Đáp số:
a) $5,63$
b) $3,32$
Giải thích các bước giải:
a) ${n_{N{H_4}Cl}} = \dfrac{{1,07}}{{53,5}} = 0,02mol$
$ \Rightarrow {C_{M(N{H_4}Cl)}} = \dfrac{{0,02}}{2} = 0,01M$
$N{H_4}Cl\xrightarrow{{}}NH_4^ + + C{l^ - }$
$0,01$ $→$ $0,01$
${K_b} = 1,{8.10^{ - 5}} \Rightarrow {K_a} = \dfrac{{{{10}^{ - 14}}}}{{1,{{8.10}^{ - 5}}}} = 5,{56.10^{ - 10}}$
$NH_4^ + \overset {} \leftrightarrows N{H_3} + {H^ + }$ ${K_a} = 5,{56.10^{ - 10}}$
$0,01 – x$ $x$ $x$
$ \Rightarrow {K_a} = \dfrac{{\left[ {N{H_3}} \right]\left[ {{H^ + }} \right]}}{{\left[ {NH_4^ + } \right]}} = \dfrac{{{x^2}}}{{0,01 - x}} = 5,{56.10^{ - 10}} \\\Rightarrow x = 2,{36.10^{ - 6}}$
$ \Rightarrow pH = - \log \left( {\left[ {{H^ + }} \right]} \right) = 5,63$
b) ${V_{dd}} = 2 + 0,1 = 2,1(l)$
$ \Rightarrow {C_{M(N{H_4}Cl)}} = \dfrac{{0,02}}{{2,1}} = 9,{52.10^{ - 3}}M;\\{C_{M(HCl)}} = \dfrac{{0,001}}{{2,1}} = 4,{76.10^{ - 4}}M$
$HCl \to {H^ + } + C{l^ - }$
$4,{76.10^{ - 4}}$ $4,{76.10^{ - 4}}$
$NH_4^ + \overset {} \leftrightarrows N{H_3} + {H^ + }$
$9,{52.10^{ - 3}}$ – x x x + $4,{76.10^{ - 4}}$
$ \Rightarrow {K_a} = \dfrac{{x.(x + 4,{{76.10}^{ - 4}})}}{{9,{{52.10}^{ - 3}} - x}} = 5,{56.10^{ - 10}}$
$ \Rightarrow x = 1,{11.10^{ - 8}}$
$ \Rightarrow \left[ {{H^ + }} \right] = 4,{76.10^{ - 4}}M$
$ \Rightarrow pH = 3,32$
