ANSWER
[tex]\begin{gathered} [\text{ H}^+\text{ \rbrack = 1}\times\text{ 10}^{-5}\text{ M} \\ [\text{ OH}^-\text{ \rbrack = 1 }\times\text{ 10}^{-9}\text{ M} \\ \text{ pOH = 9} \end{gathered}[/tex]
EXPLANATION
Given that;
pH is 5
Follow the steps below
Firstly, we need to find the H^+ of the sample
[tex]\text{ pH = -log \lparen H}^+)[/tex]
Recall, that pH is 5
[tex]\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}\times\text{ 10}^{-5}\text{ M} \end{gathered}[/tex]
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
[tex]\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 }\times\text{ 10}^{-9}M \end{gathered}[/tex]
