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.