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, we need to determine the nature of the resulting solution when 100 ml of \( \frac{M}{10} \) \( H_2SO_4 \) is mixed with 500 ml of \( \frac{M}{10} \) \( NaOH \) and calculate the normality of the excess reactant left.

### Step 1: Calculate the gram-equivalent of \( H_2SO_4 \)

1. **Normality of \( H_2SO_4 \)**:
\[ \text{Normality} = \text{Molarity} \times \text{N factor} \]
For \( H_2SO_4 \), the molarity is \( \frac{1}{10} \) and the N factor is 2 (since it can donate 2 protons).

\[ \text{Normality of } H_2SO_4 = \frac{1}{10} \times 2 = \frac{2}{10} = \frac{1}{5} \, N \]

2. **Volume in liters**:
\[ \text{Volume} = 100 \, ml = \frac{100}{1000} = 0.1 \, L \]

3. **Gram-equivalent of \( H_2SO_4 \)**:
\[ \text{Gram-equivalent} = \text{Normality} \times \text{Volume in L} = \frac{1}{5} \times 0.1 = 0.02 \, \text{gram-equivalents} \]

### Step 2: Calculate the gram-equivalent of \( NaOH \)

1. **Normality of \( NaOH \)**:
The molarity is \( \frac{1}{10} \) and the N factor is 1 (since it can donate 1 hydroxide ion).

\[ \text{Normality of } NaOH = \frac{1}{10} \times 1 = \frac{1}{10} \, N \]

2. **Volume in liters**:
\[ \text{Volume} = 500 \, ml = \frac{500}{1000} = 0.5 \, L \]

3. **Gram-equivalent of \( NaOH \)**:
\[ \text{Gram-equivalent} = \text{Normality} \times \text{Volume in L} = \frac{1}{10} \times 0.5 = 0.05 \, \text{gram-equivalents} \]

### Step 3: Determine the limiting reactant and nature of the solution

- **Reacting equivalents**:
- \( H_2SO_4 \) provides 0.02 equivalents.
- \( NaOH \) provides 0.05 equivalents.

Since \( NaOH \) is in excess (0.05 > 0.02), the resulting solution will be basic.

### Step 4: Calculate the normality of the excess \( NaOH \)

1. **Excess \( NaOH \)**:
\[ \text{Excess} = 0.05 - 0.02 = 0.03 \, \text{gram-equivalents} \]

2. **Total volume of the solution**:
\[ \text{Total Volume} = 100 \, ml + 500 \, ml = 600 \, ml = \frac{600}{1000} = 0.6 \, L \]

3. **Normality of excess \( NaOH \)**:
\[ \text{Normality} = \frac{\text{Gram-equivalent}}{\text{Volume in L}} = \frac{0.03}{0.6} = 0.05 \, N \quad \text{or} \quad \frac{5}{100} = \frac{1}{20} \, N \]

### Conclusion

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