The normility of a solution that results from mixing `4g` of `NaOH, 500 mL` of `1 M HCl`, and `10.0mL` of `H_(2)O_(4)` (specific gravity 1.149% `H_(2) SO_(4)` by weight) is
The total volume of solution was made to `1 L` with water)

Options

A. 0.51
B. 0.71
C. 1.02
D. 0.45

Explanation

To find the normality of the solution resulting from mixing 4 g of NaOH, 500 mL of 1 M HCl, and 10 mL of H2SO4 (with specific gravity 1.149 and 49% by weight), we will follow these steps:

### Step 1: Calculate the moles of NaOH
- The molar mass of NaOH is approximately 40 g/mol.
- Moles of NaOH = mass (g) / molar mass (g/mol) = 4 g / 40 g/mol = 0.1 mol.

### Step 2: Convert moles of NaOH to milliequivalents
- Since NaOH is a strong base and provides one hydroxide ion (OH⁻) per molecule, the milliequivalents of NaOH = moles × 1000 = 0.1 mol × 1000 = 100 milliequivalents.

### Step 3: Calculate the milliequivalents of HCl
- The volume of HCl solution is 500 mL and its concentration is 1 M.
- Milliequivalents of HCl = volume (L) × normality (eq/L) = 0.5 L × 1 eq/L = 500 milliequivalents.

### Step 4: Calculate the normality of H2SO4
- Specific gravity = 1.149, and the solution is 49% H2SO4 by weight.
- Density of the solution = 1.149 g/mL.
- The mass of 10 mL of the solution = 10 mL × 1.149 g/mL = 11.49 g.
- Mass of H2SO4 in 10 mL = 49% of 11.49 g = 0.49 × 11.49 g = 5.63 g.
- Molar mass of H2SO4 = 98 g/mol, so moles of H2SO4 = 5.63 g / 98 g/mol = 0.0574 mol.
- H2SO4 is a diprotic acid, so it provides 2 equivalents per mole. Therefore, milliequivalents of H2SO4 = 0.0574 mol × 2 × 1000 = 114.8 milliequivalents.

### Step 5: Calculate total milliequivalents of acid
- Total milliequivalents of acid = milliequivalents of HCl + milliequivalents of H2SO4 = 500 + 114.8 = 614.8 milliequivalents.

### Step 6: Calculate the net milliequivalents of acid after reaction with NaOH
- Since NaOH will neutralize some of the acid, we subtract the milliequivalents of NaOH from the total acid:
- Net milliequivalents of acid = total acid - milliequivalents of NaOH = 614.8 - 100 = 514.8 milliequivalents.

### Step 7: Calculate the normality of the final solution
- The total volume of the solution is 1 L (or 1000 mL).
- Normality = milliequivalents / volume (L) = 514.8 milliequivalents / 1000 mL = 0.5148 N.

### Final Answer
The normality of the solution is approximately **0.515 N**.

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