How much moles of sodium propionate should be added to `1L` of an aqueous solution containing `0.020 mol` of propionic acid to obtain a buffer solution of `pH 4.75 `? What will be the `pH` if `0.010 mol` of `HCI` is dissolved in the above buffer solution. Compare the last `pH` value with the `pH of 0.010 M HCI` solution. Dissociation constant of propionic acid, `K_(a)`, at `25^(@)C` is `1.34 xx 10^(-5)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate pKa of Propionic Acid
The dissociation constant \( K_a \) of propionic acid is given as \( 1.34 \times 10^{-5} \). We can calculate \( pK_a \) using the formula:
\[ pK_a = -\log(K_a) \]
Substituting the value of \( K_a \):
\[ pK_a = -\log(1.34 \times 10^{-5}) \approx 4.87 \]

### Step 2: Use the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation for a buffer solution is given by:
\[ pH = pK_a + \log\left(\frac{[Salt]}{[Acid]}\right) \]
We want the pH to be 4.75, and we know:
- \( pK_a = 4.87 \)
- Concentration of propionic acid \([Acid] = 0.020 \, \text{mol/L}\)

Let \( x \) be the moles of sodium propionate added. Therefore, the concentration of the salt \([Salt] = \frac{x}{1} = x \, \text{mol/L}\).

Substituting into the Henderson-Hasselbalch equation:
\[ 4.75 = 4.87 + \log\left(\frac{x}{0.020}\right) \]

### Step 3: Rearranging the Equation
Rearranging the equation gives:
\[ \log\left(\frac{x}{0.020}\right) = 4.75 - 4.87 = -0.12 \]
This can be rewritten as:
\[ \frac{x}{0.020} = 10^{-0.12} \]
Calculating \( 10^{-0.12} \):
\[ 10^{-0.12} \approx 0.758 \]
Thus:
\[ x = 0.020 \times 0.758 \approx 0.01516 \, \text{mol} \]

### Step 4: Final Calculation of Sodium Propionate
To obtain a buffer solution with pH 4.75, approximately \( 0.0152 \, \text{mol} \) of sodium propionate should be added.

### Step 5: Calculate pH After Adding HCl
Now, we need to find the pH after adding \( 0.010 \, \text{mol} \) of HCl to the buffer solution.

1. **New concentration of propionic acid**:
\[ [Acid] = 0.020 + 0.010 = 0.030 \, \text{mol/L} \]
2. **New concentration of sodium propionate**:
\[ [Salt] = 0.0152 - 0.010 = 0.0052 \, \text{mol/L} \]

### Step 6: Apply the Henderson-Hasselbalch Equation Again
Using the new concentrations in the Henderson-Hasselbalch equation:
\[ pH = 4.87 + \log\left(\frac{0.0052}{0.030}\right) \]
Calculating the log term:
\[ \frac{0.0052}{0.030} \approx 0.1733 \]
Thus:
\[ \log(0.1733) \approx -0.760 \]
Substituting back into the equation:
\[ pH = 4.87 - 0.760 \approx 4.11 \]

### Step 7: Compare with pH of HCl Solution
The pH of a \( 0.010 \, \text{M} \) HCl solution is calculated as:
\[ pH = -\log(0.010) = 2 \]

### Conclusion
- The amount of sodium propionate to be added is approximately \( 0.0152 \, \text{mol} \).
- The pH after adding \( 0.010 \, \text{mol} \) of HCl is approximately \( 4.11 \).
- The pH of \( 0.010 \, \text{M} \) HCl is \( 2 \), which is significantly lower than the pH of the buffer solution.