Em tham khảo nha:
\(\begin{array}{l}
1)\\
Axit:HCl,{H_2}S{O_4}\\
HCl \to {H^ + } + C{l^ - }\\
{H_2}S{O_4} \to 2{H^ + } + S{O_4}^{2 - }\\
Bazo:\,NaOH,KOH,Mg{(OH)_2}\\
NaOH \to N{a^ + } + O{H^ - }\\
KOH \to {K^ + } + O{H^ - }\\
Mg{(OH)_2} \leftrightarrows M{g^{2 + }} + 2O{H^ - }\\
\text{ Hidroxit lưỡng tính}:Al{(OH)_3}\\
Al{(OH)_3} \leftrightarrows A{l^{3 + }} + 3O{H^ - }\\
Al{(OH)_3} \leftrightarrows Al{O_2}^ - + {H^ + } + {H_2}O\\
\text{ Muối}:CuS{O_4},KHS,FeC{l_2},Al{(N{O_3})_3}\\
CuS{O_4} \to C{u^{2 + }} + S{O_4}^{2 - }\\
KHS \to {K^ + } + H{S^ - }\\
FeC{l_2} \to F{e^{2 + }} + 2C{l^ - }\\
Al{(N{O_3})_3} \to A{l^{3 + }} + 3N{O_3}^ - \\
2)\\
a)\\
C{u^{2 + }} + 2C{l^ - } + 2N{a^ + } + 2O{H^ - } \to Cu{(OH)_2} + 2N{a^ + } + 2C{l^ - }\\
C{u^{2 + }} + 2O{H^ - } \to Cu{(OH)_2}\\
b)\\
2N{a^ + } + C{O_3}^{2 - } + 2{H^ + } + S{O_4}^{2 - } \to 2N{a^ + } + S{O_4}^{2 - } + C{O_2} + {H_2}O\\
C{O_3}^{2 - } + 2{H^ + } \to C{O_2} + {H_2}O\\
3)\\
{\rm{[}}{H^ + }{\rm{]}} = {C_M}HCl = 0,1M\\
{\rm{[}}O{H^ - }{\rm{]}} = \dfrac{{{{10}^{ - 14}}}}{{0,1}} = {10^{ - 13}}\\
pH = - \log (0,1) = 1\\
\text{ Quỳ tím hóa đỏ}\\
4)\\
{\rm{[}}{H^ + }{\rm{]}} = {10^{ - 11}}(M)\\
{\rm{[}}O{H^ - }{\rm{]}} = \dfrac{{{{10}^{ - 14}}}}{{{{10}^{ - 11}}}} = {10^{ - 3}}(M)\\
\text{ Quỳ tím hóa xanh}
\end{array}\)
