Question

Calculate the pH, pOH, [H3O+] and [OH–] of each of the following:a) [H3O+] = 3.46 × 10-6 mol•L-1b) pOH = 2.30

Answer 1

Explanation and Answer:

a) We use [H3O+] to find pH:

`\begin{gathered} pH\text{ = -log\lbrack H3O}^+] \\ \\ \text{ = -log\lparen3.46}*10^(-6)) \\ \\ \text{ = 5.46} \end{gathered}`

`\begin{gathered} pH\text{ + pOH = 14} \\ \\ \therefore pOH\text{ = 14 - pH} \\ \\ \text{ = 14 - 5.46} \\ \\ \text{ =8.54} \end{gathered}`

`\begin{gathered} pOH\text{ = -log\lbrack OH}^-] \\ \\ \therefore[OH^-]\text{ = 10}^(-pOH) \\ \\ \text{ = 10}^(-8.54) \\ \\ \text{ = 2.88}*10^(-9) \end{gathered}`

b) We use pOH to find pH:

`\begin{gathered} pOH\text{ + pH = 14} \\ \\ \therefore pH\text{ = 14-pH} \\ \\ \text{ = 14-2.30} \\ \\ \text{ = 11.7} \end{gathered}`

`\begin{gathered} pH\text{ = -log\lbrack H}_3O^+] \\ \\ [H_3O^+]\text{ = 10}^(-pH) \\ \\ [H_3O^+]\text{ = 10}^(-11.7) \\ \\ \text{ = 2}*10^(-12) \end{gathered}`

`\begin{gathered} pOH\text{ = -log\lbrack OH}^-] \\ \\ \therefore[OH^-]\text{ = 10}^(-pOH) \\ \\ \text{ = 10}^(-2.30) \\ \\ \text{ = 5.01}*10^(-3) \end{gathered}`