Em tham khảo nha :
\(\begin{array}{l}
1)\\
a)\\
{\rm{[}}{H^ + }{\rm{] = }}{{\rm{C}}_{{M_{HCl}}}} = 0,1M\\
pH = - \log (0,1) = 1\\
[O{H^ - }] = {C_{{M_{NaOH}}}} = 0,01M\\
pOH = - \log (0,01) = 2\\
pH = 14 - 2 = 12\\
b)\\
HCl + NaOH \to NaCl + {H_2}O\\
{n_{HCl}} = 0,5 \times 0,1 = 0,05mol\\
{n_{NaOH}} = {n_{HCl}} = 0,05mol\\
{V_{NaOH}} = \dfrac{{0,05}}{{0,01}} = 5l\\
2)\\
a)\\
P + 5HN{O_3} \to {H_2}O + 5N{O_2} + {H_3}P{O_4}\\
b)\\
{n_P} = \dfrac{{6,2}}{{31}} = 0,2mol\\
{n_{N{O_2}}} = 5{n_P} = 1mol\\
{V_{N{O_2}}} = 1 \times 22,4 = 22,4l
\end{array}\)
