Calulate the normality of an `NaOH` solution, 21.5mL of which is required `0.240g` of `NaH_(2)PO_(4)` in a solution to monohydrogen phosphate.

Options

A. 1.093N
B. 0.093N
C. 0.048N
D. 0.93N

Answer

To calculate the normality of the NaOH solution, we will follow these steps:

### Step 1: Determine the Equivalent Weight of NaH₂PO₄
The formula for calculating the equivalent weight of an acid or base is:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{n}} \]
where \( n \) is the number of protons (H⁺ ions) that can be donated by the acid or the number of hydroxide ions (OH⁻) that can be accepted by the base.

For NaH₂PO₄, it can donate 1 H⁺ ion, so \( n = 1 \).

**Molar Mass of NaH₂PO₄**:
- Na: 22.99 g/mol
- H: 1.01 g/mol × 2 = 2.02 g/mol
- P: 30.97 g/mol
- O: 16.00 g/mol × 4 = 64.00 g/mol

Calculating the total:
\[ \text{Molar Mass} = 22.99 + 2.02 + 30.97 + 64.00 = 119.98 \text{ g/mol} \]
Thus, the equivalent weight of NaH₂PO₄ is:
\[ \text{Equivalent Weight} = \frac{119.98 \text{ g/mol}}{1} = 119.98 \text{ g/equiv} \]

### Step 2: Calculate the Number of Equivalents of NaH₂PO₄
Using the mass of NaH₂PO₄ given (0.240 g):
\[ \text{Number of Equivalents} = \frac{\text{Mass}}{\text{Equivalent Weight}} = \frac{0.240 \text{ g}}{119.98 \text{ g/equiv}} \approx 0.00200 \text{ equiv} \]

### Step 3: Calculate the Normality of NaOH
Normality (N) is defined as the number of equivalents of solute per liter of solution. Since we have the number of equivalents and the volume of the NaOH solution in mL, we can calculate the normality:
\[ \text{Normality (N)} = \frac{\text{Number of Equivalents}}{\text{Volume of Solution in Liters}} \]
First, convert 21.5 mL to liters:
\[ 21.5 \text{ mL} = 0.0215 \text{ L} \]
Now, substituting the values:
\[ \text{Normality (N)} = \frac{0.00200 \text{ equiv}}{0.0215 \text{ L}} \approx 0.0930 \text{ N} \]

### Final Answer
The normality of the NaOH solution is approximately **0.0930 N**.

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