Question

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:

• A. 0.7; 2.0
• B. 1.1; 2.2
• C. 1.2; 2.2
• D. 1.2; 3.0

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Part (a) is a simple weak acid pH calculation before any base is added.
Part (b) describes a partial neutralization. Since we add exactly half the volume of base needed to fully neutralize the acid, we create an equimolar buffer solution of the weak acid and its conjugate base.
Step 2: Key Formula or Approach:
For a weak acid with \(\alpha \ll 1\): \([\text{H}^+] = \sqrt{K_a \cdot C}\) and \(\text{pH} = \frac{1}{2}(\text{pK}_a - \log C)\).
For a buffer solution, Henderson-Hasselbalch equation: \(\text{pH} = \text{pK}_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}\).
Step 3: Detailed Explanation:
(a) At the start of the titration:
Concentration of HX, \(C = 0.2 \text{ M}\).
Calculate the \([\text{H}^+]\) concentration:
\[ [\text{H}^+] = \sqrt{K_a \cdot C} = \sqrt{5 \times 10^{-4} \times 0.2} = \sqrt{1.0 \times 10^{-4}} = 10^{-2} \text{ M} \] \[ \text{pH} = -\log(10^{-2}) = 2.0 \] (b) After adding 10 mL of 0.2 M NaOH:
Initial millimoles of HX = \(20.0 \text{ mL} \times 0.2 \text{ M} = 4.0 \text{ mmol}\).
Millimoles of NaOH added = \(10.0 \text{ mL} \times 0.2 \text{ M} = 2.0 \text{ mmol}\).
The strong base reacts completely with the weak acid:
\(\text{HX} + \text{NaOH} \rightarrow \text{NaX} + \text{H}_2\text{O}\)
Remaining HX = \(4.0 - 2.0 = 2.0 \text{ mmol}\).
Formed NaX (salt) = \(2.0 \text{ mmol}\).
Since the millimoles of the weak acid and its conjugate base are equal (\([\text{Salt}] = [\text{Acid}]\)), this is the half-equivalence point.
Using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{2.0}{2.0}\right) = 3.3 + \log(1) = 3.3 + 0 = 3.3 \] Step 4: Final Answer:
The pH values are 2.0 and 3.3 respectively.