Two buffer, (X) and (Y) of pH `4.0` and `6.0` respectively are prepared from acid HA and the salt NaA. Both the buffers are `0.50` M in HA. What would be the pH of the solution obtained by mixing equal volumes of the two buffers ? `(K_(HA)=1.0xx10^(-5))`

Explanation

To find the pH of the solution obtained by mixing equal volumes of two buffers (X and Y) with pH values of 4.0 and 6.0 respectively, we will follow these steps:

### Step 1: Determine the concentrations of the salt (A-) in each buffer.

**For Buffer X (pH = 4.0):**
Using the Henderson-Hasselbalch equation:
\[ \text{pH} = -\log K_a + \log \left( \frac{[A^-]}{[HA]} \right) \]
Given that \(K_a = 1.0 \times 10^{-5}\), we can substitute:
\[ 4.0 = -\log(1.0 \times 10^{-5}) + \log \left( \frac{[A^-]}{0.50} \right) \]
Calculating \(-\log(1.0 \times 10^{-5})\):
\[ -\log(1.0 \times 10^{-5}) = 5 \]
Now substituting back into the equation:
\[ 4.0 = 5 + \log \left( \frac{[A^-]}{0.50} \right) \]
Rearranging gives:
\[ \log \left( \frac{[A^-]}{0.50} \right) = 4.0 - 5 = -1 \]
Thus,
\[ \frac{[A^-]}{0.50} = 10^{-1} \implies [A^-] = 0.10 \times 0.50 = 0.05 \, \text{M} \]

**For Buffer Y (pH = 6.0):**
Using the same Henderson-Hasselbalch equation:
\[ 6.0 = -\log(1.0 \times 10^{-5}) + \log \left( \frac{[A^-]}{0.50} \right) \]
Again, calculating \(-\log(1.0 \times 10^{-5})\):
\[ -\log(1.0 \times 10^{-5}) = 5 \]
Substituting into the equation:
\[ 6.0 = 5 + \log \left( \frac{[A^-]}{0.50} \right) \]
Rearranging gives:
\[ \log \left( \frac{[A^-]}{0.50} \right) = 6.0 - 5 = 1 \]
Thus,
\[ \frac{[A^-]}{0.50} = 10^{1} \implies [A^-] = 10 \times 0.50 = 5.0 \, \text{M} \]

### Step 2: Mix the two buffers.

When equal volumes of the two buffers are mixed, the concentrations of HA and A- will change.

**Concentration of A- in the mixed solution:**
\[ [A^-]_{\text{mixed}} = \frac{0.05 \, \text{M} + 5.0 \, \text{M}}{2} = \frac{5.05 \, \text{M}}{2} = 2.525 \, \text{M} \]

**Concentration of HA in the mixed solution:**
\[ [HA]_{\text{mixed}} = \frac{0.50 \, \text{M} + 0.50 \, \text{M}}{2} = \frac{1.0 \, \text{M}}{2} = 0.50 \, \text{M} \]

### Step 3: Calculate the pH of the mixed buffer.

Using the Henderson-Hasselbalch equation again:
\[ \text{pH} = -\log K_a + \log \left( \frac{[A^-]}{[HA]} \right) \]
Substituting the values:
\[ \text{pH} = -\log(1.0 \times 10^{-5}) + \log \left( \frac{2.525}{0.50} \right) \]
Calculating:
\[ \text{pH} = 5 + \log(5.05) \]
Calculating \(\log(5.05)\):
\[ \log(5.05) \approx 0.7033 \]
Thus:
\[ \text{pH} = 5 + 0.7033 = 5.7033 \]

### Final Answer:
The pH of the solution obtained by mixing equal volumes of the two buffers is approximately **5.70**.

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