Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
Bài 5:
\(\begin{array}{l}
a,\\
C{M_{HCl}} = \dfrac{{0,0014}}{{0,1}} = 0,014M\\
\to C{M_{{H^ + }}} = C{M_{HCl}} = 0,014M\\
\to pH = - \log [C{M_{{H^ + }}}{\rm{]}} = 1,85\\
b,\\
C{M_{{H_2}S{O_4}}} = 0,002M\\
\to C{M_{{H^ + }}} = 2C{M_{{H_2}S{O_4}}} = 0,004M\\
\to pH = - \log [C{M_{{H^ + }}}{\rm{]}} = 2,4\\
c,\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaOH}} = 0,01mol\\
{n_{HCl}} = 0,005mol\\
\to {n_{NaOH}} > {n_{HCl}}
\end{array}\)
Suy ra NaOH dư.
\(\begin{array}{l}
\to {n_{NaOH}}dư= 0,01 - 0,005 = 0,005mol\\
\to C{M_{NaOH}}dư= \dfrac{{0,005}}{{0,1}} = 0,05M\\
\to pOH = - \log [C{M_{NaOH}}dư{\rm{]}} = 1,3\\
\to pH = 14 - pOH = 12,7
\end{array}\)
d,
\(\begin{array}{l}
{H_2}S{O_4} + 2KOH \to {K_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = 0,00025mol\\
{n_{KOH}} = 0,000125mol\\
\to \dfrac{{{n_{KOH}}}}{2} < {n_{{H_2}S{O_4}}}
\end{array}\)
Suy ra \({H_2}S{O_4}\) dư
\(\begin{array}{l}
\to {n_{{H_2}S{O_4}}}dư = 0,00025 - \dfrac{1}{2} \times 0,000125 = 0,0001875mol\\
\to C{M_{{H_2}S{O_4}}}dư= \dfrac{{0,0001875}}{{0,1}} = 0,001875M\\
\to C{M_{{H^ + }}}dư = 2C{M_{{H_2}S{O_4}}}dư = 0,00375M\\
\to pH = - \log [C{M_{{H^ + }}}dư{\rm{]}} = 2,43
\end{array}\)
e,
\(\begin{array}{l}
{H^ + } + O{H^ - } \to {H_2}O\\
{n_{O{H^ - }}} = 0,0004mol\\
{n_{{H^ + }}} = 0,0009mol\\
\to {n_{{H^ + }}} > {n_{O{H^ - }}}
\end{array}\)
Suy ra \({H^ + }\) dư
\(\begin{array}{l}
\to {n_{{H^ + }}}dư= 0,0009 - 0,0004 = 0,0005mol\\
\to C{M_{{H^ + }}}dư= \dfrac{{0,0005}}{{0,5}} = 0,001M\\
\to pH = - \log {\rm{[}}C{M_{{H^ + }}}{\rm{]}}dư= 3
\end{array}\)
f,
\(\begin{array}{l}
{H_2}S{O_4} + Ba{(OH)_2} \to BaS{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = 0,1 \times {10^{ - 2}} = 0,001mol\\
{n_{Ba{{(OH)}_2}}} = 0,4 \times {10^{ - 2}} = 0,004mol\\
\to {n_{Ba{{(OH)}_2}}} > {n_{{H_2}S{O_4}}}
\end{array}\)
Suy ra \(Ba{(OH)_2}\) dư
\(\begin{array}{l}
\to {n_{Ba{{(OH)}_2}}}dư= 0,004 - 0,001 = 0,003mol\\
\to C{M_{Ba{{(OH)}_2}}}dư= \dfrac{{0,003}}{{0,5}} = 0,006M\\
\to C{M_{O{H^ - }}}dư= 2C{M_{Ba{{(OH)}_2}}}dư= 0,012M\\
\to pOH = - \log [C{M_{O{H^ - }}}dư{\rm{]}} = 1,92\\
\to pH = 14 - pOH = 12,08
\end{array}\)
