Đáp án:

 a) 1M và 0,4M

b) 0,56l

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2NaOH + {H_2}S{O_4} \to N{a_2}S{O_4} + 2{H_2}O(1)\\
Cu{(N{O_3})_2} + 2NaOH \to Cu{(OH)_2} + 2NaN{O_3}(2)\\
Cu{(OH)_2} \to CuO + {H_2}O(3)\\
{n_{CuO}} = \dfrac{{1,6}}{{80}} = 0,02\,mol\\
{m_{{\rm{dd}}NaOH}} = 31,25 \times 1,12 = 35g\\
{n_{NaOH}} = \dfrac{{35 \times 16\% }}{{40}} = 0,14\,mol\\
{n_{NaOH}}(2) = 0,02 \times 2 = 0,04\,mol\\
 \Rightarrow {n_{NaOH}}(1) = 0,14 - 0,04 = 0,1\,mol\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{C_M}({H_2}S{O_4}) = \dfrac{{0,05}}{{0,05}} = 1M\\
{C_M}(Cu{(N{O_3})_2}) = \dfrac{{0,02}}{{0,05}} = 0,4M\\
b)\\
{n_{Cu}} = \dfrac{{2,4}}{{64}} = 0,0375\,mol\\
{n_{{H^ + }}} = 2{n_{{H_2}S{O_4}}} = 0,1\,mol\\
{n_{N{O_3}^ - }} = 2{n_{Cu{{(N{O_3})}_2}}} = 0,04\,mol\\
Cu + 4{H^ + } + N{O_3}^ -  \to C{u^{2 + }} + NO + 2{H_2}O\\
\dfrac{{0,04}}{1} > \dfrac{{0,0375}}{1} > \dfrac{{0,1}}{4} \Rightarrow\text{ Tính theo ion H+} \\
{n_{NO}} = \frac{{0,1}}{4} = 0,025\,mol \Rightarrow {V_{NO}} = 0,025 \times 22,4 = 0,56l
\end{array}\)