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:

$\begin{gathered} pH\text{ = -log\lbrack H3O}^+] \\ \\ \text{ = -log\lparen3.46}\times10^{-6}) \\ \\ \text{ = 5.46} \end{gathered}$

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

$\begin{gathered} pOH\text{ = -log\lbrack OH}^-] \\ \\ \therefore[OH^-]\text{ = 10}^{-pOH} \\ \\ \text{ = 10}^{-8.54} \\ \\ \text{ = 2.88}\times10^{-9} \end{gathered}$

b) We use pOH to find pH:

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

$\begin{gathered} pH\text{ = -log\lbrack H}_3O^+] \\ \\ [H_3O^+]\text{ = 10}^{-pH} \\ \\ [H_3O^+]\text{ = 10}^{-11.7} \\ \\ \text{ = 2}\times10^{-12} \end{gathered}$

$\begin{gathered} pOH\text{ = -log\lbrack OH}^-] \\ \\ \therefore[OH^-]\text{ = 10}^{-pOH} \\ \\ \text{ = 10}^{-2.30} \\ \\ \text{ = 5.01}\times10^{-3} \end{gathered}$