The concentration of hydrogen ions in a `0.2M` solution of formic acid is `6.4 xx 10^(-3) mol L^(-1)`. To this solution, sodium formate is added so as to adjust the concentration of sodium formate to `1 mol L^(-1)`. What will be the `pH` of this solution? The dissociation constant of formic acid is `2.4 xx 10^(-4)` and the degree of dissociation fo sodium formate is `0.75`.

Explanation

To find the pH of the solution after adding sodium formate to the formic acid solution, we can follow these steps:

### Step 1: Determine the initial concentration of hydrogen ions
Given that the concentration of hydrogen ions \([H^+]\) in a \(0.2 \, M\) solution of formic acid is \(6.4 \times 10^{-3} \, mol \, L^{-1}\).

### Step 2: Calculate the degree of dissociation (\(\alpha\)) of formic acid
The dissociation of formic acid can be represented as:

\[ \text{HCOOH} \rightleftharpoons \text{H}^+ + \text{HCOO}^- \]

At equilibrium, the concentration of \(H^+\) ions is given by:

\[ [H^+] = C \cdot \alpha \]

Where:
- \(C = 0.2 \, M\) (initial concentration of formic acid)
- \([H^+] = 6.4 \times 10^{-3} \, mol \, L^{-1}\)

Now, substituting the values:

\[ 6.4 \times 10^{-3} = 0.2 \cdot \alpha \]

Solving for \(\alpha\):

\[ \alpha = \frac{6.4 \times 10^{-3}}{0.2} = 0.032 \]

### Step 3: Understand the effect of adding sodium formate
When sodium formate (\(HCOONa\)) is added, it dissociates completely into \(HCOO^-\) and \(Na^+\). The concentration of \(HCOO^-\) will be \(1 \, M\) as given in the problem.

### Step 4: Calculate the new pH using the Henderson-Hasselbalch equation
The solution now acts as a buffer, and we can use the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

Where:
- \(\text{Salt} = [HCOO^-] = 1 \, M\)
- \(\text{Acid} = [HCOOH]\)

To find \([\text{HCOOH}]\), we can use the dissociation constant \(K_a\) of formic acid, which is given as \(2.4 \times 10^{-4}\).

### Step 5: Calculate pKa
\[ \text{pKa} = -\log(K_a) = -\log(2.4 \times 10^{-4}) \approx 3.619 \]

### Step 6: Substitute values into the Henderson-Hasselbalch equation
Now, we can substitute the values into the equation:

\[ \text{pH} = 3.619 + \log\left(\frac{1}{6.4 \times 10^{-3}}\right) \]

Calculating the logarithm:

\[ \log\left(\frac{1}{6.4 \times 10^{-3}}\right) = -\log(6.4 \times 10^{-3}) \approx 2.193 \]

### Step 7: Final pH calculation
Now substituting back into the pH equation:

\[ \text{pH} = 3.619 + 2.193 = 5.812 \]

### Conclusion
The pH of the solution after adding sodium formate is approximately **5.812**.