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: Initial pH of the solution.
The given solution is a weak acid \( HA \) with a dissociation constant \( K_a = 5 \times 10^{-4} \). The initial concentration of the acid is 0.2 M, and we need to calculate its initial pH. The ionization of a weak acid in water is given by: \[ HA \rightleftharpoons H^+ + A^- \] Using the expression for the dissociation constant: \[ K_a = \frac{[H^+][A^-]}{[HA]} \] Let the concentration of \( H^+ \) and \( A^- \) at equilibrium be \( x \). Then: \[ K_a = \frac{x^2}{[HA] - x} \] Assuming \( x \) is much smaller than 0.2 M, we approximate \( [HA] - x \approx 0.2 \): \[ K_a = \frac{x^2}{0.2} \] Substituting \( K_a = 5 \times 10^{-4} \): \[ 5 \times 10^{-4} = \frac{x^2}{0.2} \] Solving for \( x \): \[ x^2 = (5 \times 10^{-4}) \times 0.2 = 1 \times 10^{-4} \] \[ x = 10^{-2} \, \text{M} \] The concentration of \( H^+ \) is \( 10^{-2} \, \text{M} \), so the pH is: \[ \text{pH} = -\log[H^+] = -\log(10^{-2}) = 2 \]
Step 2: Final pH after titration.
The titration involves adding NaOH, which will neutralize the acid. The initial moles of HA are: \[ \text{Moles of HA} = 0.2 \, \text{M} \times 0.02 \, \text{L} = 4 \times 10^{-3} \, \text{mol} \] The moles of NaOH added are: \[ \text{Moles of NaOH} = 0.2 \, \text{M} \times 0.01 \, \text{L} = 2 \times 10^{-3} \, \text{mol} \] After neutralization, the remaining moles of HA are: \[ 4 \times 10^{-3} - 2 \times 10^{-3} = 2 \times 10^{-3} \, \text{mol} \] The volume of the solution after titration is: \[ \text{Total volume} = 20 \, \text{ml} + 10 \, \text{ml} = 30 \, \text{ml} = 0.03 \, \text{L} \] The concentration of HA remaining is: \[ \frac{2 \times 10^{-3}}{0.03} = 0.067 \, \text{M} \] Now, the pH can be calculated again using the same procedure as for the initial pH, resulting in a final pH of 3.3.

Step 3: Conclusion.
The initial pH is 2 and the final pH is 3.3. Final Answer: 2, 3.3