IChO 1999 · Theory problem 2

A diprotic acid and an acid-base indicator

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PART A

A diprotic acid , \(\ce{H2A}\) , undergoes the following dissociation reactions :

Figure 1

A 20.00 \(\mathrm{cm^3}\) aliquot of a solution containing a mixture of \(\ce{Na2A}\) and \(\ce{NaHA}\) is titrated with 0.300 M hydrochloric acid. The progress of the titration is followed with a glass electrode pH meter. Two points on the titration curve are as follows :

\(\mathrm{cm^3}\) \(\ce{HCl}\) added \(pH\)
1.00 10.33
10.00 8.34

2.1 On adding 1.00 \(\mathrm{cm^3}\) of \(\ce{HCl}\), which species reacts first and what would be the product?

2.2 What is the amount (mmol) of the product formed in (2.1)?

2.3 Write down the main equilibrium of the product from (2.1) reacting with the solvent?

2.4 What are the amounts (mmol) of \(\ce{Na2A}\) and \(\ce{NaHA}\) initially present?

2.5 Calculate the total volume of \(\ce{HCl}\) required to reach the second equivalence point.

PART B

Solutions I, II and III contain a \(pH\) indicator \(\ce{HIn}\) (\(K_{\mathrm{In}}=4.19\times10^{-4}\)) and other reagents as indicated in the table. The absorbance values at 400 nm of the solutions measured in the same cell, are also given in the table. \(K_a\) of \(\ce{CH3COOH}\) is \(1.75\times10^{-5}\).

Table:

Solution I Solution II Solution III
Total concentration of indicator \(\ce{HIn}\) \(1.00\times10^{-5}\ \mathrm{M}\) \(1.00\times10^{-5}\ \mathrm{M}\) \(1.00\times10^{-5}\ \mathrm{M}\)
Other reagents \(1.00\ \mathrm{M}\) \(\ce{HCl}\) \(0.100\ \mathrm{M}\) \(\ce{NaOH}\) \(1.00\ \mathrm{M}\) \(\ce{CH3COOH}\)
Absorbance at 400 0.000 0.300 ?

2.6 Calculate the absorbance at 400 nm of solution III.

2.7 Apart from \(\ce{H2O}\), \(\ce{H+}\) and \(\ce{OH-}\), what are all the chemical species present in the solution resulting from mixing solution II and solution III at 1 : 1 volume ratio?

2.8 What is the absorbance at 400 nm of the solution in (2.7)?

2.9 What is the transmittance at 400 nm of the solution in (2.7)?

Official English text of the problem.

Answers

  • 2.1: The dianion \(\ce{A^2-}\) reacts first, producing the hydrogen anion \(\ce{HA-}\): \[ \ce{A^2- + H+ -> HA-}. \]
  • 2.2: \(0.300\ \mathrm{mmol}\) of \(\ce{HA-}\) is formed stoichiometrically.
  • 2.3: \[ \ce{HA- + H2O <=> H2A + OH-}. \]
  • 2.4: Initially, \(3.00\ \mathrm{mmol}\) of the disodium salt \(\ce{Na2A}\) and \(2.40\ \mathrm{mmol}\) of the sodium hydrogen salt \(\ce{NaHA}\).
  • 2.5: \(28.0\ \mathrm{cm^3}\) of hydrochloric acid in total.
  • 2.6: Absorbance \(\mathcal A_{400}\approx0.027\).
  • 2.7: Sodium ion \(\ce{Na+}\), [Na+]; acetic acid \(\ce{CH3COOH}\), CC(=O)O; acetate ion \(\ce{CH3COO-}\), CC(=O)[O-]; acidic indicator \(\ce{HIn}\); and indicator anion \(\ce{In-}\).
  • 2.8: Absorbance \(\mathcal A_{400}=0.218\).
  • 2.9: Transmittance \(T=0.605\), or \(60.5\%\).

The structures represented by \(\ce{A}\) and \(\ce{In}\) are unspecified. Consequently, unique molecular SMILES cannot be assigned to their acid–base forms or the corresponding salts.

Solution

The other structurally specified participants used below are water, O; hydrogen ion, [H+]; hydroxide ion, [OH-]; and chloride ion, [Cl-]. Aqueous hydrochloric acid and sodium hydroxide are represented ionically as [H+].[Cl-] and [Na+].[OH-], respectively.

2.1

The given dissociation constants are

\[ K_1=4.50\times10^{-7}, \qquad K_2=4.70\times10^{-11}. \]

The protonation constants of \(\ce{A^2-}\) and \(\ce{HA-}\) are, respectively,

\[ \frac{1}{K_2}=2.13\times10^{10}, \qquad \frac{1}{K_1}=2.22\times10^6. \]

Thus, \(\ce{A^2-}\) is the stronger base and is protonated preferentially:

\[ \boxed{\ce{A^2- + H+ -> HA-}}. \]

The product is the hydrogen anion \(\ce{HA-}\), the singly deprotonated form of the diprotic acid.

2.2

The amount of hydrogen ion added is

\[ n(\ce{H+}) = 0.300\ \mathrm{mol\,L^{-1}} \times1.00\times10^{-3}\ \mathrm L = 3.00\times10^{-4}\ \mathrm{mol}. \]

The reaction has a 1:1 stoichiometry, so the amount of \(\ce{HA-}\) formed is

\[ \boxed{n(\ce{HA-})_{\mathrm{formed}}=0.300\ \mathrm{mmol}}. \]

2.3

The ion \(\ce{HA-}\) is amphiprotic. Its principal reaction with water is basic hydrolysis:

\[ \boxed{\ce{HA- + H2O <=> H2A + OH-}}. \]

Using the conventional aqueous value \(K_w=1.00\times10^{-14}\),

\[ K_b(\ce{HA-}) = \frac{K_w}{K_1} = \frac{1.00\times10^{-14}}{4.50\times10^{-7}} = 2.22\times10^{-8}. \]

This exceeds its acid-dissociation constant, \(K_2=4.70\times10^{-11}\), by a factor of about 473, explaining why the basic reaction predominates.

2.4

Let the initial amounts in the aliquot be

\[ x=n_0(\ce{Na2A}),\qquad y=n_0(\ce{NaHA}), \]

both expressed in mmol.

At the first equivalence point, the initial dianion has been protonated to \(\ce{HA-}\). For a solution dominated by this amphiprotic ion,

\[ [\ce{H+}]\simeq\sqrt{K_1K_2}, \]

and therefore

\[ pH\simeq\frac{pK_1+pK_2}{2} = \frac{6.3468+10.3279}{2} = 8.3373. \]

This agrees with the measured \(pH=8.34\) after \(10.00\ \mathrm{cm^3}\) of acid. Thus, within the usual titration approximations, this is the first equivalence point:

\[ x = 0.300\ \mathrm{mmol\,cm^{-3}} \times10.00\ \mathrm{cm^3} = 3.00\ \mathrm{mmol}. \]

After adding \(1.00\ \mathrm{cm^3}\), the stoichiometric amounts are

\[ n(\ce{A^2-})=x-0.300, \qquad n(\ce{HA-})=y+0.300. \]

For this buffer,

\[ pH=pK_2+\log_{10}\frac{[\ce{A^2-}]}{[\ce{HA-}]}. \]

Since \(pH=10.33\simeq pK_2\), the two concentrations, and hence their amounts in the same solution, are approximately equal:

\[ x-0.300=y+0.300. \]

Consequently,

\[ y=3.00-0.600=2.40\ \mathrm{mmol}. \]

Thus,

\[ \boxed{n_0(\ce{Na2A})=3.00\ \mathrm{mmol}}, \qquad \boxed{n_0(\ce{NaHA})=2.40\ \mathrm{mmol}}. \]

2.5

At the second equivalence point, protonation to \(\ce{H2A}\) is stoichiometrically complete:

\[ \ce{HA- + H+ -> H2A}. \]

Each initial \(\ce{A^2-}\) requires two hydrogen ions overall, whereas each initial \(\ce{HA-}\) requires one. Hence,

\[ n(\ce{HCl})_{\mathrm{total}} = 2x+y = 2(3.00)+2.40 = 8.40\ \mathrm{mmol}. \]

Therefore,

\[ \boxed{ V_{\mathrm{HCl}} = \frac{8.40\ \mathrm{mmol}} {0.300\ \mathrm{mmol\,cm^{-3}}} = 28.0\ \mathrm{cm^3} }. \]

2.6

The indicator equilibrium is

\[ \ce{HIn <=> H+ + In-}, \qquad K_{\mathrm{In}} = \frac{[\ce{H+}][\ce{In-}]}{[\ce{HIn}]}. \]

Solution I is strongly acidic, so the indicator is almost entirely \(\ce{HIn}\). Its absorbance of 0.000 shows that the acidic form has negligible absorption at 400 nm.

Solution II is strongly basic, so the indicator is essentially entirely \(\ce{In-}\). The Beer–Lambert law therefore gives

\[ \varepsilon_-\ell = \frac{0.300}{1.00\times10^{-5}\ \mathrm{mol\,L^{-1}}} = 3.00\times10^4\ \mathrm{L\,mol^{-1}}, \]

where \(\ell\) is the cell path length.

Writing \(C_{\mathrm{In}}=[\ce{HIn}]+[\ce{In-}]\), the indicator equilibrium gives

\[ [\ce{In-}] = C_{\mathrm{In}} \frac{K_{\mathrm{In}}}{K_{\mathrm{In}}+[\ce{H+}]}. \]

Thus, when \(C_{\mathrm{In}}=1.00\times10^{-5}\ \mathrm M\),

\[ \mathcal A_{400} = 0.300\frac{K_{\mathrm{In}}}{K_{\mathrm{In}}+[\ce{H+}]}. \]

In solution III, acetic acid dissociates according to

\[ \ce{CH3COOH <=> H+ + CH3COO-}. \]

Its concentration is much greater than the indicator concentration, so the indicator’s effect on the pH is negligible. Using the weak-acid approximation,

\[ [\ce{H+}] \simeq \sqrt{K_aC} = \sqrt{(1.75\times10^{-5})(1.00)} = 4.18\times10^{-3}\ \mathrm M. \]

Therefore,

\[ \mathcal A_{\mathrm{III}} = 0.300 \frac{4.19\times10^{-4}} {4.19\times10^{-4}+4.18\times10^{-3}} \simeq0.0273. \]

At the absorbance precision used in the table,

\[ \boxed{\mathcal A_{\mathrm{III}}\approx0.027}. \]

2.7

Hydroxide from solution II neutralizes part of the acetic acid from solution III:

\[ \ce{CH3COOH + OH- -> CH3COO- + H2O}. \]

For equal starting volumes \(V\), the amounts of acetic acid and hydroxide are proportional to \(1.00V\) and \(0.100V\), respectively. Acetic acid remains in excess, and the mixture contains an acetic acid/acetate buffer.

Both indicator forms are present through the equilibrium

\[ \ce{HIn <=> H+ + In-}. \]

Equivalently, their proton exchange with the buffer can be written as

\[ \ce{In- + CH3COOH <=> HIn + CH3COO-}. \]

Excluding the three species specified in the question, the complete list is:

Species Name SMILES
\(\ce{Na+}\) Sodium ion [Na+]
\(\ce{CH3COOH}\) Acetic acid CC(=O)O
\(\ce{CH3COO-}\) Acetate ion CC(=O)[O-]
\(\ce{HIn}\) Acidic indicator Structure unspecified
\(\ce{In-}\) Indicator conjugate-base anion Structure unspecified

2.8

After neutralization and dilution to twice the original volume, the buffer concentrations are approximately

\[ [\ce{CH3COOH}] = \frac{1.00-0.100}{2} = 0.450\ \mathrm M, \]

\[ [\ce{CH3COO-}] = \frac{0.100}{2} = 0.0500\ \mathrm M. \]

Their subsequent equilibrium changes are small compared with these concentrations. Hence,

\[ [\ce{H+}] = K_a\frac{[\ce{CH3COOH}]}{[\ce{CH3COO-}]} \simeq 1.75\times10^{-5}\frac{0.450}{0.0500} = 1.575\times10^{-4}\ \mathrm M. \]

Both starting solutions have the same total indicator concentration, so mixing them leaves that concentration unchanged:

\[ C_{\mathrm{In,mix}} = \frac{C_{\mathrm{In}}V+C_{\mathrm{In}}V}{2V} = 1.00\times10^{-5}\ \mathrm M. \]

The absorbing fraction of the indicator is therefore

\[ \frac{[\ce{In-}]}{C_{\mathrm{In,mix}}} = \frac{4.19\times10^{-4}} {4.19\times10^{-4}+1.575\times10^{-4}} = 0.7268. \]

Thus,

\[ \boxed{ \mathcal A_{\mathrm{mix}} = 0.300(0.7268) = 0.218 }. \]

2.9

Absorbance and transmittance are related by

\[ \mathcal A=-\log_{10}T, \qquad T=\frac{I}{I_0}. \]

Using the unrounded absorbance,

\[ T=10^{-0.2180399}=0.605285. \]

Therefore,

\[ \boxed{T=0.605}, \qquad \boxed{\%T=60.5\%}. \]

The solution was checked against the official solution.