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 solve the problem step by step, we will follow these steps:

### Step 1: Calculate the Molarity of Solution A (NaOH)
1. **Find the number of moles of NaOH in Solution A:**
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \, \text{g}}{40 \, \text{g/mol}} = 0.1 \, \text{mol} \]

2. **Calculate the molarity of Solution A:**
\[ \text{Molarity (M)} = \frac{\text{number of moles}}{\text{volume in liters}} = \frac{0.1 \, \text{mol}}{100 \, \text{L}} = 10^{-3} \, \text{M} \]

### Step 2: Calculate the Molarity of Solution B (H2SO4)
1. **Find the number of moles of H2SO4 in Solution B:**
\[ \text{Number of moles} = \frac{9.8 \, \text{g}}{98 \, \text{g/mol}} = 0.1 \, \text{mol} \]

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

### Step 3: Calculate Normality of Solutions
1. **Normality of Solution A (NaOH):**
- Since NaOH is a strong base, its normality is equal to its molarity:
\[ N_A = M_A = 10^{-3} \, \text{N} \]

2. **Normality of Solution B (H2SO4):**
- H2SO4 is a diprotic acid, so its normality is twice its molarity:
\[ N_B = 2 \times M_B = 2 \times 10^{-3} \, \text{N} = 2 \times 10^{-3} \, \text{N} \]

### Step 4: Calculate the Total Normality after Mixing
1. **Calculate the total normality using the formula:**
\[ N_{total} = \frac{N_A \cdot V_A + N_B \cdot V_B}{V_A + V_B} \]
where \( V_A = 40 \, \text{L} \) and \( V_B = 10 \, \text{L} \):
\[ N_{total} = \frac{(10^{-3} \, \text{N}) \cdot 40 \, \text{L} + (2 \times 10^{-3} \, \text{N}) \cdot 10 \, \text{L}}{40 + 10} \]
\[ N_{total} = \frac{(0.04 + 0.02)}{50} = \frac{0.06}{50} = 0.0012 \, \text{N} \]

### Step 5: Calculate the Concentration of OH⁻ Ions
Since NaOH is a base, we need to find the concentration of OH⁻ ions:
\[ \text{Concentration of OH}^- = N_{total} = 0.0012 \, \text{N} \]

### Step 6: Calculate pOH
1. **Calculate pOH:**
\[ pOH = -\log[\text{OH}^-] = -\log(0.0012) \approx 2.92 \]

### Step 7: Calculate pH
1. **Use the relationship between pH and pOH:**
\[ pH = 14 - pOH = 14 - 2.92 = 11.08 \]

### Final Answer
The pH of the mixed solution is approximately **11.08**.

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