50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK$$_b$$ of ammonia solution is 4.75, the pH of the mixture will be:

We have been given 50 mL of an ammonia solution of molarity 0.2 M and 25 mL of a hydrochloric-acid solution of molarity 0.2 M. First we convert every volume to litres, because molarity is defined as $$\text{moles} / \text{litre}\,.$$ Thus $$50\ \text{mL}=0.050\ \text{L}\quad\text{and}\quad25\ \text{mL}=0.025\ \text{L}\,.$$

Now we calculate the number of moles present before any reaction takes place, using the formula $$n = M \times V\,.$$

For ammonia, $$n_{\text{NH}_3}^{\text{initial}} = 0.2\ \text{M} \times 0.050\ \text{L} = 0.010\ \text{mol}\,.$$

For hydrochloric acid, $$n_{\text{HCl}}^{\text{initial}} = 0.2\ \text{M} \times 0.025\ \text{L} = 0.005\ \text{mol}\,.$$

The reaction taking place is a simple acid-base neutralisation

$$\text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4^+ + \text{Cl}^-\,.$$

Because HCl is a strong acid, it reacts completely with an equivalent amount of NH3. The stoichiometric coefficients are 1:1, so the change in moles is straightforward.

HCl is the limiting reagent because it has the smaller number of moles (0.005 mol). All of this HCl will be consumed, using an equal 0.005 mol of NH3, and producing 0.005 mol of NH4+.

After the reaction:

$$ \begin{aligned} n_{\text{NH}_3}^{\text{final}} &= 0.010 - 0.005 = 0.005\ \text{mol},\\[4pt] n_{\text{NH}_4^+}^{\text{final}} &= 0 + 0.005 = 0.005\ \text{mol}\,. \end{aligned} $$

The total volume of the solution is now

$$V_{\text{total}} = 0.050\ \text{L} + 0.025\ \text{L} = 0.075\ \text{L}\,.$$

We therefore find the molar concentrations of the conjugate-base and conjugate-acid components:

$$ \begin{aligned} [\text{NH}_3] &= \frac{0.005\ \text{mol}}{0.075\ \text{L}} = 0.0667\ \text{M},\\[4pt] [\text{NH}_4^+] &= \frac{0.005\ \text{mol}}{0.075\ \text{L}} = 0.0667\ \text{M}\,. \end{aligned} $$

Thus the mixture is a buffer in which the concentrations of the base (NH3) and its conjugate acid (NH4+) are equal.

To obtain the pH we employ the Henderson-Hasselbalch equation for a buffer based on a weak base and its conjugate acid, written in its pH form:

$$\boxed{\,\text{pH} = \text{p}K_a + \log\!\left(\frac{[\text{base}]}{[\text{acid}]}\right)}\,.$$

Because we are given the basicity constant pKb of ammonia, we first convert it to the acidity constant pKa of ammonium ion. For a conjugate acid-base pair, the relationship is

$$\text{p}K_a + \text{p}K_b = 14\,.$$

Therefore

$$\text{p}K_a = 14 - 4.75 = 9.25\,.$$

Now we substitute every value into the Henderson-Hasselbalch formula:

$$ \begin{aligned} \text{pH} &= 9.25 + \log\!\left(\frac{0.0667}{0.0667}\right) \\[6pt] &= 9.25 + \log(1) \\[6pt] &= 9.25 + 0 \\[6pt] &= 9.25\,. \end{aligned} $$

Hence, the correct answer is Option B.