Question

20 ml of 0.2 M HA (\( K_a = 5 \times 10^{-4} \)) is titrated with 10 ml of 0.2 M NaOH solution. Calculate the initial and final value of pH of the solution. [Given \( \log 5 = 0.7 \)]

• A. 2, 3.3
• B. 1.65, 2
• C. 3.3, 2
• D. 2, 3.6

Hint:

For weak acid-strong base titrations, calculate the pH at the start using the acid dissociation constant. After neutralization, calculate the remaining acid concentration and its pH in the final solution.

Answer 1

The Correct Option is A

Step 1: Calculate the initial pH of the weak acid solution.
A weak acid \( HA \) has a dissociation constant \( K_a = 5 \times 10^{-4} \) and an initial concentration of \( 0.2 \, \text{M} \). The ionization of the weak acid in water is represented as: \[ HA \rightleftharpoons H^+ + A^- \] The equilibrium expression for the dissociation constant is: \[ K_a = \frac{[H^+][A^-]}{[HA]} \] Let the concentration of \( H^+ \) formed at equilibrium be \( x \). Then: \[ K_a = \frac{x^2}{0.2 - x} \] Since the value of \( K_a \) is relatively small, we assume \( x \ll 0.2 \), so: \[ 0.2 - x \approx 0.2 \] Thus, the equation simplifies to: \[ K_a = \frac{x^2}{0.2} \] Substituting the value of \( K_a \): \[ 5 \times 10^{-4} = \frac{x^2}{0.2} \] \[ x^2 = 1 \times 10^{-4} \] \[ x = 1 \times 10^{-2} \, \text{M} \] Therefore, the hydrogen ion concentration is \( 10^{-2} \, \text{M} \), and the initial pH is: \[ \text{pH} = -\log [H^+] = -\log (10^{-2}) = 2 \]
Step 2: Determine the pH after partial neutralization with NaOH.
When sodium hydroxide (NaOH) is added, it reacts with the weak acid according to: \[ HA + OH^- \rightarrow A^- + H_2O \] First, calculate the initial moles of the acid: \[ \text{Moles of } HA = 0.2 \times 0.02 = 4 \times 10^{-3} \, \text{mol} \] Next, calculate the moles of NaOH added: \[ \text{Moles of NaOH} = 0.2 \times 0.01 = 2 \times 10^{-3} \, \text{mol} \] After neutralization: \[ \text{Remaining moles of } HA = 4 \times 10^{-3} - 2 \times 10^{-3} = 2 \times 10^{-3} \, \text{mol} \] \[ \text{Moles of } A^- \text{ formed} = 2 \times 10^{-3} \, \text{mol} \] The total volume of the solution becomes: \[ 20 \, \text{mL} + 10 \, \text{mL} = 30 \, \text{mL} = 0.03 \, \text{L} \] Since both \( HA \) and \( A^- \) are present in equal amounts, the solution forms a buffer. Therefore, we use the Henderson–Hasselbalch equation: \[ \text{pH} = \text{p}K_a + \log \frac{[A^-]}{[HA]} \] Here, \[ \text{p}K_a = -\log (5 \times 10^{-4}) \approx 3.3 \] \[ \frac{[A^-]}{[HA]} = 1 \] Thus, \[ \text{pH} = 3.3 + \log (1) \] \[ \text{pH} = 3.3 \]
Step 3: Conclusion.
The weak acid initially has a low pH due to partial ionization. After adding NaOH, a buffer solution is formed with equal concentrations of acid and its conjugate base, resulting in a higher pH.
Final Answer: Initial pH \( = 2 \), Final pH \( = 3.3 \).