ANSWER
![\begin{gathered} [\text{ H}^+\text{ \rbrack = 1}*\text{ 10}^(-5)\text{ M} \\ [\text{ OH}^-\text{ \rbrack = 1 }*\text{ 10}^(-9)\text{ M} \\ \text{ pOH = 9} \end{gathered}]()
Step-by-step explanation
Given that;
pH is 5
Follow the steps below
Firstly, we need to find the H^+ of the sample
![\text{ pH = -log \lparen H}^+)](https://img.qammunity.org/2023/formulas/chemistry/college/2cya0n49j6er7dels6guyvqle2i7kbn6in.png)
Recall, that pH is 5
![\begin{gathered} \text{ 5 = -log }\lbrack\text{ H}^+\text{ }\rbrack \\ \text{ }\lbrack\text{ H}^+\rbrack\text{ = 10}^(-5) \\ \text{ }\lbrack\text{ H}^+\rbrack\text{ = 1}*\text{ 10}^(-5)\text{ M} \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/xrbsgaokg47ktjca0luu8qufj87nk4zcqh.png)
Therefore, [H^+] = 1 x 10^-5 M
Find pOH?
pH + pOH = 14
pH = 5
5 + pOH = 14
subtract 5 from both sides of the equation
5 - 5 + pOH = 14 - 5
pOH = 9
Find OH^- using the below formula
![\begin{gathered} \text{ pOH = -log \lbrack OH}^-\text{ \rbrack} \\ \text{ 9 = -log \lbrack OH}^(-1)\text{ \rbrack} \\ \text{ \lbrack OH}^-\text{ \rbrack= 10}^(-9) \\ \text{ \lbrack OH}^-\text{ \rbrack= 1 }*\text{ 10}^(-9)M \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/h61s653z10ki3v5r0mym160hrlijo3jkh6.png)