IChO 1998 · Theory problem 1

EDTA titration of tin and lead in an alloy

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The following 8 steps describe the procedure for analysing a sample of alloy that contains both tin and lead.

  1. A 0.4062 g sample of alloy was dissolved by heating it in a small beaker with a mixture of 11 M hydrochloric and 16 M nitric acid. The beaker was heated until the entire alloy dissolved. In this procedure, lead is oxidised to Pb(II) and tin becomes Sn(IV).

  2. After 5 minutes of heating to expel oxides of nitrogen and chlorine, some acid remained. When the solution was cooled, a precipitate of some tin compounds and a lead compound appeared.

  3. A 25.00 \(\mathrm{cm^3}\) aliquot of 0.2000 M \(\ce{Na2H2EDTA}\) solution was added. The precipitate dissolved and a clear, colourless solution was obtained.

  4. This solution was quantitatively transferred to a 250.0 \(\mathrm{cm^3}\) volumetric flask and made up to the mark with distilled water.

  5. A 25.00 \(\mathrm{cm^3}\) aliquot of this solution was treated with 15 \(\mathrm{cm^3}\) of a 30 % w/v solution of hexamine (hexamethylenetetramine), some water and two drops of Xylenol Orange solution. The pH of each aliquot was 6.

  6. The clear, yellow solution from Step 5 was titrated with standard 0.009970 M lead nitrate solution until the colour just changed from yellow to red. The titre at this endpoint was 24.05 \(\mathrm{cm^3}\).

  7. 2.0 g of solid \(\ce{NaF}\) was added to the titration flask. The solution turned back to yellow.

  8. The solution was titrated with more standard 0.009970 M lead nitrate solution until the colour changed to red again. The titre at this endpoint was 15.00 \(\mathrm{cm^3}\).

Hexamine and Xylenol Orange have the structures shown below. The \(pK_b\) of hexamine is 9.5. Xylenol Orange is red below \(pH\) 4, yellow above \(pH\) 5.

Figure 1

\(K'_{\mathrm{MY}}\) is the conditional formation constant \(=\alpha K_{\mathrm{MY}}\).

\(K'_{\mathrm{MY}}\) for the formation of the EDTA complexes of Pb(II) and Sn(IV), in the presence and absence of fluoride, are shown in the following Figure.

Figure 2

1.1 What is the lead compound that precipitates in Step 2?

1.2 Write a balanced ionic equation that explains the disappearance of the precipitate in Step 3 (at \(pH\) 6).

1.3 What is the purpose of hexamine in Step 5 of the analysis?

1.4 What is the purpose of Xylenol Orange in the analysis?

1.5 Write balanced ionic equations for the reactions occurring during the titration and responsible for the colour change at the endpoint in Step 6 of the analysis

1.6 What is the purpose of \(\ce{NaF}\) in Step 7 of the analysis?

1.7 Write a balanced ionic equation for the reaction that occurs in Step 7.

1.8 Write a balanced ionic equation that explains why the colour changed from red to yellow in Step 7 of the analysis.

1.9 Write a balanced ionic equation that explains why the lines on the graph below of \(\log K_{\mathrm{MY}}\) vs \(pH\) for \(\ce{Pb}\) + EDTA and \(\ce{Pb}\) + EDTA + \(\ce{F-}\) are coincident below \(pH\) 2.

1.10 Calculate the percentage by weight of \(\ce{Sn}\) and \(\ce{Pb}\) in the alloy.

Official English text of the problem.

Answers

Let \(\ce{Y^{4-}}\) denote fully deprotonated EDTA. For charge-explicit indicator equations, use the formal XO representation \(\ce{X^{4-}}\) and its lead complex \(\ce{[PbX]^{2-}}\). This indicator convention does not identify the unique microscopic protonation state at pH 6; see the discussion and SMILES below.

  • 1.1 Lead(II) chloride, \(\ce{PbCl2}\).
  • 1.2 \[ \ce{PbCl2(s) + H2Y^{2-} -> [PbY]^{2-} + 2Cl- + 2H+} \]
  • 1.3 Hexamine neutralizes residual acid and, with its conjugate acid, helps maintain the solution at pH 6.
  • 1.4 Xylenol Orange is a metallochromic indicator: free indicator is yellow at pH 6, whereas its lead complex is red.
  • 1.5 \[ \ce{Pb^{2+} + H2Y^{2-} -> [PbY]^{2-} + 2H+} \] \[ \ce{Pb^{2+} + X^{4-} <=> [PbX]^{2-}} \]
  • 1.6 Fluoride selectively masks tin(IV), releasing its bound EDTA for a second titration.
  • 1.7 \[ \ce{[SnY] + 6F- + 2H+ -> [SnF6]^{2-} + H2Y^{2-}} \]
  • 1.8 \[ \ce{[PbX]^{2-} + H2Y^{2-} -> [PbY]^{2-} + X^{4-} + 2H+} \]
  • 1.9 \[ \ce{H+ + F- <=> HF} \] Protonation suppresses fluoride’s competition for lead.
  • 1.10 \[ \boxed{w(\ce{Sn})=43.70\%,\qquad w(\ce{Pb})=56.45\%} \] The supplied measurements give a total of \(100.15\%\), a small mass-balance discrepancy.

Solution

1.1

The precipitate is lead(II) chloride, \(\ce{PbCl2}\), SMILES [Cl-].[Cl-].[Pb+2].

The solution contains lead(II) ions and chloride from hydrochloric acid. Lead(II) chloride is sparingly soluble, and its solubility decreases on cooling:

\[ \ce{Pb^{2+} + 2Cl- <=> PbCl2(s)} \]

Lead(II) nitrate would remain soluble. The disconnected SMILES describes the ionic composition of the solid, not its crystal lattice.

1.2

EDTA forms a sufficiently stable, soluble lead complex to drive dissolution of the chloride precipitate. Writing the added ligand as \(\ce{H2Y^{2-}}\),

\[ \boxed{\ce{PbCl2(s) + H2Y^{2-} -> [PbY]^{2-} + 2Cl- + 2H+}} \]

This is the sum of dissolution and complex formation:

\[ \ce{PbCl2(s) <=> Pb^{2+} + 2Cl-} \]

\[ \ce{Pb^{2+} + H2Y^{2-} <=> [PbY]^{2-} + 2H+} \]

Complexation lowers the free lead-ion concentration and therefore draws more solid into solution.

Tin(IV) is also strongly complexed:

\[ \ce{Sn^{4+} + H2Y^{2-} <=> [SnY] + 2H+} \]

The problem does not identify the tin precipitates individually. If a precipitate is represented as tin(IV) oxide, its corresponding dissolution equation is

\[ \ce{SnO2(s) + H2Y^{2-} + 2H+ -> [SnY] + 2H2O} \]

This last equation illustrates the principle; it does not establish the identity of every tin solid present.

1.3

Hexamine is a weak base, as indicated by its \(pK_b=9.5\). It accepts protons from the residual acid:

\[ \ce{B + H+ <=> BH+} \]

Here \(\ce{B}\) is hexamine and \(\ce{BH+}\) its conjugate acid. Their mixture provides acid–base buffering and accommodates protons released during complex formation.

Maintaining pH 6 serves two purposes:

  • The conditional formation constants of both metal–EDTA complexes are large enough for quantitative complexation.
  • Uncomplexed XO is yellow, allowing a distinct red endpoint when lead binds the indicator.

1.4

XO acts as a metallochromic endpoint indicator.

While unbound EDTA remains, added lead(II) is captured by EDTA. Once that EDTA has been consumed, the first slight excess of lead forms the red lead–XO complex.

Because the pH is maintained at 6, this yellow-to-red endpoint signals metal binding, rather than acidification into the indicator’s low-pH red range.

1.5

During the first titration, lead(II) consumes the excess EDTA:

\[ \boxed{\ce{Pb^{2+} + H2Y^{2-} -> [PbY]^{2-} + 2H+}} \]

Each mole of added lead consumes one mole of unbound EDTA.

At the endpoint, lead binds XO. In the stated formal indicator convention,

\[ \boxed{\ce{Pb^{2+} + X^{4-} <=> [PbX]^{2-}}} \]

The free-indicator population is yellow; the lead-bound population is red.

The exact microscopic proton balance of XO cannot be determined from the supplied acid–base information. More generally, if binding removes \(r\) protons from a free indicator species of charge \(q-\), the fully balanced form is

\[ \ce{Pb^{2+} + H_rIn^{q-} <=> [PbIn]^{(q+r-2)-} + rH+} \]

The simpler equation above is a formal representation of this indicator chemistry at fixed pH.

1.6

Sodium fluoride, SMILES [Na+].[F-], supplies fluoride ions that preferentially complex tin(IV).

The graph shows that, at pH 6:

  • The tin–EDTA conditional formation constant becomes extremely small in the presence of fluoride.
  • The lead–EDTA conditional formation constant remains very large.

Consequently, fluoride displaces EDTA from tin while leaving lead effectively bound to EDTA. The released EDTA can then be measured with additional standard lead nitrate.

This is selective masking of tin(IV).

1.7

Tin(IV) forms hexafluoridostannate(IV), \(\ce{[SnF6]^{2-}}\), releasing one EDTA ligand per tin atom:

\[ \boxed{\ce{[SnY] + 6F- -> [SnF6]^{2-} + Y^{4-}}} \]

Including protonation of the released EDTA gives the equivalent equation

\[ \boxed{\ce{[SnY] + 6F- + 2H+ -> [SnF6]^{2-} + H2Y^{2-}}} \]

Thus, the second titre measures tin in a 1:1 molar ratio:

\[ n(\text{lead added in second titration})=n(\text{tin in aliquot}) \]

1.8

The newly released EDTA removes lead from the red indicator complex, regenerating the yellow free indicator:

\[ \boxed{\ce{[PbX]^{2-} + H2Y^{2-} -> [PbY]^{2-} + X^{4-} + 2H+}} \]

Using fully deprotonated EDTA, the equivalent formal equation is

\[ \ce{[PbX]^{2-} + Y^{4-} -> [PbY]^{2-} + X^{4-}} \]

For the general protonation convention introduced in 1.5, the explicitly balanced version is

\[ \ce{[PbIn]^{(q+r-2)-} + H2Y^{2-} + rH+ -> [PbY]^{2-} + H_rIn^{q-} + 2H+} \]

Further addition of lead nitrate consumes the released EDTA. Once it is exhausted, red lead–XO complex forms again.

1.9

In strongly acidic solution, fluoride is protonated:

\[ \boxed{\ce{H+ + F- <=> HF}} \]

Hydrogen fluoride is much less effective than free fluoride ion at competing for lead. Below pH 2, this suppresses the fluoride side reaction sufficiently that the conditional lead–EDTA formation constants with and without added fluoride coincide on the graph.

1.10

Both metals form 1:1 EDTA complexes. The dilution factor from the aliquot to the original sample is

\[ D=\frac{250.0\ \mathrm{cm^3}}{25.00\ \mathrm{cm^3}}=10.00 \]

The 15.00 \(\mathrm{cm^3}\) second titre is the additional volume added after fluoride treatment.

The original EDTA amount is

\[ n_{Y,0} =(0.2000\ \mathrm{mol\,L^{-1}}) (25.00\times10^{-3}\ \mathrm L) =5.000\times10^{-3}\ \mathrm{mol} \]

Therefore, each aliquot contains

\[ n_{Y,\mathrm{aliquot}}=5.000\times10^{-4}\ \mathrm{mol} \]

of total EDTA, including bound and unbound ligand.

The first titre measures the EDTA initially left unbound:

\[ \begin{aligned} n_{Y,\mathrm{excess}} &=(0.009970\ \mathrm{mol\,L^{-1}}) (24.05\times10^{-3}\ \mathrm L)\\ &=2.397785\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Hence, retaining guard digits,

\[ \begin{aligned} n_{\ce{Sn},a}+n_{\ce{Pb},a} &=5.000\times10^{-4}-2.397785\times10^{-4}\\ &=2.602215\times10^{-4}\ \mathrm{mol} \end{aligned} \]

The second titre measures EDTA released from tin:

\[ \begin{aligned} n_{\ce{Sn},a} &=(0.009970\ \mathrm{mol\,L^{-1}}) (15.00\times10^{-3}\ \mathrm L)\\ &=1.49550\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Therefore,

\[ \begin{aligned} n_{\ce{Pb},a} &=2.602215\times10^{-4}-1.49550\times10^{-4}\\ &=1.106715\times10^{-4}\ \mathrm{mol} \end{aligned} \]

Scaling to the whole sample,

\[ n_{\ce{Sn}}=1.49550\times10^{-3}\ \mathrm{mol} \]

\[ n_{\ce{Pb}}=1.106715\times10^{-3}\ \mathrm{mol} \]

Using \(M_{\ce{Sn}}=118.7\ \mathrm{g\,mol^{-1}}\) and \(M_{\ce{Pb}}=207.2\ \mathrm{g\,mol^{-1}}\),

\[ m_{\ce{Sn}} =(1.49550\times10^{-3})(118.7) =0.17751585\ \mathrm g \]

\[ m_{\ce{Pb}} =(1.106715\times10^{-3})(207.2) =0.229311348\ \mathrm g \]

Thus,

\[ \boxed{ w(\ce{Sn}) =\frac{0.17751585}{0.4062}\times100\% =43.70\% } \]

\[ \boxed{ w(\ce{Pb}) =\frac{0.229311348}{0.4062}\times100\% =56.45\% } \]

The reconstructed metal mass is \(0.4068272\ \mathrm g\), compared with the stated \(0.4062\ \mathrm g\). The two independently calculated percentages have therefore been reported without forcing their sum to \(100\%\).

Names, SMILES, and structural conventions

The following representations complete the structural identification of species used above. Dative arrows in the complex SMILES represent donor-to-metal bonds; they do not specify three-dimensional geometry.

Species Name and SMILES
\(\ce{Y^{4-}}\) Ethylenediaminetetraacetate, EDTA(4−): [O-]C(=O)CN(CC(=O)[O-])CCN(CC(=O)[O-])CC(=O)[O-]
\(\ce{H2Y^{2-}}\) Dihydrogen ethylenediaminetetraacetate, shown with both amine nitrogens protonated: [O-]C(=O)C[NH+](CC(=O)[O-])CC[NH+](CC(=O)[O-])CC(=O)[O-]
\(\ce{[PbY]^{2-}}\) Lead(II)–EDTA complex: [Pb+2]12345<-[O-]C(=O)CN->1(CC(=O)[O-]->2)CCN->3(CC(=O)[O-]->4)CC(=O)[O-]->5
\(\ce{[SnY]}\) Tin(IV)–EDTA complex: [Sn+4]12345<-[O-]C(=O)CN->1(CC(=O)[O-]->2)CCN->3(CC(=O)[O-]->4)CC(=O)[O-]->5
\(\ce{B}\) Hexamine, or hexamethylenetetramine: C1N2CN3CN1CN(C2)C3
\(\ce{BH+}\) Monoprotonated hexamine: C1[NH+]2CN3CN1CN(C2)C3
\(\ce{[SnF6]^{2-}}\) Hexafluoridostannate(IV): [F-]->[Sn+4](<-[F-])(<-[F-])(<-[F-])(<-[F-])<-[F-]
\(\ce{X^{4-}}\) Xylenol Orange, formal tetracarboxylate representation: C1(c2cc(CN(CC(=O)[O-])CC(=O)[O-])c(O)c(C)c2)(c2cc(CN(CC(=O)[O-])CC(=O)[O-])c(O)c(C)c2)OS(=O)(=O)c2ccccc21
\(\ce{[PbX]^{2-}}\) Lead(II)–XO, schematic 1:1 chelate representation: [Pb+2]12<-[O-]C(=O)CN->1(CC(=O)[O-]->2)Cc1cc(C2(c3cc(CN(CC(=O)[O-])CC(=O)[O-])c(O)c(C)c3)OS(=O)(=O)c3ccccc32)cc(C)c1O
\(\ce{SnO2}\) Tin(IV) oxide, formula-unit representation: O=[Sn]=O
\(\ce{HF}\) Hydrogen fluoride: [H]F
\(\ce{H2O}\) Water: O

The simple ions are lead(II), [Pb+2]; tin(IV), [Sn+4]; chloride, [Cl-]; fluoride, [F-]; and hydrogen ion, [H+]. Aqueous hydrogen ions are solvated; [H+] is the conventional ionic-equation representation.

The EDTA complex models show coordination through two nitrogen and four oxygen donors. Hexafluoridostannate(IV) has octahedral coordination. The precise lead coordination geometry, additional hydration, and XO donor arrangement are not established by this problem. In particular, the Pb–XO SMILES is a schematic connectivity model, not a determination of its colour-bearing microscopic structure. Solid tin(IV) oxide has an extended lattice rather than discrete molecules of the displayed formula-unit form.

The solution was checked against the official solution.