IChO 1970 · Theory problem 2

Water of crystallisation in two samples of crystalline soda

A sample of crystalline soda (A) with a mass of 1.287 g was allowed to react with an excess of hydrochloric acid and 100.8 cm³ of a gas was liberated (measured at STP).

Another sample of different crystalline soda (B) with a mass of 0.715 g was decomposed by 50 cm³ of 0.2 N sulphuric acid.

After total decomposition of soda, the excess of the sulphuric acid was neutralized which required 50 cm³ of 0.1 N sodium hydroxide solution (by titration on methyl orange indicator).

Problems:

2.1 How many molecules of water in relation to one molecule of \(\ce{Na2CO3}\) are contained in the first sample of soda?

2.2 Have both samples of soda the same composition?

Relative atomic masses: \(A_{\mathrm r}(\ce{Na}) = 23;\quad A_{\mathrm r}(\ce{H}) = 1;\quad A_{\mathrm r}(\ce{C}) = 12;\quad A_{\mathrm r}(\ce{O}) = 16.\)

Official English text of the problem.

Answers

  • 2.1 Sample A contains 10 molecules of water (SMILES O) per formula unit of \(\ce{Na2CO3}\). It is sodium carbonate decahydrate, \(\ce{Na2CO3.10H2O}\); component SMILES: [Na+].[Na+].[O-]C(=O)[O-].O.O.O.O.O.O.O.O.O.O.
  • 2.2 Yes. Both samples have the composition \(\ce{Na2CO3.10H2O}\).

The hydrate SMILES specifies its constituent ions and water molecules; it does not represent their arrangement in the crystal lattice.

Solution

2.1

Write the composition of sample A as \(\ce{Na2CO3.xH2O}\). The given atomic masses yield

\[ M(\ce{Na2CO3})=2(23)+12+3(16)=106\ \mathrm{g\,mol^{-1}}, \] \[ M(\ce{H2O})=2(1)+16=18\ \mathrm{g\,mol^{-1}}. \]

Thus the hydrate has molar mass

\[ M_A=(106+18x)\ \mathrm{g\,mol^{-1}}. \]

The carbonate component reacts with hydrochloric acid according to

\[ \ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O}. \]

The water of crystallisation enters the solution. The liberated gas is carbon dioxide, and one mole of hydrate produces one mole of \(\ce{CO2}\).

Using the conventional STP molar gas volume, \(22.4\ \mathrm{dm^3\,mol^{-1}}\),

\[ n(\ce{CO2}) =\frac{0.1008\ \mathrm{dm^3}} {22.4\ \mathrm{dm^3\,mol^{-1}}} =4.50\times10^{-3}\ \mathrm{mol}. \]

Therefore,

\[ M_A=\frac{1.287\ \mathrm g}{4.50\times10^{-3}\ \mathrm{mol}} =286\ \mathrm{g\,mol^{-1}}, \]

and

\[ 106+18x=286 \quad\Longrightarrow\quad \boxed{x=10}. \]

Sample A is consequently \(\boxed{\ce{Na2CO3.10H2O}}\).

2.2

The carbonate component of sample B reacts with sulphuric acid:

\[ \ce{Na2CO3 + H2SO4 -> Na2SO4 + CO2 + H2O}. \]

The excess sulphuric acid is then neutralized:

\[ \ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}. \]

Normality expresses reactive equivalents per litre. The initial amount of acid, in equivalents, was

\[ E_{\mathrm{initial}} =(0.2\ \mathrm{eq\,L^{-1}})(0.050\ \mathrm L) =0.010\ \mathrm{eq}. \]

The sodium hydroxide titre measures the remaining acid:

\[ E_{\mathrm{excess}} =(0.1\ \mathrm{eq\,L^{-1}})(0.050\ \mathrm L) =0.0050\ \mathrm{eq}. \]

Thus sample B consumed

\[ E_B=0.010-0.0050=0.0050\ \mathrm{eq}. \]

Each mole of sodium carbonate consumes two acid equivalents, so

\[ n_B(\ce{Na2CO3}) =\frac{0.0050\ \mathrm{eq}}{2\ \mathrm{eq\,mol^{-1}}} =2.5\times10^{-3}\ \mathrm{mol}. \]

Writing sample B as \(\ce{Na2CO3.yH2O}\), its molar mass is

\[ M_B=\frac{0.715\ \mathrm g}{2.5\times10^{-3}\ \mathrm{mol}} =286\ \mathrm{g\,mol^{-1}}. \]

Hence

\[ y=\frac{286-106}{18}=10. \]

Therefore both samples have the same composition, \(\boxed{\ce{Na2CO3.10H2O}}\).

The names and SMILES of the other species appearing in the reactions are:

Species Formula SMILES
Sodium carbonate \(\ce{Na2CO3}\) [Na+].[Na+].[O-]C(=O)[O-]
Hydrogen chloride, used as aqueous hydrochloric acid \(\ce{HCl}\) Cl
Sodium chloride \(\ce{NaCl}\) [Na+].[Cl-]
Carbon dioxide \(\ce{CO2}\) O=C=O
Water \(\ce{H2O}\) O
Sulphuric acid \(\ce{H2SO4}\) OS(=O)(=O)O
Sodium sulphate \(\ce{Na2SO4}\) [Na+].[Na+].[O-]S(=O)(=O)[O-]
Sodium hydroxide \(\ce{NaOH}\) [Na+].[OH-]

The solution was checked against the official solution.