Em tham khảo nha:
\(\begin{array}{l}
1)\\
{\rm{[}}O{H^ - }{\rm{]}} = {C_M}KOH = 0,005M\\
pOH = - \log (0,005) = 2,3\\
pH = 14 - 2,3 = 11,7\\
2)\\
{\rm{[}}O{H^ - }{\rm{]}} = {C_M}NaOH = 1,2 \times {10^{ - 7}}M\\
pOH = - \log (1,2 \times {10^{ - 7}}) = 6,92\\
pH = 14 - 6,92 = 7,08\\
{\rm{[}}O{H^ - }{\rm{]}} = 2{C_M}Ba{(OH)_2} = 2 \times {10^{ - 2}} = 0,02M\\
pOH = - \log (0,02) = 1,7\\
pH = 14 - 1,7 = 12,3\\
\end{array}\)
\(\begin{array}{l}
*KOH\,\,{10^{ - 7}}M\\
KOH \to {K^ + } + O{H^ - }\\
[O{H^ - }] = {C_M}KOH = {10^{ - 7}}\\
{H_2}O \to {H^ + } + O{H^ - }\\
\Rightarrow [{H^ + }] = [O{H^ - }] = x\,mol\\
[O{H^ - }] = x + {10^{ - 7}}\\
\Rightarrow x \times (x + {10^{ - 7}}) = {10^{ - 14}} \Rightarrow x = 6,18 \times {10^{ - 8}}\\
pH = - \log (6,18 \times {10^{ - 8}}) = 7,21\\
*NaOH\,\,{10^{ - 9}}M\\
NaOH \to N{a^ + } + O{H^ - }\\
[O{H^ - }] = {C_M}NaOH = {10^{ - 9}}\\
{H_2}O \to {H^ + } + O{H^ - }\\
\Rightarrow [{H^ + }] = [O{H^ - }] = x\,mol\\
[O{H^ - }] = x + {10^{ - 9}}\\
\Rightarrow x \times (x + {10^{ - 9}}) = {10^{ - 14}} \Rightarrow x = 9,95 \times {10^{ - 8}}\\
pH = - \log (9,95 \times {10^{ - 8}}) = 7,002
\end{array}\)
