$0.4 g$ of an organic compound was treated according to Kjeldal's method. The ammonia evolved was absorbed in $50 m L$ of $0.5 M H_{3} P O_{3}$ . The residual acid required
(30~mL\) of $0.5 M C a (O H)_{2}$ . What is the percentage of $N_{2}$ in the compound?

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Solution

Here, $H_{3} P O_{3}$ is a dibasic (i.e. n = 2).
The normality of $H_{3} P O_{3} = M \times n = 0.5 \times 2 = 1 N$
$mEq. of H_{3} P O_{3} = 50 \times 0.5 \times 2 = 50$
$mEq. of C a (O H)_{2} = 30 \times 0.5 \times 2 = 30$
The acid used to neutralise $N H_{3} = 50 - 30 = 20$
Percentage of $N_{2} = \frac{1.4 \times mEq. of acid used to neutralise N H_{3}}{Weight of compound} \frac{1.4 \times 0.2}{0.4} = 70 %$

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