Question

Calculate the [OH−] of each aqueous solution with the following [H3O+].milk, 3.5×10−7M Express the molarity to two significant figures.

Answer 1

Step 1

The equations used here:

`\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}`

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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