Question

A quantity of 20 g of H₃PO₄ is dissolved in water and made up to 1 L. What is the normality of the solution, if titration against NaOH is carried only up to the second stage of neutralization?

• A. 0.408
• B. 0.204
• C. 0.612
• D. 0.102

Answer 1

Answer: (A)

0.408

Answer 2

The molar mass of $H_3PO_4$ is calculated as $(3 \times 1.008) + 30.974 + (4 \times 15.999) \approx 98.074$ g/mol.

When $H_3PO_4$ is titrated up to the second stage of neutralization with NaOH, it reacts as follows: $H_3PO_4 + 2NaOH \rightarrow Na_2HPO_4 + 2H_2O$

In this reaction, $H_3PO_4$ loses two protons, so its n-factor is 2.

The equivalent weight of $H_3PO_4$ is calculated by: Equivalent Weight = $\frac{\text{Molar Mass}}{\text{n-factor}} = \frac{98.074 \, \text{g/mol}}{2} = 49.037 \, \text{g/equivalent}$.

The normality (N) of the solution is given by: Normality = $\frac{\text{Mass of solute}}{\text{Equivalent weight of solute} \times \text{Volume of solution in liters}}$

Given: Mass of $H_3PO_4$ = 20 g Volume of solution = 1 L

Normality = $\frac{20 \, \text{g}}{49.037 \, \text{g/equivalent} \times 1 \, \text{L}}$ Normality $\approx 0.40787 \, \text{N}$

Rounding to three decimal places, the normality is approximately 0.408 N.