Step 1
The equations used here:
[tex]\begin{gathered} pH\text{ = -log }\lbrack H3O+\rbrack\text{ \lparen1\rparen} \\ pOH\text{ = - log }\lbrack OH-\rbrack\text{ \lparen2\rparen} \\ pH\text{ + pOH = 14 \lparen3\rparen} \\ \end{gathered}[/tex]---------------
Step 2
Information provided:
[OH−] = 1.6×10^−5 M (It is the concentration of OH-)
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Step 3
Firstly, we begin with pOH as follows:
pOH = - log [OH−]
pOH = - log (1.6×10^−5 M) = 4.8
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Secondly, it is needed the equation (3) for pH:
pH + pOH = 14
pH = 14 - 4.8 = 9.2
pH = 9.2
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Thirdly, From (2), the concentration of H3O+ is found as follows:
pH = - log [H3O+]
It is needed the [H3O+], so it proceeds as follows:
[H3O+] = 10^-pH
[H3O+] = 10^-(9.2) = 6.3x10^-10 M
Answer: [H3O+] = 6.3x10^-10 M
