IChO 1987 · Theory problem 2

Identification of a hydrated sodium phosphate by acid-base titration

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500 mg of a hydrated sodium salt of phosphoric acid are dissolved in 50.0 cm³ of 0.1 molar sulphuric acid. This solution is diluted with distilled water to 100.0 cm³ and 20.0 cm³ of it are titrated with 0.100 molar \(\ce{NaOH}\) solution using thymolphthalein as indicator. The average of the burette reading is 26.53 cm³. The pH at the end-point is 10.00.

Problems:

2.1 Calculate the percentage distribution by moles of all protonated \(\ce{H_{n}PO4^{n-3}}\) species at the end-point.

2.2 What is the stoichiometric formula of the salt?

The cumulative protonation constants are given by

\[ \beta_n= \frac{[\ce{H_{n}PO4^{n-3}}]} {[\ce{PO4^3-}][\ce{H+}]^n} \]

where \(\log\beta_1=11.70;\quad \log\beta_2=18.6;\quad \log\beta_3=20.6.\)

The relative atomic masses are: Na = 23.0; P = 31.0; H = 1.0; O = 16.0.

Official English text of the problem.

Answers

  • 2.1: The endpoint distribution, including unprotonated phosphate, is approximately:
Species Name SMILES Mole percentage
\(\ce{PO4^3-}\) Phosphate O=P([O-])([O-])[O-] \(1.95\%\)
\(\ce{HPO4^2-}\) Hydrogen phosphate O=P(O)([O-])[O-] \(97.97\%\)
\(\ce{H2PO4-}\) Dihydrogen phosphate O=P(O)(O)[O-] \(0.078\%\)
\(\ce{H3PO4}\) Phosphoric acid O=P(O)(O)O \(7.8\times10^{-10}\%\)
  • 2.2: Sodium dihydrogen phosphate dihydrate, \[ \boxed{\ce{NaH2PO4 * 2H2O}} \] SMILES: [Na+].O=P(O)(O)[O-].O.O. This represents the ionic composition and two waters of crystallization; crystal packing is unspecified.

Solution

2.1 — Phosphate distribution

The successive protonation equilibria are

\[ \begin{aligned} \ce{PO4^3- + H+ &<=> HPO4^2-}\\ \ce{HPO4^2- + H+ &<=> H2PO4-}\\ \ce{H2PO4- + H+ &<=> H3PO4}. \end{aligned} \]

Here the hydrogen ion, or proton, has SMILES [H+].

At pH 10.00,

\[ h=[\ce{H+}]=10^{-10.00}\ \mathrm{mol\,L^{-1}}. \]

Taking \(\beta_0=1\), the mole fraction of the species containing \(n\) bound protons is

\[ \alpha_n=\frac{\beta_nh^n}{D}, \qquad D=1+\beta_1h+\beta_2h^2+\beta_3h^3. \]

The relative amounts are therefore

\[ \begin{aligned} \ce{PO4^3-}:&\quad 1,\\ \ce{HPO4^2-}:&\quad 10^{11.70-10.00}=50.1187,\\ \ce{H2PO4-}:&\quad 10^{18.6-20.00}=0.0398107,\\ \ce{H3PO4}:&\quad 10^{20.6-30.00}=3.98107\times10^{-10}. \end{aligned} \]

Thus \(D=51.1585\). Dividing each relative amount by \(D\) and multiplying by \(100\) gives the percentages listed above.

For the titration calculation, the mean number of bound protons per phosphate group is

\[ \begin{aligned} \bar n &=\alpha_1+2\alpha_2+3\alpha_3\\ &=0.979674736+2(0.000778183)+3(7.78183\times10^{-12})\\ &=0.981231. \end{aligned} \]

2.2 — Identification of the hydrated salt

Write the salt as

\[ \ce{Na_{x}H_{3-x}PO4 * yH2O}, \qquad x=1,\ 2,\ \text{or }3. \]

The 20.0 cm³ aliquot contains

\[ m=0.500\ \mathrm{g}\times\frac{20.0}{100.0} =0.100\ \mathrm{g} \]

of the original salt, together with

\[ n_{\mathrm S} =(0.1\ \mathrm{mol\,L^{-1}})(0.0500\ \mathrm L) \frac{20.0}{100.0} =1.00\times10^{-3}\ \mathrm{mol} \]

of sulphuric acid.

At pH 10, sulphuric acid (O=S(=O)(O)O) has lost both acidic protons, forming sulfate (O=S(=O)([O-])[O-]):

\[ \ce{H2SO4 + 2OH- -> SO4^2- + 2H2O}. \]

Sodium hydroxide ([Na+].[OH-]) supplies hydroxide ions ([OH-]), which form water (O) on neutralization. The phosphate neutralization equilibria are

\[ \begin{aligned} \ce{H3PO4 + OH- &<=> H2PO4- + H2O}\\ \ce{H2PO4- + OH- &<=> HPO4^2- + H2O}\\ \ce{HPO4^2- + OH- &<=> PO4^3- + H2O}. \end{aligned} \]

The amount of sodium hydroxide added is

\[ n_{\mathrm B} =(0.100\ \mathrm{mol\,L^{-1}}) (0.02653\ \mathrm L) =2.653\times10^{-3}\ \mathrm{mol}. \]

After accounting for the two acidic protons of sulphuric acid, the remaining base consumption is

\[ n_{\mathrm B}-2n_{\mathrm S} =2.653\times10^{-3}-2.000\times10^{-3} =6.530\times10^{-4}\ \mathrm{mol}. \]

Each phosphate group initially carries \(3-x\) acidic protons and retains, on average, \(\bar n=0.981231\) at the endpoint. Neglecting free hydrogen and hydroxide ions in the endpoint balance, the net base consumption per mole of phosphate is consequently

\[ (3-x)-\bar n=2.018769-x. \]

If \(n_{\mathrm P}\) is the amount of phosphate in the aliquot,

\[ \boxed{6.530\times10^{-4} =(2.018769-x)n_{\mathrm P}}. \]

The hydrated salt’s molar mass must therefore be

\[ M=\frac{0.100\ \mathrm g}{n_{\mathrm P}} =\frac{(0.100\ \mathrm g)(2.018769-x)} {6.530\times10^{-4}\ \mathrm{mol}}. \]

The three possible sodium contents give:

\(x\) Inferred molar mass Assessment
1 \(156.0\ \mathrm{g\,mol^{-1}}\) Possible
2 \(2.874\ \mathrm{g\,mol^{-1}}\) Impossible: the anhydrous salt alone has molar mass \(142.0\ \mathrm{g\,mol^{-1}}\)
3 Would require a negative phosphate amount Impossible

Hence \(x=1\). The anhydrous part has molar mass

\[ M_{\mathrm{anhydrous}} =23.0+2(1.0)+31.0+4(16.0) =120.0\ \mathrm{g\,mol^{-1}}. \]

The hydration number is

\[ y=\frac{156.0-120.0}{18.0}=2.00. \]

Therefore,

\[ \boxed{\text{sodium dihydrogen phosphate dihydrate, }\ce{NaH2PO4 * 2H2O}}. \]

As a check, this formula predicts

\[ n_{\mathrm P}=\frac{0.100}{156.0} =6.41026\times10^{-4}\ \mathrm{mol}, \]

and hence

\[ V_{\mathrm B} =\frac{2.000\times10^{-3} +(1.018769)(6.41026\times10^{-4})} {0.100} =0.0265306\ \mathrm L =26.53\ \mathrm{cm^3}, \]

in agreement with the reported titre.

The solution was checked against the official solution.