Question

Equal volumes of 0.1 M acetic acid and 0.1 M sodium acetate are mixed to form a buffer solution. Considering that the ionization of acetic acid is occuring at dissociation constant of \(1.74 \times 10^{-5}\), what will be its pKa value? (Given: log 1.74 = 0.24)

• A. 5.24
• B. 4.76
• C. 0.024
• D. 0.5

Hint:

For any weak acid with a dissociation constant in the form of \(a \times 10^{-b}\), the \(\text{p}K_a\) can be estimated quickly as \(b - \log_{10}(a)\).
Here, \(5 - \log_{10}(1.74) = 5 - 0.24 = 4.76\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The \(\text{p}K_a\) value is defined as the negative logarithm (base 10) of the acid dissociation constant (\(K_a\)).
It serves as a quantitative measure of the strength of an acid in solution.
Key Formula or Approach:
The mathematical definition of \(\text{p}K_a\) is: \[ \text{p}K_a = -\log_{10}(K_a) \]

Step 2: Detailed Explanation:
The acid dissociation constant \(K_a\) for acetic acid is given as: \[ K_a = 1.74 \times 10^{-5} \] We calculate the \(\text{p}K_a\) as follows: \[ \text{p}K_a = -\log_{10}(1.74 \times 10^{-5}) \] Using logarithmic identity properties: \[ \text{p}K_a = -\left[ \log_{10}(1.74) + \log_{10}(10^{-5}) \right] \] \[ \text{p}K_a = -\log_{10}(1.74) - (-5) \] \[ \text{p}K_a = 5 - \log_{10}(1.74) \] Substituting the given value of \(\log_{10}(1.74) = 0.24\): \[ \text{p}K_a = 5 - 0.24 = 4.76 \] Since equal volumes of equal concentrations of weak acid and conjugate base are mixed, the pH of this buffer solution is equal to its \(\text{p}K_a\), which is 4.76.

Step 3: Final Answer:
Thus, the \(\text{p}K_a\) value is 4.76, corresponding to option (B).