Đáp án:
Bạn tham khảo lời giải ở dưới nhé!!
Giải thích các bước giải:
Phần tự luận:
Câu 1:
a,
\(\begin{array}{l}
Cu + 4HN{O_3} \to Cu{(N{O_3})_2} + 2N{O_2} + 2{H_2}O\\
Cu + 4{H^ + } + 2N{O_3}^ - \to C{u^{2 + }} + 2N{O_2} + 2{H_2}O
\end{array}\)
b,
\(\begin{array}{l}
N{a_2}C{O_3} + 2HN{O_3} \to 2NaN{O_3} + C{O_2} + {H_2}O\\
C{O_3}^{2 - } + 2{H^ + } \to C{O_2} + {H_2}O
\end{array}\)
c,
\(\begin{array}{l}
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
3Cu + 8{H^ + } + 2N{O_3}^ - \to 3C{u^{2 + }} + 2NO + 4{H_2}O
\end{array}\)
d,
\(\begin{array}{l}
CaC{O_3} + 2HN{O_3} \to Ca{(N{O_3})_2} + C{O_2} + {H_2}O\\
CaC{O_3} + 2{H^ + } \to C{a^{2 + }}+C{O_2} + {H_2}O
\end{array}\)
e,
\(\begin{array}{l}
Fe + 6HN{O_3} \to Fe{(N{O_3})_3} + 3N{O_2} + 3{H_2}O\\
Fe + 6{H^ + } + 3N{O_3}^ - \to F{e^{3 + }} + 3N{O_2} + 3{H_2}O
\end{array}\)
f,
\(\begin{array}{l}
3FeO + 10HN{O_3} \to 3Fe{(N{O_3})_3} + NO + 5{H_2}O\\
3FeO + 10{H^ + } + N{O_3}^ - \to 3F{e^{3 + }} + NO + 5{H_2}O
\end{array}\)
Câu 2:
\(\begin{array}{l}
{n_{Ba{{(OH)}_2}}} = 0,01mol\\
{n_{HCl}} = 0,01mol\\
\to {n_{O{H^ - }}} = 2{n_{Ba{{(OH)}_2}}} = 0,02mol\\
\to {n_{{H^ + }}} = {n_{HCl}} = 0,01mol\\
O{H^ - } + {H^ + } \to {H_2}O\\
\to {n_{O{H^ - }}} > {n_{{H^ + }}}
\end{array}\)
Suy ra dung dịch A có \(O{H^ - }\) dư
\(\begin{array}{l}
\to {n_{O{H^ - }}}dư= 0,02 - 0,01 = 0,01mol\\
\to C{M_{O{H^ - }}}dư= \dfrac{{0,01}}{{0,3}} = 0,033M\\
\to pOH = - \log [C{M_{O{H^ - }}}dư{\rm{]}} = 1,48\\
\to pH = 14 - pOH = 12,52
\end{array}\)
Câu 3:
\(\begin{array}{l}
{n_{NaOH}} = 0,01mol\\
{n_{{H_2}S{O_4}}} = 0,02mol\\
\to {n_{O{H^ - }}} = {n_{NaOH}} = 0,01mol\\
\to {n_{{H^ + }}} = 2{n_{{H_2}S{O_4}}} = 0,04mol\\
O{H^ - } + {H^ + } \to {H_2}O\\
{n_{O{H^ - }}} < {n_{{H^ + }}}
\end{array}\)
Suy ra có \({H^ + }\) dư
\(\begin{array}{l}
{n_{{H^ + }}}dư= 0,04 - 0,01 = 0,03mol\\
\to C{M_{{H^ + }}}dư= \dfrac{{0,03}}{{0,3}} = 0,1M\\
\to pH = - \log [C{M_{{H^ + }}}dư{\rm{]}} = 1
\end{array}\)
