Đáp án:
a) $ \left[ {{K^ + }} \right] = 0,03M;\left[ {B{a^{2 + }}} \right] = 0,01M$; $\left[ {O{H^ - }} \right] = 0,05M$
b) $ \left[ {C{l^ - }} \right] = 7,{5.10^{ - 4}}M;\left[ {SO_4^{2 - }} \right] = 9,{375.10^{ - 4}}M$; $\left[ {{H^ + }} \right] = 2,{625.10^{ - 3}}M$
Giải thích các bước giải:
a)
${n_{KOH}} = 0,3.0,05 = 0,015mol;{n_{Ba{{(OH)}_2}}} = 0,2.0,025 = {5.10^{ - 3}}mol$
$ \Rightarrow {n_{{K^ + }}} = {n_{KOH}} = 0,015mol;{n_{B{a^{2 + }}}} = {n_{Ba{{(OH)}_2}}} = 0,005mol$
$\sum {{n_{O{H^ - }}}} = {n_{KOH}} + 2{n_{Ba{{(OH)}_2}}} = 0,025mol$
$ \Rightarrow \left[ {{K^ + }} \right] = \dfrac{{0,015}}{{0,5}} = 0,03M;\left[ {B{a^{2 + }}} \right] = \dfrac{{0,005}}{{0,5}} = 0,01M$; $\left[ {O{H^ - }} \right] = \dfrac{{0,025}}{{0,5}} = 0,05M$
b)
${n_{HCl}} = 0,15.0,002 = {3.10^{ - 4}}mol;{n_{{H_2}S{O_4}}} = 0,25.0,0015 = 3,{75.10^{ - 4}}mol$
$ \Rightarrow {n_{C{l^ - }}} = {n_{HCl}} = {3.10^{ - 4}}mol;{n_{SO_4^{2 - }}} = {n_{{H_2}S{O_4}}} = 3,{75.10^{ - 4}}mol$
$\sum {{n_{{H^ + }}}} = {n_{HCl}} + 2{n_{{H_2}S{O_4}}} = 1,{05.10^{ - 3}}mol$
$ \Rightarrow \left[ {C{l^ - }} \right] = \dfrac{{{{3.10}^{ - 4}}}}{{0,4}} = 7,{5.10^{ - 4}}M;\left[ {SO_4^{2 - }} \right] = \dfrac{{3,{{75.10}^{ - 4}}}}{{0,4}} = 9,{375.10^{ - 4}}M$
$\left[ {{H^ + }} \right] = \dfrac{{1,{{05.10}^{ - 3}}}}{{0,4}} = 2,{625.10^{ - 3}}M$
