] = 0,125M$

Giải thích các bước giải:

a)

${n_{{{(N{H_4})}_2}S{O_4}}} = \dfrac{{1,65}}{{132}} = 0,0125mol;{n_{{K_2}S{O_2}}} = \dfrac{{2,61}}{{174}} = 0,015mol$

$\begin{gathered}    \Rightarrow {C_{M{{(N{H_4})}_2}S{O_4}}} = \dfrac{{0,0125}}{{0,25}} = 0,05M \hfill \\   {C_{M({K_2}S{O_4})}} = \dfrac{{0,015}}{{0,25}} = 0,06M \hfill \\ \end{gathered} $

b) Trong $50ml$ dung dịch $A$:

${n_{{{(N{H_4})}_2}S{O_4}}} = 2,{5.10^{ - 3}}mol;{n_{{K_2}S{O_2}}} = {3.10^{ - 3}}mol$

$ \Rightarrow {n_{NH_4^ + }} = 0,005mol;{n_{{K^ + }}} = 0,006mol;{n_{SO_4^{2 - }}} = 0,0055mol$

${n_{B{a^{2 + }}}} = 0,05.0,2 = 0,01mol$; ${n_{O{H^ - }}} = 0,05.0,2.2 = 0,02mol$

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

${n_{B{a^{2 + }}(pu)}} = {n_{BaS{O_4}}} = {n_{SO_4^{2 - }}} = 0,0055mol$

$ \Rightarrow {m_1} = 0,0055.233 = 1,2815g$

$ \Rightarrow {n_{B{a^{2 + }}(dư)}} = 0,01 - 0,0055 = 0,0045mol$

$O{H^ - } + NH_4^ +  \to N{H_3} + {H_2}O$

$ \Rightarrow {n_{O{H^ - }(du)}} = 0,02 - 0,005 = 0,015mol$

Dung dịch ${A_1}$ gồm:

${K^ + }(0,006{\text{ mol); B}}{{\text{a}}^{2 + }}(0,0045{\text{ mol); O}}{{\text{H}}^ - }(0,015{\text{ mol)}}$

$\begin{gathered}    \Rightarrow \left[ {{K^ + }} \right] = \dfrac{{0,006}}{{0,12}} = 0,05M \hfill \\   \left[ {B{a^{2 + }}} \right] = \dfrac{{0,0045}}{{0,12}} = 0,0375M \hfill \\   \left[ {O{H^ - }} \right] = \dfrac{{0,015}}{{0,12}} = 0,125M \hfill \\ \end{gathered} $