IChO 2003 · Theory problem 7

Titration curves of phosphate solutions

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Solutions containing \(\ce{H3PO4}\) and/or \(\ce{NaH2PO4}\) are titrated with a strong base standard solution. Associate the contents of these solutions with the titration curves (\(pH\) vs. volume of titrant) shown in the figure:

(For \(\ce{H3PO4}\): \(pK_1 = 2.1,\ pK_2 = 7.2,\ pK_3 = 12.0\))

Figure 1

a) The sample contains \(\ce{H3PO4}\) only.

Curve A ( ), Curve B ( ), Curve C ( ), Curve D ( )

b) The sample contains both in a mole ratio \(\ce{H3PO4} : \ce{NaH2PO4} = 2 : 1\).

Curve A ( ), Curve B ( ), Curve C ( ), Curve D ( )

c) The sample contains both in a mole ratio \(\ce{H3PO4} : \ce{NaH2PO4} = 1 : 1\).

Curve A ( ), Curve B ( ), Curve C ( ), Curve D ( )

Official English text of the problem.

Answers

  • a) Curve A.
  • b) Curve B.
  • c) Curve D.

Solution

The chemical species needed for the explanation are:

Name Formula SMILES
Phosphoric acid \(\ce{H3PO4}\) O=P(O)(O)O
Sodium dihydrogen phosphate \(\ce{NaH2PO4}\) [Na+].O=P([O-])(O)O
Dihydrogen phosphate ion \(\ce{H2PO4^-}\) O=P([O-])(O)O
Hydrogen phosphate ion \(\ce{HPO4^2-}\) O=P([O-])([O-])O
Hydroxide ion \(\ce{OH^-}\) [OH-]
Water \(\ce{H2O}\) O

Let the initial amounts of phosphoric acid and sodium dihydrogen phosphate be \(a\) and \(b\) mol, respectively. The two observed equivalence points correspond to the successive neutralizations

\[ \ce{H3PO4 + OH^- -> H2PO4^- + H2O} \]

and

\[ \ce{H2PO4^- + OH^- -> HPO4^2- + H2O}. \]

Reaching the first equivalence point requires \(a\) mol of hydroxide. At that point, the solution contains \(a+b\) mol of dihydrogen phosphate, so reaching the second equivalence point requires a further \(a+b\) mol of hydroxide.

Thus, for base concentration \(c_{\mathrm B}\) in \(\mathrm{mol\,L^{-1}}\) and cumulative equivalence-point volumes \(V_1,V_2\) in litres,

\[ c_{\mathrm B}V_1=a, \qquad c_{\mathrm B}V_2=2a+b. \]

Consequently,

\[ \boxed{\frac{V_2}{V_1}=2+\frac{b}{a}} \qquad\text{and}\qquad \boxed{a:b=V_1:(V_2-2V_1)}. \]

The widely separated \(pK_i\) values allow these neutralization stages to be distinguished. The approximate pH values at the first and second equivalence points are, respectively,

\[ \frac{pK_1+pK_2}{2}=4.65, \qquad \frac{pK_2+pK_3}{2}=9.6, \]

consistent with the two steep rises shown.

a) Phosphoric acid only

Here \(b=0\), giving

\[ V_2=2V_1. \]

Curve A has equivalence points at approximately \(2\ \mathrm{cm^3}\) and \(4\ \mathrm{cm^3}\), so it satisfies this condition. Mark Curve A.

b) Mole ratio \(\ce{H3PO4}:\ce{NaH2PO4}=2:1\)

Here \(b/a=1/2\), so

\[ \frac{V_2}{V_1}=2+\frac12=2.5. \]

Curve B has equivalence points at approximately \(2\ \mathrm{cm^3}\) and \(5\ \mathrm{cm^3}\). Equivalently,

\[ a:b = 2:(5-2\times2) = 2:1. \]

Mark Curve B.

c) Mole ratio \(\ce{H3PO4}:\ce{NaH2PO4}=1:1\)

Here \(b/a=1\), so

\[ \frac{V_2}{V_1}=3. \]

Curve D has equivalence points at approximately \(2\ \mathrm{cm^3}\) and \(6\ \mathrm{cm^3}\). Hence,

\[ a:b = 2:(6-2\times2) = 2:2 = 1:1. \]

Mark Curve D.

The remaining curve, C, corresponds to sodium dihydrogen phosphate alone. It begins near the amphiprotic dihydrogen phosphate pH, approximately \(4.65\), and has no initial phosphoric-acid neutralization step. Its prominent rise corresponds to conversion into hydrogen phosphate.

The answers were checked against the official answers published by the organisers.