Each of solution A and B of 100 L containing 4 g NaOH and 9.8 g H2SO4. Find pH of solution which is obtain by mixing 40 L solution of A and 10 L solution of B.

Answer

To find the pH of the solution obtained by mixing 40 L of solution A (containing NaOH) and 10 L of solution B (containing H2SO4), we can follow these steps:

### Step 1: Calculate the Molarity of NaOH in Solution A
1. **Given**:
- Mass of NaOH = 4 g
- Molecular mass of NaOH = 40 g/mol
- Volume of solution A = 100 L

2. **Calculate moles of NaOH**:
\[ \text{Moles of NaOH} = \frac{\text{Mass}}{\text{Molecular Mass}} = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ mol} \]

3. **Calculate molarity (M)**:
\[ \text{Molarity of NaOH} = \frac{\text{Moles}}{\text{Volume in L}} = \frac{0.1 \text{ mol}}{100 \text{ L}} = 10^{-3} \text{ M} \]

### Step 2: Calculate the Molarity of H2SO4 in Solution B
1. **Given**:
- Mass of H2SO4 = 9.8 g
- Molecular mass of H2SO4 = 98 g/mol
- Volume of solution B = 100 L

2. **Calculate moles of H2SO4**:
\[ \text{Moles of H2SO4} = \frac{9.8 \text{ g}}{98 \text{ g/mol}} = 0.1 \text{ mol} \]

3. **Calculate molarity (M)**:
\[ \text{Molarity of H2SO4} = \frac{0.1 \text{ mol}}{100 \text{ L}} = 10^{-3} \text{ M} \]

### Step 3: Calculate the Amount of NaOH and H2SO4 in the Mixed Solution
1. **For solution A (40 L)**:
\[ \text{Moles of NaOH in 40 L} = 10^{-3} \text{ M} \times 40 \text{ L} = 0.04 \text{ mol} \]

2. **For solution B (10 L)**:
\[ \text{Moles of H2SO4 in 10 L} = 10^{-3} \text{ M} \times 10 \text{ L} = 0.01 \text{ mol} \]

### Step 4: Determine the Reaction Between NaOH and H2SO4
1. **Reaction**:
\[ \text{H2SO4} + 2 \text{NaOH} \rightarrow \text{Na2SO4} + 2 \text{H2O} \]

2. **Stoichiometry**:
- 1 mole of H2SO4 reacts with 2 moles of NaOH.
- From 0.01 mol of H2SO4, we need 0.02 mol of NaOH.
- We have 0.04 mol of NaOH available, which is sufficient to react with 0.01 mol of H2SO4.

3. **Calculate the remaining moles**:
- Moles of NaOH after reaction = 0.04 mol - 0.02 mol = 0.02 mol (remaining)

### Step 5: Calculate the Concentration of OH⁻ Ions in the Mixed Solution
1. **Total Volume of the Mixed Solution**:
\[ \text{Total Volume} = 40 \text{ L} + 10 \text{ L} = 50 \text{ L} \]

2. **Concentration of OH⁻**:
\[ \text{Concentration of OH⁻} = \frac{\text{Remaining moles of NaOH}}{\text{Total Volume}} = \frac{0.02 \text{ mol}}{50 \text{ L}} = 0.0004 \text{ M} = 4 \times 10^{-4} \text{ M} \]

### Step 6: Calculate pOH and pH
1. **Calculate pOH**:
\[ \text{pOH} = -\log[OH⁻] = -\log(4 \times 10^{-4}) \approx 3.397 \]

2. **Calculate pH**:
\[ \text{pH} = 14 - \text{pOH} = 14 - 3.397 \approx 10.603 \]

### Final Answer:
The pH of the resulting solution is approximately **10.603**.

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