When 100 ml of `(M)/(10) H_(2)SO_(4)` is mixed with 500 ml of `(M)/(10) NaOH` then nature of resulting solution and normality of excess of reactant left is

Options

A. Acidic, `(N)/(5)`
B. Basic `(N)/(5)`
C. Basic `(N)/(20)`
D. Acidic `(N)/(10)`

Answer

To solve the problem of determining the nature of the resulting solution and the normality of the excess reactant when 100 ml of \( \frac{M}{10} H_2SO_4 \) is mixed with 500 ml of \( \frac{M}{10} NaOH \), we can follow these steps:

### Step 1: Calculate the Gram Equivalent of \( H_2SO_4 \)

1. **Identify the Normality and Volume**:
- Normality of \( H_2SO_4 = \frac{M}{10} = 0.1 \, M \)
- Volume of \( H_2SO_4 = 100 \, ml = 0.1 \, L \)

2. **Determine the n-factor**:
- For \( H_2SO_4 \), it can donate 2 protons (H⁺ ions), so the n-factor = 2.

3. **Calculate the Gram Equivalent**:
\[ \text{Gram Equivalent of } H_2SO_4 = \text{Normality} \times \text{Volume (L)} \times n \]
\[ = 0.1 \times 0.1 \times 2 = 0.02 \, \text{equivalents} \]

### Step 2: Calculate the Gram Equivalent of \( NaOH \)

1. **Identify the Normality and Volume**:
- Normality of \( NaOH = \frac{M}{10} = 0.1 \, M \)
- Volume of \( NaOH = 500 \, ml = 0.5 \, L \)

2. **Determine the n-factor**:
- For \( NaOH \), it can donate 1 proton (H⁺ ion), so the n-factor = 1.

3. **Calculate the Gram Equivalent**:
\[ \text{Gram Equivalent of } NaOH = \text{Normality} \times \text{Volume (L)} \times n \]
\[ = 0.1 \times 0.5 \times 1 = 0.05 \, \text{equivalents} \]

### Step 3: Determine the Nature of the Resulting Solution

1. **Compare the Gram Equivalents**:
- \( H_2SO_4 \) has 0.02 equivalents.
- \( NaOH \) has 0.05 equivalents.

2. **Determine the Excess Reactant**:
- Since \( NaOH \) has more equivalents than \( H_2SO_4 \), the solution will be basic.

### Step 4: Calculate the Normality of the Excess Reactant

1. **Calculate the Excess of \( NaOH \)**:
\[ \text{Excess} = \text{Equivalents of } NaOH - \text{Equivalents of } H_2SO_4 = 0.05 - 0.02 = 0.03 \, \text{equivalents} \]

2. **Calculate the Total Volume of the Solution**:
\[ \text{Total Volume} = 100 \, ml + 500 \, ml = 600 \, ml = 0.6 \, L \]

3. **Calculate the Normality of the Excess \( NaOH \)**:
\[ \text{Normality} = \frac{\text{Gram Equivalent of Excess}}{\text{Total Volume (L)}} \]
\[ = \frac{0.03}{0.6} = 0.05 \, N = \frac{1}{20} \, N \]

### Final Answer

- The nature of the resulting solution is **basic**.
- The normality of the excess reactant left is **\( \frac{1}{20} \, N \)**.