Question

Calculate the [H3O+] value of each aqueous solution.NaOH with [OH−]=1.6×10−2MExpress your answer using two significant figures.

Answer 1

Step 1

It is known that pH and pOH are calculated as follows:

`\begin{gathered} pH\text{ = - log }\lbrack H3O+\rbrack\text{ \lparen1\rparen} \\ pOH\text{ = - log }\lbrack OH-\rbrack\text{ \lparen2\rparen} \\ Both,\text{ pH and pOH are related to the next equation:} \\ pH\text{ + pOH = 14 \lparen3\rparen} \end{gathered}`

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Step 2

Information provided:

[OH−] = 1.6×10^−2 M (concentration of OH-)

NaOH is a strong base:

NaOH => Na+ + OH-

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Step 3

Procedure:

Firstly, pOH is calculated because it is provided a base:

From (2),

pOH = - log (1.6x10^-2 M) = 1.8

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Secondly, pH is calculated from (3) as follows:

pH = 14 - pOH = 14 - 1.8 = 12.2

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Thirdly, [H3O+] is calculated from (1) as follows:

pH = - log [H3O+] => [H3O+] = 10^(-pH) = 10^-12.2 = 6.3x10^-13 M

[H3O+] is calculated from the pH equation as:

`\begin{gathered} pH\text{ = -log }\lbrack H3O+\rbrack \\ 10^(-pH)=\text{ }\lbrack H3O+\rbrack \end{gathered}`

Answer: [H3O+] = 6.3x10^-13 M