Calculate the `pH` of solution obtained by mixing `10 ml` of `0.1 M HCl` and `40 ml` of `0.2 M H_(2)SO_(4)`

Explanation

To calculate the pH of the solution obtained by mixing 10 ml of 0.1 M HCl and 40 ml of 0.2 M H₂SO₄, we can follow these steps:

### Step 1: Calculate the Normality of Each Acid
1. **HCl**:
- Molarity (M) = 0.1 M
- n-factor for HCl = 1 (since it donates 1 H⁺ ion)
- Normality (N) = Molarity × n-factor = 0.1 × 1 = 0.1 N

2. **H₂SO₄**:
- Molarity (M) = 0.2 M
- n-factor for H₂SO₄ = 2 (since it donates 2 H⁺ ions)
- Normality (N) = Molarity × n-factor = 0.2 × 2 = 0.4 N

### Step 2: Calculate the Total Normality of the Mixture
Using the formula for the final normality (N₃) of the mixture:
\[ N_3 = \frac{N_1V_1 + N_2V_2}{V_1 + V_2} \]

Where:
- \( N_1 = 0.1 \, \text{N} \) (normality of HCl)
- \( V_1 = 10 \, \text{ml} \) (volume of HCl)
- \( N_2 = 0.4 \, \text{N} \) (normality of H₂SO₄)
- \( V_2 = 40 \, \text{ml} \) (volume of H₂SO₄)

Substituting the values:
\[ N_3 = \frac{(0.1 \times 10) + (0.4 \times 40)}{10 + 40} \]
\[ N_3 = \frac{1 + 16}{50} \]
\[ N_3 = \frac{17}{50} = 0.34 \, \text{N} \]

### Step 3: Calculate the Hydrogen Ion Concentration
Since both acids contribute H⁺ ions, the concentration of H⁺ ions in the solution is equal to the normality of the mixture:
- \( [H^+] = N_3 = 0.34 \, \text{M} \)

### Step 4: Calculate the pH of the Solution
Using the formula for pH:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of H⁺ concentration:
\[ \text{pH} = -\log(0.34) \]

Calculating:
1. Find the logarithm:
- \( \log(0.34) \approx -0.468 \)
2. Therefore,
\[ \text{pH} = -(-0.468) = 0.468 \]

### Final Answer:
The pH of the solution obtained by mixing 10 ml of 0.1 M HCl and 40 ml of 0.2 M H₂SO₄ is approximately **0.468**.

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