Chemistry - Ionic Equilibrium Question with Solution | TestHub

Solution:

(P) For $0.5\mathrm{M}{\mathrm{H}}_2{\mathrm{CO}}_3$, $K_{a_1}=10^{-10}$.

$[{\mathrm{H}}^{+}]=\sqrt{C\cdot K_{a_1}}=\sqrt{0.5\cdot 10^{-10}}=\frac{1}{\sqrt{2}}\times 10^{-5}\mathrm{M}$.

$\mathrm{pH}=-{\log }_{10}(\frac{1}{\sqrt{2}}\times 10^{-5})=5+{\log }_{10}\sqrt{2}=5+0.15=5.15$.

Thus, (P) matches (3) and (5) (since $\mathrm{pH}+\mathrm{pOH}=14$ at $25^{\circ }\mathrm{C}$).

Also, at $\mathrm{pH}=5.15$, $[{\mathrm{H}}^{+}]>[{\mathrm{OH}}^{-}]$, so (P) matches (4).

(Q) For $10^{-3}\mathrm{M}{\mathrm{NH}}_4\mathrm{OH}$ with $\alpha =\frac{0.01}{\sqrt{2}}$.

$[{\mathrm{OH}}^{-}]=C\cdot \alpha =10^{-3}\cdot \frac{0.01}{\sqrt{2}}=\frac{10^{-5}}{\sqrt{2}}\mathrm{M}$.

$\mathrm{pOH}=-{\log }_{10}(\frac{10^{-5}}{\sqrt{2}})=5+{\log }_{10}\sqrt{2}=5+0.15=5.15$.

$\mathrm{pH}=14-5.15=8.85$.

Thus, (Q) matches (1) and (5).

(R) Total moles of ${\mathrm{H}}^{+}$:

From $\mathrm{HCl}$: $100\times 10^{-4}=10^{-3}$ mmol.

From ${\mathrm{H}}_2{\mathrm{SO}}_4$: $200\times 2\times 10^{-4}\times 2=8\times 10^{-2}$ mmol. (Note: ${\mathrm{H}}_2{\mathrm{SO}}_4$ is diprotic, so $2\times$ concentration).

Total moles ${\mathrm{H}}^{+}=10^{-3}+8\times 10^{-2}=0.001+0.08=0.081$ mmol.

Total volume = $100+200=300$ mL.

$[{\mathrm{H}}^{+}]=\frac{0.081mmol}{300mL}=\frac{8.1\times 10^{-2}}{300}=2.7\times 10^{-4}\mathrm{M}$.

$\mathrm{pH}=-{\log }_{10}(2.7\times 10^{-4})=4-{\log }_{10}2.7\approx 4-0.43=3.57$.

At $62^{\circ }\mathrm{C}$, $K_w=10^{-13}$, so $\mathrm{pH}+\mathrm{pOH}=13$.

$\mathrm{pOH}=13-3.57=9.43$.

Thus, (R) matches (2) and (4) (since $\mathrm{pH}=3.57<6.5$, $[{\mathrm{H}}^{+}]>[{\mathrm{OH}}^{-}]$)

(S) Equal volumes of $2\mathrm{M}{\mathrm{NH}}_4\mathrm{OH}$ and $2\times 10^{-3}\mathrm{M}\mathrm{NaOH}$ are mixed.

The strong base $\mathrm{NaOH}$ will dominate the $\mathrm{pH}$.

After mixing equal volumes, the concentration of $\mathrm{NaOH}$ becomes half:

$[\mathrm{NaOH}]=\frac{2\times 10^{-3}}{2}=10^{-3}\mathrm{M}$.

$[{\mathrm{OH}}^{-}]=10^{-3}\mathrm{M}$.

$\mathrm{pOH}=-{\log }_{10}(10^{-3})=3$.

$\mathrm{pH}=14-3=11$.

Thus, (S) matches (5) and (1) (since $\mathrm{pH}=11$, $\mathrm{pOH}=3$, so $\mathrm{pH}\ne 8.85$).

Also, at $\mathrm{pH}=11$, $[{\mathrm{OH}}^{-}]>[{\mathrm{H}}^{+}]$, so (S) does NOT match (4).

Matching:

(P) $\to$ (3), (4), (5)

(Q) $\to$ (1), (5)

(R) $\to$ (2), (4)

(S) $\to$ (5)

The correct option is (C).