Question:
Equal volumes of \(0.5 \, N\) acetic acid and \(0.5 \, N\) sodium acetate are mixed. What is the pH of resultant solution?
\[
(pK_a \text{ of acetic acid } = 4.75)
\]
Equal volumes of \(0.5 \, N\) acetic acid and \(0.5 \, N\) sodium acetate are mixed. What is the pH of resultant solution?
\[ (pK_a \text{ of acetic acid } = 4.75) \]
\[ (pK_a \text{ of acetic acid } = 4.75) \]
Show Hint
For a buffer solution:
\[
pH=pK_a+\log\frac{[\text{Salt}]}{[\text{Acid}]}
\]
If concentrations of acid and salt are equal, then:
\[
pH=pK_a
\]
Updated On: Jun 15, 2026
- \(4.85\)
- \(4.65\)
- \(4.75\)
- \(7.0\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Identify the type of solution formed.
Acetic acid is a weak acid and sodium acetate is its conjugate salt.
Mixing them forms a buffer solution.
For an acidic buffer, pH is calculated using Henderson-Hasselbalch equation.
Step 2: Write Henderson-Hasselbalch equation.
\[ pH=pK_a+\log\frac{[\text{Salt}]}{[\text{Acid}]} \]
Step 3: Compare concentrations after mixing.
Equal volumes of \(0.5\,N\) acetic acid and \(0.5\,N\) sodium acetate are mixed.
Hence, concentration ratio becomes
\[ \frac{[\text{Salt}]}{[\text{Acid}]}=1 \]
Step 4: Substitute the values.
\[ pH=4.75+\log(1) \] Since,
\[ \log(1)=0 \] therefore,
\[ pH=4.75 \]
Step 5: Final conclusion.
Hence, the pH of the resultant solution is
\[ \boxed{4.75} \]
Acetic acid is a weak acid and sodium acetate is its conjugate salt.
Mixing them forms a buffer solution.
For an acidic buffer, pH is calculated using Henderson-Hasselbalch equation.
Step 2: Write Henderson-Hasselbalch equation.
\[ pH=pK_a+\log\frac{[\text{Salt}]}{[\text{Acid}]} \]
Step 3: Compare concentrations after mixing.
Equal volumes of \(0.5\,N\) acetic acid and \(0.5\,N\) sodium acetate are mixed.
Hence, concentration ratio becomes
\[ \frac{[\text{Salt}]}{[\text{Acid}]}=1 \]
Step 4: Substitute the values.
\[ pH=4.75+\log(1) \] Since,
\[ \log(1)=0 \] therefore,
\[ pH=4.75 \]
Step 5: Final conclusion.
Hence, the pH of the resultant solution is
\[ \boxed{4.75} \]
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