Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!!
Giải thích các bước giải:
2,
Axit: \({\rm{[}}{H^ + }{\rm{]}} = \dfrac{{{n_{axit}}}}{{{V_{axit}}}}\), \(pH = - \log ({H^ + })\)
Bazo: \({10^{ - 14}} - \dfrac{{{n_{bazo}}}}{{{V_{bazo}}}}\), \(pH = - \log ({10^{ - 14}} - \dfrac{{{n_{bazo}}}}{{{V_{bazo}}}})\)
Trung tính: \({\rm{[}}{H^ + }{\rm{]}} = {10^{ - 7}},pH = 7\)
3,
a,
\(\begin{array}{l}
HCl \to {H^ + } + C{l^ - }\\
\to C{M_{{H^ + }}} = C{M_{HCl}} = 0,01M\\
\to pH = - \log ({H^ + }) = 2
\end{array}\)
b,
\(\begin{array}{l}
{H_2}S{O_4} \to 2{H^ + } + 2S{O_4}^{2 - }\\
\to C{M_{{H^ + }}} = 2C{M_{{H_2}S{O_4}}} = 0,001M\\
\to pH = - \log ({H^ + }) = 3
\end{array}\)
c,
\(\begin{array}{l}
NaOH \to N{a^ + } + O{H^ - }\\
\to C{M_{O{H^ - }}} = C{M_{NaOH}} = 0,001M\\
\to pH = 14 - pOH = 14 - 3 = 11
\end{array}\)
d,
\(\begin{array}{l}
Ba{(OH)_2} \to B{a^{2 + }} + 2O{H^ - }\\
\to C{M_{O{H^ - }}} = 2C{M_{Ba{{(OH)}_2}}} = 0,01M\\
\to pH = 14 - pOH = 12
\end{array}\)
e,
\(\begin{array}{l}
HCl \to {H^ + } + C{l^ - }\\
\to C{M_{{H^ + }}} = C{M_{HCl}} = 0,15M\\
\to pH = - \log ({H^ + }) = 0,8
\end{array}\)
g,
\(\begin{array}{l}
KOH \to {K^ + } + O{H^ - }\\
\to C{M_{O{H^ - }}} = C{M_{KOH}} = 0,25M\\
\to pH = 14 - pOH = 13,4
\end{array}\)
