Question Stem (Q17 & Q18): Consider the following reaction sequence in which J, K, L and M are the major products.

Given: Atomic mass (in amu): H : 1, C : 12, N : 14, O : 16, S : 32, Br : 80, Ba : 137

The volume of 1 M aqueous $$\mathrm{H_2SO_4}$$ required to completely neutralize the ammonia evolved from 5.72 g of L in Kjeldahl's method of nitrogen estimation is ___ mL.

In the previous question (Q 17) we have already identified the structure of compound $$L$$. It contains exactly one nitrogen atom per molecule and its molar mass is $$286 \,\text{g mol}^{-1}$$.

Moles of $$L$$ taken in Kjeldahl’s estimation:
$$n_L=\frac{5.72\ \text{g}}{286\ \text{g mol}^{-1}}=0.020\ \text{mol}$$

Because each molecule of $$L$$ has a single nitrogen atom, the moles of nitrogen (and therefore of the ammonia that will be liberated) are also

$$n_{\mathrm N}=n_{\mathrm{NH_3}}=0.020\ \text{mol}$$

The neutralisation step in Kjeldahl’s method follows

$$2\mathrm{NH_3}+ \mathrm{H_2SO_4}\;\longrightarrow\;(\mathrm{NH_4})_2\mathrm{SO_4}$$

Thus, $$2$$ mol of $$\mathrm{NH_3}$$ are neutralised by $$1$$ mol of $$\mathrm{H_2SO_4}$$. Therefore, moles of $$\mathrm{H_2SO_4}$$ required are

$$n_{\mathrm{H_2SO_4}}=\frac{n_{\mathrm{NH_3}}}{2} =\frac{0.020}{2}=0.010\ \text{mol}$$

The acid supplied is $$1\ \text{M}$$, i.e. $$1\ \text{mol L}^{-1}$$. Volume needed:

$$V=\frac{n_{\mathrm{H_2SO_4}}}{C} =\frac{0.010\ \text{mol}}{1\ \text{mol L}^{-1}} =0.010\ \text{L}=10.0\ \text{mL}$$

Hence the volume of $$1\ \text{M}\;\mathrm{H_2SO_4}$$ required is 10.00 mL.