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 can follow these steps:

### Step 1: Calculate the number of moles of each acid
1. **Acetic Acid (CH₃COOH)**:
- Volume = 100 mL = 0.1 L
- Concentration = 0.1 M
- Moles of CH₃COOH = Concentration × Volume = 0.1 M × 0.1 L = 0.01 moles

2. **Formic Acid (HCOOH)**:
- Volume = 100 mL = 0.1 L
- Concentration = 0.1 M
- Moles of HCOOH = Concentration × Volume = 0.1 M × 0.1 L = 0.01 moles

### Step 2: Calculate the total volume of the mixed solution
- Total Volume = Volume of CH₃COOH + Volume of HCOOH = 100 mL + 100 mL = 200 mL = 0.2 L

### Step 3: Calculate the new concentrations after mixing
1. **Concentration of CH₃COOH after mixing**:
- New Concentration (C₁') = Moles / Total Volume = 0.01 moles / 0.2 L = 0.05 M

2. **Concentration of HCOOH after mixing**:
- New Concentration (C₂') = Moles / Total Volume = 0.01 moles / 0.2 L = 0.05 M

### Step 4: Use the formula to calculate [H⁺]
The formula to calculate the hydrogen ion concentration [H⁺] from the two weak acids is:
\[ [H^+] = \sqrt{K_{a1} \cdot C_1' + K_{a2} \cdot C_2'} \]

Where:
- \( K_{a1} \) for CH₃COOH = \( 2 \times 10^{-5} \)
- \( K_{a2} \) for HCOOH = \( 6 \times 10^{-5} \)

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

### Step 5: Calculate the values inside the square root
1. Calculate each term:
- For CH₃COOH: \( 2 \times 10^{-5} \cdot 0.05 = 1 \times 10^{-6} \)
- For HCOOH: \( 6 \times 10^{-5} \cdot 0.05 = 3 \times 10^{-6} \)

2. Add the two results:
- Total = \( 1 \times 10^{-6} + 3 \times 10^{-6} = 4 \times 10^{-6} \)

### Step 6: Calculate [H⁺]
\[ [H^+] = \sqrt{4 \times 10^{-6}} = 2 \times 10^{-3} \, \text{M} \]

### Step 7: Calculate the pH
Using the formula for pH:
\[ \text{pH} = -\log[H^+] \]
\[ \text{pH} = -\log(2 \times 10^{-3}) \]

Using logarithmic properties:
\[ \text{pH} = -\log(2) - \log(10^{-3}) \]
\[ \text{pH} = -0.301 - (-3) \]
\[ \text{pH} = 3 - 0.301 \]
\[ \text{pH} \approx 2.699 \]

### 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 **2.70**.

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