Đáp án:
a. $pH = 2$
b. $pH = 11$
Giải thích các bước giải:
a.
${H_2}S{O_4} \to 2{H^ + } + SO_4^{2 - } $
$\,0,005 \to \,\,\,0,01\,\,\,\,\,\,\,\,0,005M $
$\to {\text{[}}{H^ + }{\text{]}} = 0,01M $
${\text{[}}SO_4^{2 - }{\text{]}} = 0,005M $
$\to pH = - \log [{H^ + }{\text{]}} = - \log 0,01 = 2 $
b.
${n_{NaOH}} = 4,{5.10^{ - 3}}\,\,mol $
${n_{HCl}} = {4.10^{ - 3}}\,\,mol $
$\,\,\,\,\,\,\,\,\,\,{H^ + }\,\, + \,\,\,O{H^ - }\,\,\, \to \,\,\,{H_2}O$
$\,\,\,\,{4.10^{ - 3}}\, \to {4.10^{ - 3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol $
$\to {n_{O{H^ - }\,\,du}} = {5.10^{ - 4}}\,\,mol $
$pOH = - \log [O{H^ - }{\text{]}} = - \log \frac{{{{5.10}^{ - 4}}}}{{0,5}} = 3$
$\to pH = 14 - pOH = 14 - 3 = 11 $
