Respuesta :
Step 1
It is known that pH and pOH are calculated as follows:
[tex]\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}[/tex]----------------
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:
[tex]\begin{gathered} pH\text{ = -log }\lbrack H3O+\rbrack \\ 10^{-pH}=\text{ }\lbrack H3O+\rbrack \end{gathered}[/tex]Answer: [H3O+] = 6.3x10^-13 M
