Question

Calculate the [H3O+] of the following solutions: [OH-] = 1.0 x 10^-12 M

Answer 1

Step 1

Formulas used here:

`\begin{gathered} pOH\text{ = -log }\lbrack OH-\rbrack\text{ \lparen1\rparen} \\ pH\text{ + pOH = 14 \lparen2\rparen} \\ pH\text{ = -log}\lbrack H3O+\rbrack\text{ => }\lbrack H3O+\rbrack=10^(-pH)(3) \end{gathered}`

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Step 2

From (1):

pOH = -log [OH-] = -log(1.0 x 10^-12 M) = 12

pOH = 12

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Step 3

From (2):

pH + pOH = 14 => pH = 14 - pOH = 14 - 12 = 2

pH = 2

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Step 4

From (3):

[H3O+] = 10^-pH = 10^-2 = 0.01 M or 1x10^-2 M

Answer: [H3O+] = 0.01 M or 1x10^-2 M