The Kjeldahl analysis in Section 11-8 is used to measure the nitrogen content of organic compounds, which are digested in boiling sulfuric acid to decompose to ammonia, which, in turn, is distilled into standard acid. The remaining acid is then back-titrated with base. Kjeldahl himself had difficulty in 1880 discerning by lamplight the methyl red indicator end point in the back titration. He could have refrained from working at night, but instead he chose to complete the analysis differently. After distilling the ammonia into standard sulfuric acid, he added a mixture of ${\mathrm{KIO}}_3$and KI to the acid. The liberated iodine was then titrated with thiosulfate, using starch for easy end-point detection even by lamplight. Explain how the thiosulfate titration is related to the nitrogen content of the unknown. Derive a relationship between moles of${\mathrm{NH}}_3$ liberated in the digestion and moles of thiosulfate required for titration of iodine.

• The mole concept is a simple way to express the amount of a substance. Any measurement is divided into two parts: the numerical magnitude and the units in which the magnitude is expressed. For example, if the mass of a ball is 2 kilogrammes, the magnitude is '2' and the unit is 'kilogramme.'
• When dealing with particles at the atomic (or molecular) level, even one gramme of a pure element is known to contain a large number of atoms. This is a common application of the mole concept. It primarily focuses on the unit known as a'mole,' which is a count of a very large number of particles.

In Kjeldahl analysis, $1{\mathrm{molNH}}_3$of is equal to$1\mathrm{molofH}$:

$\mathrm{n}\left({\mathrm{NH}}_3\right)=\mathrm{n}\left({\mathrm{H}}^{+}\right)$

Since there are 2 moles of ${\mathrm{H}}^{+}$in${\mathrm{H}}_2{\mathrm{SO}}_4$they moles are doubled:

$\mathrm{n}\left({\mathrm{NH}}_3\right)=\mathrm{n}\left({\mathrm{H}}^{+}\right)$

After the initial role="math" localid="1654863625697" ${\mathrm{H}}_2{\mathrm{SO}}_4$reacts with ${\mathrm{NH}}_3$ the excess ${\mathrm{H}}_2{\mathrm{SO}}_4$reacts with KI and KI03 and it is titrated with thiosulfate.To determine the ratio of moles of ${\mathrm{H}}_2{\mathrm{SO}}_4$and thiosulfate we have to use reactions 16-18 and 16-19:

$$\begin{gathered} I{O}_3^{-}+6H^{+}+8I^{-}\rightleftharpoons 3I_3^{-}+3H_{2}O\left(1\right) \\ {I}_3^{-}+2S_2{O_3}^{2-}\rightleftharpoons S_4{O_6}^{2-}+3I^{-}\left(2\right) \end{gathered}$$

By using reaction (1) we can determine the ratio of moles of sulfuric acid and triiodide:

$$\begin{gathered} 6\mathrm{n}\left({\mathrm{H}}^{+}\right)=3\mathrm{n}(\mathrm{excess}{\mathrm{H}}_2{\mathrm{SO}}_4)=3\mathrm{n}\left({\mathrm{l}}_3^{-}\right)\mathrm{l}:3 \\ 6\mathrm{n}\left({\mathrm{H}}^{+}\right)=\mathrm{n}(\mathrm{excess}{\mathrm{H}}_2{\mathrm{SO}}_4)=\mathrm{n}\left({\mathrm{l}}_3^{-}\right) \end{gathered}$$

By using reaction (2) we can determine the ratio of moles of triiodide and thiosulfate:

$\mathrm{n}\left({\mathrm{l}}_3^{-}\right)=\frac{1}{2}\mathrm{n}\left({\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}\right)$

Therefore, the ratio of moles of sulfuric acid and thiosulfate are:

role="math" localid="1654864150133" $\mathrm{n}(\mathrm{excess}{\mathrm{H}}_2{\mathrm{SO}}_4)-\frac{1}{2}\mathrm{n}\left({\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}\right)$

Moles of${\mathrm{NH}}_3$are:

$$\begin{gathered} \mathrm{n}({\mathrm{NH}}_3)=2\mathrm{n}\left({\mathrm{H}}_2{\mathrm{SO}}_4\right) \\ \mathrm{n}\left({\mathrm{NH}}_3\right)=2(\mathrm{n}(\mathrm{excess}{\mathrm{H}}_2{\mathrm{SO}}_4)-\mathrm{n}(\mathrm{excess}{\mathrm{H}}_2{\mathrm{SO}}_4\left)\right) \end{gathered}$$

By replacing excess moles of sulfuric acid with moles of thiosulfate we get:

$n\left(N{H}_3\right)=2\left(\begin{array}{c}n(initial{H}_{2}S{O}_4)-\frac{1}{2}n\left(S_2{O_3}^{2-}\right)\end{array}\right)$

Hence the relationship between moles of ${\mathrm{NH}}_3$liberated in the digestion and moles of thiosulfate required for titration of iodine is derived. And how thiosulfate titration is related to the nitrogen content of the unknown is explained.