Đáp án:
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Giải thích các bước giải:
a,
\(\begin{array}{l}
{m_{HCl}}{\rm{dd = V}} \times {\rm{d = 1,1g}}\\
\to {{\rm{n}}_{HCl}} = \dfrac{{1,1 \times 25}}{{100 \times 36,5}} = \dfrac{{11}}{{1460}}mol\\
\to C{M_{HCl}} = \dfrac{{\dfrac{{11}}{{1460}}}}{1} = \dfrac{{11}}{{1460}}M\\
HCl \to {H^ + } + C{l^ - }\\
\to C{M_{{H^ + }}} = C{M_{C{l^ - }}} = C{M_{HCl}} = \dfrac{{11}}{{1460}}M
\end{array}\)
b,
\(\begin{array}{l}
{m_{HN{O_3}}}{\rm{dd = V}} \times {\rm{d = 1,054g}}\\
\to {{\rm{n}}_{HN{O_3}}} = \dfrac{{1,054 \times 10}}{{100 \times 63}} = 0,0016mol\\
\to C{M_{HN{O_3}}} = \dfrac{{0,0016}}{1} = 0,0016M\\
HN{O_3} \to {H^ + } + N{O_3}^ - \\
\to C{M_{{H^ + }}} = C{M_{N{O_3}^ - }} = C{M_{HN{O_3}}} = 0,0016M
\end{array}\)
c,
\(\begin{array}{l}
{m_{{H_2}S{O_4}}}{\rm{dd = V}} \times {\rm{d = 1,025g}}\\
\to {{\rm{n}}_{{H_2}S{O_4}}} = \dfrac{{1,025 \times 3,92}}{{100 \times 98}} = 0,00041mol\\
\to C{M_{{H_2}S{O_4}}} = \dfrac{{0,00041}}{1} = 0,00041M\\
{H_2}S{O_4} \to 2{H^ + } + S{O_4}^{2 - }\\
\to C{M_{{H^ + }}} = 2C{M_{{H_2}S{O_4}}} = 0,00082M\\
\to C{M_{S{O_4}^{2 - }}} = C{M_{{H_2}S{O_4}}} = 0,00041M
\end{array}\)
