Two solutions, A and B, each of 100 L was made by dissolving 4g of NaOH and 9.8 g of H_(2) SO_(4)` in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is_______.

Explanation

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

### Step 1: Calculate the number of moles of NaOH and H₂SO₄ in their respective solutions.

**For NaOH:**
- Given mass = 4 g
- Molar mass of NaOH = 40 g/mol
- Number of moles of NaOH = mass / molar mass = 4 g / 40 g/mol = 0.1 moles

**For H₂SO₄:**
- Given mass = 9.8 g
- Molar mass of H₂SO₄ = 98 g/mol
- Number of moles of H₂SO₄ = mass / molar mass = 9.8 g / 98 g/mol = 0.1 moles

### Step 2: Calculate the molarity of solutions A and B.

**For Solution A (NaOH):**
- Volume = 100 L
- Molarity of NaOH = moles / volume = 0.1 moles / 100 L = 0.001 M or 10⁻³ M

**For Solution B (H₂SO₄):**
- Volume = 100 L
- Molarity of H₂SO₄ = moles / volume = 0.1 moles / 100 L = 0.001 M or 10⁻³ M

### Step 3: Calculate the number of moles of NaOH and H₂SO₄ in the mixed solutions.

**For 40 L of Solution A (NaOH):**
- Moles of NaOH = Molarity × Volume = 0.001 M × 40 L = 0.04 moles

**For 10 L of Solution B (H₂SO₄):**
- Moles of H₂SO₄ = Molarity × Volume = 0.001 M × 10 L = 0.01 moles

### Step 4: Determine the reaction between NaOH and H₂SO₄.

The balanced reaction is:
\[ \text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]

From the reaction:
- 1 mole of H₂SO₄ reacts with 2 moles of NaOH.

### Step 5: Calculate the limiting reagent and remaining moles after reaction.

- Moles of NaOH needed for 0.01 moles of H₂SO₄ = 0.01 moles × 2 = 0.02 moles
- Moles of NaOH available = 0.04 moles

Since we have excess NaOH:
- Moles of NaOH left = 0.04 moles - 0.02 moles = 0.02 moles
- Moles of H₂SO₄ left = 0.01 moles - 0.01 moles = 0 moles

### Step 6: Calculate the concentration of OH⁻ ions in the resultant solution.

Total volume after mixing = 40 L + 10 L = 50 L

Concentration of OH⁻ ions:
- Moles of OH⁻ from NaOH = 0.02 moles (since 1 mole of NaOH gives 1 mole of OH⁻)
- Concentration of OH⁻ = moles / volume = 0.02 moles / 50 L = 0.0004 M or 4 × 10⁻⁴ M

### Step 7: Calculate the concentration of H⁺ ions using the ion product of water.

At 25°C, \( K_w = [H^+][OH^-] = 10^{-14} \)

Given:
- \([OH^-] = 4 \times 10^{-4} M\)

Calculate \([H^+]\):
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{4 \times 10^{-4}} = 2.5 \times 10^{-11} M \]

### Step 8: Calculate the pH of the resultant solution.

\[ \text{pH} = -\log[H^+] = -\log(2.5 \times 10^{-11}) \]

Calculating this gives:
\[ \text{pH} \approx 10.6 \]

### Final Answer:
The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution B is approximately **10.6**.

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