Os valores de \mathrm{pH}, \mathrm{pOH},\left[\mathrm{H}_{3} \mathrm{O}^{+}\right], e \left[\mathrm{OH}^{-}\right]são relacionados. Conforme dado o valor de um deles, complete a tabela abaixo:
\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \left[\mathrm{OH}^{-}\right] pH pOH
A 2,5 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}
B 6,24 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}
C 4,10
D 10,50

Passo 1

O enunciado envolve as relações fundamentais de equilíbrio ácido–base em água pura a 25^\circ\text{C}:
\mathrm{pH} + \mathrm{pOH} = 14,
\mathrm{pH} = -\log\bigl[\mathrm{H_{3}O^{+}}\bigr], \quad \mathrm{pOH} = -\log\bigl[\mathrm{OH^{-}}\bigr],
\bigl[\mathrm{H_{3}O^{+}}\bigr]\times\bigl[\mathrm{OH^{-}}\bigr] = 10^{-14}.
Usaremos essas fórmulas para completar cada linha da tabela.

Passo 2

Na linha A temos \bigl[\mathrm{H_{3}O^{+}}\bigr]=2,5\times10^{-3}.
Calculemos a pH:
\mathrm{pH} = -\log\bigl(2,5\times10^{-3}\bigr) = -\bigl(\log2{,}5 - 3\bigr) = 2{,}60.
Em seguida
\mathrm{pOH} = 14 - 2{,}60 = 11{,}40,
e
\bigl[\mathrm{OH^{-}}\bigr] = 10^{-11{,}40} = 3{,}98\times10^{-12}\,\mathrm{mol\,L^{-1}}.

Passo 3

Na linha B temos \bigl[\mathrm{OH^{-}}\bigr]=6,24\times10^{-4}.
Calculemos a pOH:
\mathrm{pOH} = -\log\bigl(6,24\times10^{-4}\bigr) = -\bigl(\log6{,}24 - 4\bigr) = 3{,}20.
Então
\mathrm{pH} = 14 - 3{,}20 = 10{,}80,
e
\bigl[\mathrm{H_{3}O^{+}}\bigr] = 10^{-10{,}80} = 1{,}58\times10^{-11}\,\mathrm{mol\,L^{-1}}.

Passo 4

Na linha C a pH é dada como 4{,}10.
Logo
\bigl[\mathrm{H_{3}O^{+}}\bigr] = 10^{-4{,}10} = 7{,}94\times10^{-5}\,\mathrm{mol\,L^{-1}},
\mathrm{pOH} = 14 - 4{,}10 = 9{,}90,
\bigl[\mathrm{OH^{-}}\bigr] = 10^{-9{,}90} = 1{,}26\times10^{-10}\,\mathrm{mol\,L^{-1}}.

Passo 5

Na linha D a pOH é 10{,}50.
Portanto
\bigl[\mathrm{OH^{-}}\bigr] = 10^{-10{,}50} = 3{,}16\times10^{-11}\,\mathrm{mol\,L^{-1}},
\mathrm{pH} = 14 - 10{,}50 = 3{,}50,
\bigl[\mathrm{H_{3}O^{+}}\bigr] = 10^{-3{,}50} = 3{,}16\times10^{-4}\,\mathrm{mol\,L^{-1}}.

Resposta

A tabela completa fica:
A) \bigl[\mathrm{H_{3}O^{+}}\bigr]=2,5\times10^{-3},
\bigl[\mathrm{OH^{-}}\bigr]=3,98\times10^{-12},
\mathrm{pH}=2,60,
\mathrm{pOH}=11,40.
B) \bigl[\mathrm{H_{3}O^{+}}\bigr]=1,58\times10^{-11},
\bigl[\mathrm{OH^{-}}\bigr]=6,24\times10^{-4},
\mathrm{pH}=10,80,
\mathrm{pOH}=3,20.
C) \bigl[\mathrm{H_{3}O^{+}}\bigr]=7,94\times10^{-5},
\bigl[\mathrm{OH^{-}}\bigr]=1,26\times10^{-10},
\mathrm{pH}=4,10,
\mathrm{pOH}=9,90.
D) \bigl[\mathrm{H_{3}O^{+}}\bigr]=3,16\times10^{-4},
\bigl[\mathrm{OH^{-}}\bigr]=3,16\times10^{-11},
\mathrm{pH}=3,50,
\mathrm{pOH}=10,50.

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