Calcualte the pH of solution when 100 mL, 0.1 M `CH_(3)COOH` and 100 mL, 0.1 M `HCOOH` are mixed together. `("Given : "K_(a)(CH_(3)COOH)=2xx10^(-5)), K_(a)(HCOOH)=6xx10^(-5)`

Answer

To calculate the pH of the solution when 100 mL of 0.1 M acetic acid (CH₃COOH) and 100 mL of 0.1 M formic acid (HCOOH) are mixed together, we will follow these steps:

### Step 1: Calculate the number of moles of each acid
1. **For Acetic Acid (CH₃COOH)**:
- Concentration (C₁) = 0.1 M
- Volume (V₁) = 100 mL = 0.1 L
- Number of moles (n₁) = C₁ × V₁ = 0.1 mol/L × 0.1 L = 0.01 mol (or 10 millimoles)

2. **For Formic Acid (HCOOH)**:
- Concentration (C₂) = 0.1 M
- Volume (V₂) = 100 mL = 0.1 L
- Number of moles (n₂) = C₂ × V₂ = 0.1 mol/L × 0.1 L = 0.01 mol (or 10 millimoles)

### Step 2: Calculate the total volume after mixing
- Total volume (V_total) = V₁ + V₂ = 100 mL + 100 mL = 200 mL = 0.2 L

### Step 3: Calculate the new concentrations after mixing
1. **Concentration of Acetic Acid (C₁')**:
- C₁' = n₁ / V_total = 0.01 mol / 0.2 L = 0.05 M

2. **Concentration of Formic Acid (C₂')**:
- C₂' = n₂ / V_total = 0.01 mol / 0.2 L = 0.05 M

### Step 4: Use the formula for hydrogen ion concentration
The hydrogen ion concentration [H⁺] can be calculated using the formula:
\[ [H^+] = \sqrt{K_{a1} \cdot C_1' + K_{a2} \cdot C_2'} \]
Where:
- \( K_{a1} \) (for acetic acid) = \( 2 \times 10^{-5} \)
- \( K_{a2} \) (for formic acid) = \( 6 \times 10^{-5} \)

Substituting the values:
\[ [H^+] = \sqrt{(2 \times 10^{-5} \cdot 0.05) + (6 \times 10^{-5} \cdot 0.05)} \]
Calculating each term:
- \( 2 \times 10^{-5} \cdot 0.05 = 1 \times 10^{-6} \)
- \( 6 \times 10^{-5} \cdot 0.05 = 3 \times 10^{-6} \)

Adding these together:
\[ [H^+] = \sqrt{1 \times 10^{-6} + 3 \times 10^{-6}} = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3} \text{ M} \]

### Step 5: Calculate the pH
Using the formula for pH:
\[ pH = -\log[H^+] \]
Substituting the value of [H⁺]:
\[ pH = -\log(2 \times 10^{-3}) \approx 2.3 \]

### Final Answer:
The pH of the solution when 100 mL of 0.1 M CH₃COOH and 100 mL of 0.1 M HCOOH are mixed together is approximately **2.3**.

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