100 ml of 0.1 N NaOH is mixed with 50 ml of 0.1 N `H_2SO_4` . The pH of the resulting solution is

Options

A. `lt 7`
B. `gt 7`
C. `=7`
D. Cannot be predicted

Answer

To solve the problem of finding the pH of the resulting solution when 100 ml of 0.1 N NaOH is mixed with 50 ml of 0.1 N H2SO4, we can follow these steps:

### Step 1: Determine the number of moles of NaOH and H2SO4
- **For NaOH:**
- Normality (N) = 0.1 N
- Volume (V) = 100 ml = 0.1 L
- Moles of NaOH = Normality × Volume = 0.1 N × 0.1 L = 0.01 moles = 10 millimoles

- **For H2SO4:**
- Normality (N) = 0.1 N
- Volume (V) = 50 ml = 0.05 L
- The N factor for H2SO4 is 2 (since it can donate 2 H+ ions).
- Molarity = Normality / N factor = 0.1 N / 2 = 0.05 M
- Moles of H2SO4 = Molarity × Volume = 0.05 M × 0.05 L = 0.0025 moles = 2.5 millimoles

### Step 2: Write the balanced chemical equation for the reaction
The reaction between NaOH and H2SO4 is as follows:
\[ \text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]

### Step 3: Determine the limiting reagent
From the balanced equation, we see that 1 mole of H2SO4 reacts with 2 moles of NaOH.

- H2SO4: 2.5 millimoles will require 2.5 × 2 = 5 millimoles of NaOH.
- NaOH available: 10 millimoles.

Since we have enough NaOH to react with H2SO4, H2SO4 is the limiting reagent.

### Step 4: Calculate the remaining moles after the reaction
- Moles of NaOH that will react with 2.5 millimoles of H2SO4 = 5 millimoles.
- Remaining moles of NaOH = Initial moles of NaOH - Moles of NaOH that reacted
\[ \text{Remaining NaOH} = 10 \text{ millimoles} - 5 \text{ millimoles} = 5 \text{ millimoles} \]

### Step 5: Determine the pH of the resulting solution
Since there are still 5 millimoles of NaOH left in the solution, the solution will be basic.

- To find the concentration of NaOH in the total volume of the solution:
- Total volume = 100 ml + 50 ml = 150 ml = 0.15 L
- Concentration of NaOH = Remaining moles / Total volume = 5 millimoles / 0.15 L = 0.0333 M

### Step 6: Calculate the pOH and then the pH
- pOH = -log[OH⁻] = -log(0.0333) ≈ 1.47
- pH + pOH = 14
- pH = 14 - pOH = 14 - 1.47 ≈ 12.53

### Conclusion
The pH of the resulting solution is greater than 7 due to the excess NaOH present.

### Final Answer
The pH of the resulting solution is greater than 7.

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