## Q. 18.3

Calculating $[H_3O^+], pH, [OH^−]$, and pOH for Strong Acids and Bases

Problem Calculate $[H_3O^+], pH, [OH^−]$, and pOH for each solution at 25°C:

(a) 0.30 M $HNO_3$, used for etching copper metal

(b) 0.0042 M $Ca(OH)_2$, used in leather tanning to remove hair from hides

## Verified Solution

Plan We know that $HNO_3$ is a strong acid and dissociates completely; thus, $[H_3O^+] = [HNO_3]_{\text{init}}$. Similarly, $Ca(OH)_2$ is a strong base and dissociates completely; the molar ratio is 1 mol $Ca(OH)_2/2 \text{ mol }OH^−$ so $[OH^−] = 2[Ca(OH)_2]_{\text{init}}$. We use these concentrations and the value of $K_w$ at 25°C $(1.0×10^{−14})$ to find $[OH^−]$ or $[H_3O^+]$, which we then use to calculate pH and pOH.

Solution (a) For 0.30 M $HNO_3$:
$[H_3O^+] = 0.30 M$
$pH = −\log [H_3O^+] = −\log 0.30 = 0.52$
$[OH^−] = \frac{K_w}{[H_3O^+]} = \frac{1.0×10^{−14}}{0.30} = 3.3×10^{−14} M$
$pOH = −\log [OH^−] = −\log (3.3×10^{−14}) = 13.48$
(b) For 0.0042 M $Ca(OH)_2: Ca(OH)_2(aq) ⟶ Ca^{2+} (aq) + 2OH^−(aq)$
$[OH^−] = 2(0.0042 M) = 0.0084 M$
$pOH = −\log [OH^−] = −\log (0.0084) = 2.08$
$[H_3O^+] = \frac{K_w}{[OH^−]} = \frac{1.0×10^{−14}}{0.0084} = 1.2×10^{−12} M$
$pH = −\log [H_3O^+] = −\log (1.2×10^{−12}) = 11.92$
Check The strong acid has pH < 7 and the strong base has a pH > 7, as expected. In each case, pH + pOH = 14, so the arithmetic seems correct.