Step 1
The equations used here:
[tex]\begin{gathered} pH\text{ = - log }\lbrack H3O+\rbrack\text{ \lparen1\rparen} \\ \lbrack H3O+\rbrack\text{ = concentration in M} \\ ------ \\ pOH\text{ = -log }\lbrack OH-\rbrack\text{ \lparen2\rparen} \\ ------ \\ pH\text{ + pOH = 14 \lparen3\rparen} \end{gathered}[/tex]------------
Step 2
Information provided:
[H3O+] = 3.5×10^−7 M
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Step 3
Procedure:
Firstly, from (1):
pH = -log [H3O+]
pH = -log (3.5×10^−7 M) = 6.5
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From (3):
pH + pOH = 14
pOH = 14 - pH
pOH = 14 - 6.5
pOH = 7.5
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From (2)
pOH = - log [OH-] => [OH-] = 10^-(pOH)
[OH-] = 10^-(pOH) = 10^-(7.5) = 3.2x10^-8 M
Answer: [OH-] = 3.2x10^-8 M