In Kjeldahl's method for the estimation of nitrogen present in an agricultural soil sample, the ammo — Qualitative and Quantitative Analysis Chemistry Question
Question
In Kjeldahl's method for the estimation of nitrogen present in an agricultural soil sample, the ammonia evolved from exactly $0.75 \text{ g}$ of the sample completely neutralized $10 \text{ mL}$ of a $1.0 \text{ M}$ $H_2SO_4$ solution (no back titration required). What is the percentage of nitrogen in the soil sample? (Calculate to two decimal places).
💡 Solution & Explanation
Milli-equivalents of $H_2SO_4$ used = $\text{Volume(mL)} \times \text{Molarity} \times \text{Basicity} = 10 \text{ mL} \times 1.0 \text{ M} \times 2 = 20 \text{ meq}$. Since $H_2SO_4$ was completely neutralized, milli-equivalents of $NH_3$ = 20 meq. This means $20 \text{ mmol}$ of Nitrogen atoms are present. Mass of Nitrogen = $20 \times 10^{-3} \text{ mol} \times 14 \text{ g/mol} = 0.28 \text{ g}$. Percentage of $N = (\frac{0.28}{0.75}) \times 100 = 37.33\%$.