Which one of these solution has the highest normality?

Options

A. 8 gr KOH per 100 ml
B. 0.5 M `H_(2)SO_(4)`
C. 6 gr of `NaOH` per 100 ml
D. 1 N `H_(3)PO_(4)`

Explanation

To determine which solution has the highest normality, we need to calculate the normality for each of the given solutions. Normality (N) is defined as the number of equivalents of solute per liter of solution. The formula for normality is:

\[ \text{Normality (N)} = \text{Molarity (M)} \times \text{Acidity or Basicity} \]

Where:
- Molarity (M) = \(\frac{\text{moles of solute}}{\text{volume of solution in liters}}\)
- Acidity or Basicity refers to the number of ionizable H+ ions for acids or OH- ions for bases.

Let's calculate the normality for each option step by step.

### Step 1: Calculate Normality for KOH Solution
**Given:** 8 g KOH in 100 ml

1. **Find the molecular weight of KOH:**
- K: 39 g/mol
- O: 16 g/mol
- H: 1 g/mol
- Molecular weight of KOH = 39 + 16 + 1 = 56 g/mol

2. **Calculate moles of KOH:**
\[ \text{Moles of KOH} = \frac{\text{Weight}}{\text{Molecular Weight}} = \frac{8 \text{ g}}{56 \text{ g/mol}} = 0.1429 \text{ moles} \]

3. **Convert volume from ml to liters:**
\[ 100 \text{ ml} = 0.1 \text{ L} \]

4. **Calculate Molarity:**
\[ \text{Molarity (M)} = \frac{0.1429 \text{ moles}}{0.1 \text{ L}} = 1.429 \text{ M} \]

5. **Determine Acidity (ACDT) for KOH:**
- KOH releases 1 OH- ion, so ACDT = 1.

6. **Calculate Normality:**
\[ \text{Normality} = 1.429 \text{ M} \times 1 = 1.429 \text{ N} \approx 1.4 \text{ N} \]

### Step 2: Calculate Normality for H2SO4 Solution
**Given:** 0.5 m H2SO4

1. **Molarity is given as 0.5 M.**
2. **Determine Basicity for H2SO4:**
- H2SO4 can donate 2 H+ ions, so Basicity = 2.

3. **Calculate Normality:**
\[ \text{Normality} = 0.5 \text{ M} \times 2 = 1 \text{ N} \]

### Step 3: Calculate Normality for NaOH Solution
**Given:** 6 g NaOH in 100 ml

1. **Find the molecular weight of NaOH:**
- Na: 23 g/mol
- O: 16 g/mol
- H: 1 g/mol
- Molecular weight of NaOH = 23 + 16 + 1 = 40 g/mol

2. **Calculate moles of NaOH:**
\[ \text{Moles of NaOH} = \frac{6 \text{ g}}{40 \text{ g/mol}} = 0.15 \text{ moles} \]

3. **Convert volume from ml to liters:**
\[ 100 \text{ ml} = 0.1 \text{ L} \]

4. **Calculate Molarity:**
\[ \text{Molarity (M)} = \frac{0.15 \text{ moles}}{0.1 \text{ L}} = 1.5 \text{ M} \]

5. **Determine Acidity (ACDT) for NaOH:**
- NaOH releases 1 OH- ion, so ACDT = 1.

6. **Calculate Normality:**
\[ \text{Normality} = 1.5 \text{ M} \times 1 = 1.5 \text{ N} \]

### Step 4: Given Normality for H3PO4 Solution
**Given:** 1 N H3PO4

### Step 5: Compare Normalities
- KOH: 1.4 N
- H2SO4: 1 N
- NaOH: 1.5 N
- H3PO4: 1 N

### Conclusion
The solution with the highest normality is the NaOH solution, which has a normality of **1.5 N**.