Chemistry - Ionic Equilibrium Question with Solution | TestHub

Solution:

We need to identify solutions with $\mathrm{pH}<7$ at $25^{\circ }\mathrm{C}$.

I. $10^{-4}\mathrm{M}\mathrm{HCl}$: Strong acid, $[{\mathrm{H}}^{+}]=10^{-4}\mathrm{M}$, $\mathrm{pH}=4$. (Acidic)

II. $10^{-8}\mathrm{M}\mathrm{HCl}$: Strong acid, but water's autoionization is significant. $[{\mathrm{H}}^{+}]=\sqrt{K_w+C_{acid}{K}_w}\approx 10^{-7}\mathrm{M}$. More precisely, $[{\mathrm{H}}^{+}{]}^2-10^{-8}[{\mathrm{H}}^{+}]-10^{-14}=0$, giving $[{\mathrm{H}}^{+}]\approx 1.05\times 10^{-7}\mathrm{M}$, so $\mathrm{pH}\approx 6.98$. (Acidic)

III. $10^{-6}\mathrm{M}{\mathrm{CH}}_3\mathrm{COOH}$: Weak acid. $[{\mathrm{H}}^{+}]=\sqrt{K_{a}C}=\sqrt{1.8\times 10^{-5}\times 10^{-6}}=\sqrt{1.8\times 10^{-11}}\approx 1.34\times 10^{-6}\mathrm{M}$. $\mathrm{pH}<7$. (Acidic)

IV. $10^{-6}\mathrm{M}{\mathrm{NH}}_4\mathrm{OH}$: Weak base. $[{\mathrm{OH}}^{-}]=\sqrt{K_{b}C}=\sqrt{1.8\times 10^{-5}\times 10^{-6}}\approx 1.34\times 10^{-6}\mathrm{M}$. $\mathrm{pOH}\approx 5.87$, $\mathrm{pH}\approx 8.13$. (Basic)

V. $0.01\mathrm{M}{\mathrm{NH}}_4\mathrm{Cl}$: Salt of strong acid and weak base. ${\mathrm{NH}}_4^{+}$ hydrolyzes: ${\mathrm{NH}}_4^{+}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{NH}}_4\mathrm{OH}+{\mathrm{H}}^{+}$. $K_h=K_w/K_b=10^{-14}/(1.8\times 10^{-5})\approx 5.56\times 10^{-10}$. $[{\mathrm{H}}^{+}]=\sqrt{K_{h}C}=\sqrt{5.56\times 10^{-10}\times 0.01}\approx 2.36\times 10^{-6}\mathrm{M}$. $\mathrm{pH}<7$. (Acidic)

VI. $0.01\mathrm{M}{\mathrm{CH}}_3\mathrm{COONa}$: Salt of weak acid and strong base. ${\mathrm{CH}}_3{\mathrm{COO}}^{-}$ hydrolyzes: ${\mathrm{CH}}_3{\mathrm{COO}}^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{OH}}^{-}$. $K_h=K_w/K_a=10^{-14}/(1.8\times 10^{-5})\approx 5.56\times 10^{-10}$. $[{\mathrm{OH}}^{-}]=\sqrt{K_{h}C}=\sqrt{5.56\times 10^{-10}\times 0.01}\approx 2.36\times 10^{-6}\mathrm{M}$. $\mathrm{pOH}<7$, so $\mathrm{pH}>7$. (Basic)

VII. Solution of $0.01\mathrm{M}{\mathrm{CH}}_3\mathrm{COONa}+0.01\mathrm{M}{\mathrm{CH}}_3\mathrm{COOH}$: This is an acidic buffer. $\mathrm{pH}=\mathrm{p}{K}_a+\log ([salt]/[acid])$. Since $[salt]=[acid]$, $\mathrm{pH}=\mathrm{p}{K}_a=-\log (1.8\times 10^{-5})\approx 4.74$. (Acidic)

VIII. $0.01\mathrm{M}{\mathrm{CH}}_3{\mathrm{COONH}}_4$: Salt of weak acid and weak base. $K_a({\mathrm{CH}}_3\mathrm{COOH})=K_b({\mathrm{NH}}_4\mathrm{OH})$. For such salts, $\mathrm{pH}=7+\frac{1}{2}(\mathrm{p}{K}_a-\mathrm{p}{K}_b)$. Since $\mathrm{p}{K}_a=\mathrm{p}{K}_b$, $\mathrm{pH}=7$. (Neutral)

Solutions with $\mathrm{pH}<7$ are I, II, III, V, VII.

Total number of solutions = 5.