Question

Household ammonia, pH = 12.10 Calculate [H +] and [OH﹘]

Answer 1

ANSWER

$\begin{gathered} \text{ \lbrack H}^+\text{\rbrack = 7.943}\times\text{ 10}^{-13}M \\ \text{ \lbrack OH}^-\text{\rbrack = 0.01259M} \end{gathered}$

EXPLANATION

Given that;

The pH of ammonia is 12.10

Follow the steps below to find [H+] and [OH-]

$\begin{gathered} \text{ pH = -log \lbrack H}^+\text{ \rbrack} \\ \text{ \lbrack H}^+\text{ \rbrack= 10}^{-pH} \\ \text{ \lbrack H}^+\text{ \rbrack= 10}^{-12.10} \\ \text{ \lbrack H}^+\text{ \rbrack= 7.943}\times10^{-13}M \end{gathered}$$\begin{gathered} \text{ pH = -log \lbrack H}^+\text{ \rbrack} \\ \text{ \lbrack H}^+\text{ \rbrack= 10}^{-pH} \\ \text{ \lbrack H}^+\text{ \rbrack= 10}^{-12.10} \\ \text{ \lbrack H}^+\text{ \rbrack= 7.943}\times10^{-13}M \end{gathered}$

Find [OH-] ?

Recall, that

pOH + pH = 14

pH = 12.10

pOH + 12.10 = 14

pOH = 14 - 12.10

pOH = 1.9

$\begin{gathered} \text{ pOH = -log \lbrack OH}^-\text{\rbrack} \\ \text{ \lbrack OH}^-\text{\rbrack= 10}^{-pOH} \\ \text{ \lbrack OH}^-\text{\rbrack= 10}^{-1.9} \\ \text{ \lbrack OH}^-\text{\rbrack= 0.01259M} \end{gathered}$