Question 18.3: Calculating [H3O^+], pH, [OH^−], and pOH for Strong Acids an......
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
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.