A buffer solution was prepared by dissolving 0:05 mole formic acid and 0.06 mole sodium formate in enough water make 1 litre of solution. `K_a (HCOOH)= 1.8 xx 10^(-4).`
(a) Calculate the pH of the solution.
(b) If this solution were diluted to 10 times its volume, what would be the pH?

Answer

To solve the problem step by step, we will calculate the pH of the buffer solution prepared with formic acid and sodium formate, and then determine the pH after diluting the solution.

### Step-by-Step Solution:

**(a) Calculate the pH of the solution:**

1. **Identify the components of the buffer solution:**
- Weak acid: Formic acid (HCOOH)
- Conjugate base: Sodium formate (HCOONa)

2. **Write the expression for the dissociation constant (Ka) of formic acid:**
\[ K_a = 1.8 \times 10^{-4} \]

3. **Calculate the pKa:**
\[ pK_a = -\log(K_a) = -\log(1.8 \times 10^{-4}) \]
Using a calculator:
\[ pK_a \approx 3.74 \]

4. **Use the Henderson-Hasselbalch equation to calculate the pH:**
\[ pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Here, \([\text{Salt}] = 0.06 \, \text{mol/L}\) and \([\text{Acid}] = 0.05 \, \text{mol/L}\).

5. **Substituting the values into the equation:**
\[ pH = 3.74 + \log\left(\frac{0.06}{0.05}\right) \]

6. **Calculate the ratio:**
\[ \frac{0.06}{0.05} = 1.2 \]

7. **Calculate the logarithm:**
\[ \log(1.2) \approx 0.079 \]

8. **Final calculation of pH:**
\[ pH = 3.74 + 0.079 \approx 3.82 \]

**(b) Calculate the pH after diluting the solution to 10 times its volume:**

1. **Understand the effect of dilution on the buffer components:**
- When diluted, the concentrations of both the weak acid and the conjugate base will decrease by a factor of 10.
- New concentrations:
\[ [\text{Acid}] = \frac{0.05}{10} = 0.005 \, \text{mol/L} \]
\[ [\text{Salt}] = \frac{0.06}{10} = 0.006 \, \text{mol/L} \]

2. **Reapply the Henderson-Hasselbalch equation:**
\[ pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

3. **Substituting the new concentrations:**
\[ pH = 3.74 + \log\left(\frac{0.006}{0.005}\right) \]

4. **Calculate the ratio:**
\[ \frac{0.006}{0.005} = 1.2 \]

5. **Calculate the logarithm:**
\[ \log(1.2) \approx 0.079 \]

6. **Final calculation of pH after dilution:**
\[ pH = 3.74 + 0.079 \approx 3.82 \]

### Final Answers:
- (a) The pH of the solution is approximately **3.82**.
- (b) The pH after dilution remains approximately **3.82**.