100 mL of 1N `H_2SO_4` is mixed with 100 mL of 1M NaOH solution. The resulting solution will be

Options

A. highly acidic
B. neutral
C. highly basic
D. slightly acidic

Answer

To determine the resulting solution when 100 mL of 1N \( H_2SO_4 \) is mixed with 100 mL of 1M NaOH, we can follow these steps:

### Step 1: Understand the Given Solutions
- We have 100 mL of 1N \( H_2SO_4 \).
- We have 100 mL of 1M NaOH.

### Step 2: Calculate the Normality and Molarity of \( H_2SO_4 \)
- The normality (N) of \( H_2SO_4 \) is given as 1N.
- The n-factor for \( H_2SO_4 \) is 2, because it can donate 2 protons (H\(^+\)).
- To find the molarity (M) from normality, we use the formula:
\[ \text{Normality} = \text{Molarity} \times \text{n-factor} \]
Thus,
\[ M = \frac{N}{\text{n-factor}} = \frac{1N}{2} = 0.5M \]

### Step 3: Calculate the Number of Moles of \( H_2SO_4 \)
- Volume of \( H_2SO_4 \) = 100 mL = 0.1 L.
- Molarity of \( H_2SO_4 \) = 0.5 M.
- Number of moles of \( H_2SO_4 \):
\[ \text{Moles} = \text{Molarity} \times \text{Volume} = 0.5 \, \text{mol/L} \times 0.1 \, \text{L} = 0.05 \, \text{moles} \]

### Step 4: Calculate the Number of Moles of \( NaOH \)
- Volume of \( NaOH \) = 100 mL = 0.1 L.
- Molarity of \( NaOH \) = 1 M.
- Number of moles of \( NaOH \):
\[ \text{Moles} = \text{Molarity} \times \text{Volume} = 1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.1 \, \text{moles} \]

### Step 5: Determine the Reaction
The reaction between \( H_2SO_4 \) and \( NaOH \) can be represented as:
\[ H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O \]
From the stoichiometry of the reaction:
- 1 mole of \( H_2SO_4 \) reacts with 2 moles of \( NaOH \).

### Step 6: Calculate the Limiting Reactant
- We have 0.05 moles of \( H_2SO_4 \) which will require:
\[ 0.05 \, \text{moles of } H_2SO_4 \times 2 = 0.1 \, \text{moles of } NaOH \]
- We have exactly 0.1 moles of \( NaOH \), which means both reactants will completely react.

### Step 7: Determine the Resulting Solution
- After the reaction, all \( H_2SO_4 \) and \( NaOH \) will be consumed.
- Since they react in a 1:2 ratio and we have equal moles of \( H_2SO_4 \) and \( NaOH \) required, the resulting solution will be neutral.

### Conclusion
The resulting solution will be neutral.

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