Q.A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H2SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

Concept: Kjeldahl's method — In Kjeldahl's method, the nitrogen in the organic compound is converted to ammonia, which is absorbed in a known volume of standard acid. The amount of acid consumed by ammonia is found by back-titration with a standard base.

Step 1: Moles of acid taken initially

$M_1V_1 = 0.5 \times 50 = 25$ millimoles of $\text{H}_2\text{SO}_4$.

Step 2: Moles of base used for neutralisation

$M_2V_2 = 0.5 \times 60 = 30$ millimoles of NaOH.

Since 2 moles of NaOH neutralise 1 mole of $\text{H}_2\text{SO}_4$, the moles of residual acid are $30/2 = 15$ millimoles.

Step 3: Moles of acid consumed by ammonia

$25 - 15 = 10$ millimoles of $\text{H}_2\text{SO}_4$. …

Why this approach works

Kjeldahl's method is the classic way to estimate nitrogen in organic compounds. The organic sample is digested with concentrated sulphuric acid, which converts all the nitrogen into ammonium sulphate. The ammonia is then liberated by adding excess alkali and distilled into a known volume of standard acid. The key insight: the acid that remains unreacted (the "residual" acid) is titrated with standard alkali. The difference between the initial acid taken and the residual acid gives the amount of acid that actually reacted with the ammonia. From that, we can work backwards to find the mass of nitrogen originally present.

The trap most students fall into is forgetting that each mole of $H_2SO_4$ provides two moles of $H^+$ ions, so the neutralisation stoichiometry with $NH_3$ (which is a monoacidic base) is $2NH_3 : 1H_2SO_4$.

Let's work through the numbers step by step.

1. Find the total acid taken initially

   Volume of $H_2SO_4$ taken = 50 mL = 0.050 L

   Molarity of $H_2SO_4$ = 0.5 M

   Moles of $H_2SO_4$ initially = $M \times V$ = $0.5 \times 0.050$ = 0.025 moles

2. Find the residual acid after absorbing ammonia

   The residual acid was neutralised by 60 mL of 0.5 M $NaOH$.

   Moles of $NaOH$ used = $0.5 \times 0.060$ = 0.030 moles

   Since $H_2SO_4$ is diprotic, the neutralisation reaction is:

So 2 moles of $NaOH$ neutralise 1 mole of $H_2SO_4$.

Moles of $H_2SO_4$ that remained unreacted = $\frac{0.030}{2}$ = 0.015 moles

3. Find the moles of acid that reacted with ammonia

   Acid reacted = Initial acid − Residual acid

   = $0.025 - 0.015$ = 0.010 moles of $H_2SO_4$

4. Convert this to moles of ammonia (and hence nitrogen) …