At 25°C, 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when 10 mL of NaOH is added respectively, are :
Given: $$K_a = 5 \times 10^{-4}$$,
$$pK_a = 3.3$$,
$$\alpha \ll 1$$

(a) Initial pH (before adding NaOH)

Molar concentration of the weak acid $$HX$$ is $$C = 0.2\;{\rm M}$$ and its acid-dissociation constant is $$K_a = 5 \times 10^{-4}$$.
Because the acid is weak and the degree of ionisation is small (given $$\alpha \ll 1$$), we can use the approximation

$$[H^+] = \sqrt{K_a\,C}$$

Substituting the values:

$$[H^+] = \sqrt{(5 \times 10^{-4})(0.2)} = \sqrt{1.0 \times 10^{-4}} = 1.0 \times 10^{-2}\;{\rm M}$$

Hence

$$\text{pH} = -\log(1.0 \times 10^{-2}) = 2.0$$

(b) pH after adding 10.0 mL of 0.2 M NaOH

Step 1: Calculate moles before reaction.
Moles of $$HX$$ initially:

$$n_{HX}^{\,\text{initial}} = 0.2\;{\rm M} \times 20.0{\rm \,mL} = 0.2 \times 0.020 = 0.004\;{\rm mol}$$

Moles of $$NaOH$$ added:

$$n_{NaOH} = 0.2\;{\rm M} \times 10.0{\rm \,mL} = 0.2 \times 0.010 = 0.002\;{\rm mol}$$

Step 2: Perform the neutralisation reaction $$HX + OH^- \rightarrow X^- + H_2O$$.
Since $$n_{OH^-} = \frac{1}{2}\,n_{HX}^{\,\text{initial}}$$, one-half of the acid is neutralised.

After reaction:

Unreacted $$HX$$: $$0.004 - 0.002 = 0.002\;{\rm mol}$$
Formed salt $$NaX$$: $$0.002\;{\rm mol}$$

Step 3: Find concentrations in the final volume.

Final volume $$V_{\text{final}} = 20.0{\rm \,mL} + 10.0{\rm \,mL} = 30.0{\rm \,mL} = 0.030{\rm \,L}$$

$$[HX] = \frac{0.002}{0.030} = 0.0667\;{\rm M}$$
$$[X^-] = \frac{0.002}{0.030} = 0.0667\;{\rm M}$$

Step 4: Use the Henderson-Hasselbalch equation for the buffer $$HX/X^-$$:

$$\text{pH} = pK_a + \log\frac{[X^-]}{[HX]}$$

Because $$[X^-] = [HX]$$, the logarithmic term is zero:

$$\text{pH} = pK_a = 3.3$$

Results
(a) $$\text{pH} = 2.0$$; (b) $$\text{pH} = 3.3$$

Therefore the correct choice is Option B which is: (a) 2.0, (b) 3.3.