A buffer solution is prepared by dissolving 0.2 mole of sodium formate and 0.25 mole of formic acid in approximately 200 `(pm 50)` mL of water. What will be the concentration of `H^+` and OH? `K_a (HCOOH) =1.8 xx 10^(-4)`

Answer

To find the concentration of \( H^+ \) and \( OH^- \) in the buffer solution prepared by dissolving sodium formate and formic acid, we can follow these steps:

### Step 1: Calculate the concentrations of formic acid and sodium formate
We start by calculating the concentrations of the components in the buffer solution.

1. **Volume of the solution**: Approximately 200 mL = 0.2 L
2. **Moles of formic acid (HCOOH)**: 0.25 moles
3. **Moles of sodium formate (HCOONa)**: 0.2 moles

Now, we can calculate the concentrations:
- Concentration of formic acid (\( [HCOOH] \)):
\[ [HCOOH] = \frac{0.25 \text{ moles}}{0.2 \text{ L}} = 1.25 \text{ M} \]
- Concentration of sodium formate (\( [HCOONa] \)):
\[ [HCOONa] = \frac{0.2 \text{ moles}}{0.2 \text{ L}} = 1.0 \text{ M} \]

### Step 2: Calculate \( pK_a \)
Next, we need to find \( pK_a \) using the given \( K_a \) for formic acid.

1. **Given**: \( K_a (HCOOH) = 1.8 \times 10^{-4} \)
2. **Calculate \( pK_a \)**:
\[ pK_a = -\log(K_a) = -\log(1.8 \times 10^{-4}) \approx 3.74 \]

### Step 3: Calculate the pH of the buffer solution
Using the Henderson-Hasselbalch equation for the buffer solution:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
where \( [A^-] \) is the concentration of the base (sodium formate) and \( [HA] \) is the concentration of the acid (formic acid).

Substituting the values we calculated:
\[ pH = 3.74 + \log\left(\frac{1.0}{1.25}\right) \]
Calculating the logarithm:
\[ \log\left(\frac{1.0}{1.25}\right) = \log(0.8) \approx -0.0969 \]
Now substituting back into the pH equation:
\[ pH = 3.74 - 0.0969 \approx 3.64 \]

### Step 4: Calculate the concentration of \( H^+ \)
To find the concentration of \( H^+ \):
\[ [H^+] = 10^{-pH} = 10^{-3.64} \approx 2.29 \times 10^{-4} \text{ M} \]

### Step 5: Calculate the concentration of \( OH^- \)
Using the relation \( K_w = [H^+][OH^-] \) where \( K_w = 1.0 \times 10^{-14} \):
\[ [OH^-] = \frac{K_w}{[H^+]} = \frac{1.0 \times 10^{-14}}{2.29 \times 10^{-4}} \approx 4.37 \times 10^{-11} \text{ M} \]

### Final Answer
- Concentration of \( H^+ \): \( 2.29 \times 10^{-4} \text{ M} \)
- Concentration of \( OH^- \): \( 4.37 \times 10^{-11} \text{ M} \)

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