At 25°C , 20.0 mL of 0.2 M weak monoprotic acid HX is titrated...

For part (a), at the start of the titration, the solution contains only the weak acid HX.

The concentration of HX is $C = 0.2$ M.

Using the formula for the hydrogen ion concentration of a weak acid with $\alpha \ll 1$:

$[H^+] = \sqrt{K_a \times C} = \sqrt{5 \times 10^{-4} \times 0.2} = \sqrt{10^{-4}} = 10^{-2}$ M

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

For part (b), when $10$ mL of $0.2$ M NaOH is added to $20$ mL of $0.2$ M HX:

Initial millimoles of HX $= 20 \times 0.2 = 4$ mmol

Millimoles of NaOH added $= 10 \times 0.2 = 2$ mmol

The added NaOH neutralizes half of the HX to form NaX, creating an acidic buffer.

Millimoles of HX remaining $= 4 - 2 = 2$ mmol

Millimoles of NaX formed $= 2$ mmol

Using the Henderson-Hasselbalch equation:

$pH = pK_a + \log \left( \dfrac{[\text{Salt}]}{[\text{Acid}]} \right)$

Since the millimoles of salt and acid are equal, $\log(1) = 0$.

$pH = pK_a = 3.3$

Answer: $2.0 \quad 3.3$