Chemistry - Ionic Equilibrium Question with Solution | TestHub

Initial pH:

The pOH is calculated using the formula:

$$pOH=\frac{1}{2}(p{K}_b-\log C)$$

Given $p{K}_b=6$ and assuming $C=10^{-1}$ (from $\log C=-1$), we have:

$$pOH=\frac{1}{2}(6-(-1))=\frac{1}{2}(7)=3.5$$

Therefore, the initial pH is $14-pOH=14-3.5=10.5$.

Equivalence point:

At the equivalence point, 25 mL of 0.1 M HCl has been added, which means equal moles of acid and base have reacted. The salt concentration is calculated as 2.5 mmol in a total volume of 50 mL, resulting in a concentration of 0.05 M.

For the hydrolysis of the salt, the hydrogen ion concentration, $[H^{+}]$, is given by:

$$[H^{+}]=\sqrt{\frac{K_w\cdot C}{K_b}}$$

Substituting the given values, $K_w=10^{-14}$, $C=0.05$, and $K_b=10^{-6}$:

$$[H^{+}]=\sqrt{\frac{10^{-14}\times 0.05}{10^{-6}}}=\sqrt{5\times 10^{-10}}\approx 2.2\times 10^{-5}$$

Thus, the pH at the equivalence point is $-\log (2.2\times 10^{-5})\approx 4.65$ (approximately 4.7).

Based on these calculations:

(A) The equivalence point is incorrectly placed at 12.5 mL.

(B) The starting pH is incorrectly shown as 13.

(D) The equivalence point pH is incorrectly shown as 7.

(C) This option correctly shows the initial pH at 10.5 and the equivalence point at 25 mL with a pH of approximately 4.7. Therefore, (C) is the correct answer.