IChO 1994 · Theory problem 2

Kjeldahl determination of nitrogen

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Nitrogen in agricultural materials is often determined by the Kjeldahl method. The method involves a treatment of the sample with hot concentrated sulphuric acid, to convert organically bound nitrogen to ammonium ion. Concentrated sodium hydroxide is then added, and the ammonia formed is distilled into hydrochloric acid of known volume and concentration. The excess hydrochloric acid is then back-titrated with a standard solution of sodium hydroxide, to determine nitrogen in the sample.

2.1 0.2515 g of a grain sample was treated with sulphuric acid, sodium hydroxide was then added and the ammonia distilled into \(50.00\ \mathrm{cm^3}\) of 0.1010 M hydrochloric acid. The excess acid was back-titrated with \(19.30\ \mathrm{cm^3}\) of 0.1050 M sodium hydroxide.

Calculate the concentration of nitrogen in the sample, in percent by mass.

2.2 Calculate the pH of the solution which is titrated in 2.1 when \(0\ \mathrm{cm^3}\), \(9.65\ \mathrm{cm^3}\), \(19.30\ \mathrm{cm^3}\) and \(28.95\ \mathrm{cm^3}\) of sodium hydroxide have been added. Disregard any volume change during the reaction of ammonia gas with hydrochloric acid. \(K_a\) for ammonium ion is \(5.7\times10^{-10}\).

2.3 Draw the titration curve based on the calculations in b).

2.4 What is the pH transition range of the indicator which could be used for the back titration.

2.5 The Kjeldahl method can also be used to determine the molecular weight of amino acids. In a given experiment, the molecular weight of a naturally occurring amino acid was determined by digesting 0.2345 g of the pure acid and distilling ammonia released into \(50.00\ \mathrm{cm^3}\) of 0.1010 M hydrochloric acid. A titration volume of \(17.50\ \mathrm{cm^3}\) was obtained for the back titration with 0.1050 M sodium hydroxide.

Calculate the molecular weight of the amino acid based on one and two nitrogen groups in the molecule, respectively.

Official English text of the problem.

Answers

  • 2.1: \(\boxed{16.84\%}\) nitrogen by mass.
  • 2.2: At added sodium hydroxide volumes \(0,\ 9.65,\ 19.30,\ 28.95\ \mathrm{cm^3}\), respectively: \[ \boxed{\mathrm{pH}=1.39,\quad1.77,\quad5.30,\quad8.95.} \]
  • 2.3: A smooth, increasing pH–volume curve through these four points, rising sharply around \(19.30\ \mathrm{cm^3}\), then becoming less steep in the ammonium–ammonia buffer region.
  • 2.4: A transition interval approximately \(\boxed{\mathrm{pH}\ 4\text{–}6}\), encompassing the endpoint at pH \(5.30\), is suitable.
  • 2.5: The molar masses are \(\boxed{73.00\ \mathrm{g\,mol^{-1}}}\) for one nitrogen atom per molecule and \(\boxed{146.0\ \mathrm{g\,mol^{-1}}}\) for two. The corresponding relative molecular masses are \(73.00\) and \(146.0\).

Solution

2.1 — Nitrogen content

The species involved in the ionic reactions are ammonium ion, \(\ce{NH4+}\) ([NH4+]); ammonia, \(\ce{NH3}\) (N); hydroxide ion, \(\ce{OH-}\) ([OH-]); water, \(\ce{H2O}\) (O); and aqueous hydrogen ion, \(\ce{H+}\) ([H+], representing a hydrated proton).

Digestion converts the nitrogen into ammonium ion. Addition of hydroxide releases ammonia:

\[ \ce{NH4+ + OH- -> NH3 + H2O}. \]

The distilled ammonia consumes acid in a \(1:1\) molar ratio:

\[ \ce{NH3 + H+ -> NH4+}. \]

The remaining acid is determined by neutralization:

\[ \ce{H+ + OH- -> H2O}. \]

The initial amount of acid is

\[ n_{\text{acid,initial}} =(0.05000\ \mathrm{dm^3})(0.1010\ \mathrm{mol\,dm^{-3}}) =5.050\times10^{-3}\ \mathrm{mol}. \]

The amount left after ammonia absorption equals the amount of back-titrant:

\[ n_{\text{acid,remaining}} =(0.01930\ \mathrm{dm^3})(0.1050\ \mathrm{mol\,dm^{-3}}) =2.0265\times10^{-3}\ \mathrm{mol}. \]

Thus, retaining guard digits,

\[ n_{\mathrm N}=n_{\text{ammonia}} =(5.050-2.0265)\times10^{-3} =3.0235\times10^{-3}\ \mathrm{mol}. \]

Using the nitrogen atomic molar mass \(14.01\ \mathrm{g\,mol^{-1}}\),

\[ m_{\mathrm N} =(3.0235\times10^{-3})(14.01) =0.0423592\ \mathrm g. \]

Therefore,

\[ \boxed{ w_{\mathrm N} =\frac{0.0423592}{0.2515}\times100\% =16.84\%. } \]

2.2 — pH during the back titration

Before titration, the \(50.00\ \mathrm{cm^3}\) solution contains:

\[ n_{\text{excess strong acid}}=2.0265\ \mathrm{mmol}, \qquad n_{\text{ammonium}}=3.0235\ \mathrm{mmol}. \]

The strong acid dominates the pH before its neutralization. At \(19.30\ \mathrm{cm^3}\), ammonium determines the pH. Subsequent hydroxide converts some ammonium into ammonia, forming a buffer.

At \(0\ \mathrm{cm^3}\)

Ammonium dissociation is negligible in the presence of excess strong acid:

\[ [\ce{H+}] =\frac{2.0265\times10^{-3}\ \mathrm{mol}} {0.05000\ \mathrm{dm^3}} =0.04053\ \mathrm{mol\,dm^{-3}}. \]

Hence,

\[ \boxed{\mathrm{pH}=-\log_{10}(0.04053)=1.39.} \]

At \(9.65\ \mathrm{cm^3}\)

The added hydroxide neutralizes

\[ n_{\text{hydroxide}} =(0.00965)(0.1050) =1.01325\times10^{-3}\ \mathrm{mol}. \]

The remaining strong acid is therefore also \(1.01325\times10^{-3}\ \mathrm{mol}\). Including dilution,

\[ [\ce{H+}] =\frac{1.01325\times10^{-3}} {0.05000+0.00965} =0.0169866\ \mathrm{mol\,dm^{-3}}, \]

so

\[ \boxed{\mathrm{pH}=1.77.} \]

At \(19.30\ \mathrm{cm^3}\)

The excess strong acid has been neutralized. Ammonium acts as a weak acid:

\[ \ce{NH4+ <=> NH3 + H+}. \]

Its analytical concentration is

\[ C=\frac{3.0235\times10^{-3}} {0.05000+0.01930} =0.0436291\ \mathrm{mol\,dm^{-3}}. \]

Writing \(x=[\ce{H+}]\), the weak-acid equilibrium gives

\[ K_a=\frac{x^2}{C-x}. \]

Because dissociation is small, \(C-x\approx C\), and therefore

\[ [\ce{H+}] \approx\sqrt{K_aC} =\sqrt{(5.7\times10^{-10})(0.0436291)} =4.99\times10^{-6}\ \mathrm{mol\,dm^{-3}}. \]

Thus,

\[ \boxed{\mathrm{pH}=5.30.} \]

The approximation is justified because \(x/C\approx1.1\times10^{-4}\).

At \(28.95\ \mathrm{cm^3}\)

The hydroxide added beyond strong-acid neutralization reacts with ammonium:

\[ \ce{NH4+ + OH- -> NH3 + H2O}. \]

The amount reacting in this stage is

\[ n_{\text{hydroxide,additional}} =(0.02895-0.01930)(0.1050) =1.01325\times10^{-3}\ \mathrm{mol}. \]

Consequently,

\[ n_{\text{ammonia}}=1.01325\ \mathrm{mmol}, \]

\[ n_{\text{ammonium}} =3.02350-1.01325 =2.01025\ \mathrm{mmol}. \]

Rearranging the ammonium acid-dissociation expression,

\[ \mathrm{pH} =\mathrm{p}K_a+ \log_{10}\frac{[\ce{NH3}]}{[\ce{NH4+}]}. \]

Both species occupy the same volume, so their concentration ratio equals their mole ratio:

\[ \begin{aligned} \mathrm{pH} &=-\log_{10}(5.7\times10^{-10}) +\log_{10}\left(\frac{1.01325}{2.01025}\right)\\ &=9.2441-0.2975. \end{aligned} \]

Therefore,

\[ \boxed{\mathrm{pH}=8.95.} \]

2.3 — Description of the titration curve

Draw the horizontal axis as volume of 0.1050 M sodium hydroxide added, from \(0\) to approximately \(30\ \mathrm{cm^3}\), and the vertical axis as pH, from approximately \(0\) to \(10\).

Mark these points:

Added volume / \(\mathrm{cm^3}\) pH
0.00 1.39
9.65 1.77
19.30 5.30
28.95 8.95

Draw a smooth curve that:

  1. Begins at pH \(1.39\) and rises gradually through \((9.65,\ 1.77)\).
  2. Becomes increasingly steep as the excess strong acid approaches exhaustion.
  3. Passes through the steep endpoint region at \((19.30,\ 5.30)\).
  4. Bends toward a gentler rise beyond the endpoint as the ammonium–ammonia buffer forms, reaching \((28.95,\ 8.95)\).

Label \(19.30\ \mathrm{cm^3}\) as the excess strong-acid equivalence volume.

2.4 — Indicator transition range

The indicator should change colour within the steep part of the curve surrounding the equivalence pH of \(5.30\). A transition interval approximately

\[ \boxed{\mathrm{pH}\ 4\text{–}6} \]

is suitable. This detects neutralization of the excess strong acid before an appreciable fraction of the ammonium has been converted into ammonia.

2.5 — Molecular weight of the amino acid

The amount of acid initially present is again

\[ n_{\text{acid,initial}}=5.050\times10^{-3}\ \mathrm{mol}. \]

The back titration measures the remaining acid:

\[ n_{\text{acid,remaining}} =(0.01750)(0.1050) =1.8375\times10^{-3}\ \mathrm{mol}. \]

Hence the amino-acid sample produces

\[ n_{\text{ammonia}} =(5.050-1.8375)\times10^{-3} =3.2125\times10^{-3}\ \mathrm{mol}. \]

If each amino-acid molecule contains \(z\) nitrogen atoms, then

\[ n_{\text{amino acid}}=\frac{n_{\text{ammonia}}}{z}, \]

so its molar mass is

\[ M=\frac{m}{n_{\text{amino acid}}} =\frac{zm}{n_{\text{ammonia}}}. \]

For one nitrogen atom per molecule,

\[ \boxed{ M=\frac{0.2345\ \mathrm g} {3.2125\times10^{-3}\ \mathrm{mol}} =73.00\ \mathrm{g\,mol^{-1}}. } \]

For two nitrogen atoms per molecule,

\[ \boxed{ M=\frac{2(0.2345\ \mathrm g)} {3.2125\times10^{-3}\ \mathrm{mol}} =146.0\ \mathrm{g\,mol^{-1}}. } \]

The corresponding dimensionless relative molecular masses are \(73.00\) and \(146.0\), respectively.

The solution was checked against the official solution.