IChO 1981 · Theory problem 2

Dissociation of maleic acid and the choice of acid-base indicators

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Maleic acid (\(\ce{H2A}\)) is a weak dibasic acid. The correlation between the relative quantities of \(\ce{H2A}\), \(\ce{HA^-}\), \(\ce{A^{2-}}\):

\[ \alpha_0=\frac{c(\ce{H2A})}{c} \qquad \alpha_1=\frac{c(\ce{HA^-})}{c} \qquad \alpha_2=\frac{c(\ce{A^{2-}})}{c} \]

and pH values of the solution show that:

a) \(\alpha_0=\alpha_1\) for pH = 1.92

b) \(\alpha_1=\alpha_2\) for pH = 6.22

Find:

2.1 The values of the dissociation constants of maleic acid for the first (\(K_1\)) and the second (\(K_2\)) degree of dissociation.

2.2 The values of \(\alpha_0\), \(\alpha_1\), and \(\alpha_2\) for pH = 1.92 and pH = 6.22.

2.3 What is the value of pH when \(\alpha_1\) attains a maximum value? Find the maximum value of \(\alpha\).

2.4 Which of the acid-base indicators in the table are suitable for titration of a 0.1 M solution of maleic acid (as a monobasic and as a dibasic acid) with 0.1 M \(\ce{NaOH}\)?

Fill in the table 1 with the correct answers.

All the activity coefficients should be considered equal to 1.

Indicator pH interval
Methyl green 0.1 – 2.0
Tropeolin 00 1.4 – 3.2
β-Dinitrophenol 2.4 – 4.0
Bromphenol blue 3.0 – 4.6
Congo red 3.0 – 5.2
Methyl red 4.4 – 6.2
Bromphenol red 5.0 – 6.8
Bromthymol blue 6.0 – 7.6
Phenol red 6.8 – 8.0
Cresol red 7.2 – 8.8
Thymol blue 8.0 – 9.6
Phenolphthalein 8.2 – 10.0
Alizarine yellow 10.1 – 12.1
Tropeolin 0 11.0 – 13.0
1,3,5-Trinitrobenzene 12.2 – 14.0

Table 1

2.1 \(K_1=\)
\(K_2=\)
2.2 pH = 1.92 \(\alpha_0=\)
\(\alpha_1=\)
\(\alpha_2=\)
pH = 6.22 \(\alpha_0=\)
\(\alpha_1=\)
\(\alpha_2=\)
2.3 pH =
\(\alpha_1=\)
2.4 First indicator equivalence point pH =
1.
2.
3.
4.
Second indicator equivalence point pH =
1.
2.
3.
4.

Official English text of the problem.

Answers

  • 2.1: \(\boxed{K_1=1.2\times10^{-2},\quad K_2=6.0\times10^{-7}}\).

  • 2.2:

pH \(\alpha_0\) \(\alpha_1\) \(\alpha_2\)
1.92 0.49999 0.49999 \(2.5\times10^{-5}\)
6.22 \(2.5\times10^{-5}\) 0.49999 0.49999

  • 2.3: \(\boxed{\mathrm{pH}=4.07,\quad \alpha_{1,\max}=0.986\;(98.6\%)}\).

  • 2.4: Using the listed transition intervals:

  • First equivalence point: \(\boxed{\mathrm{pH}\approx4.12}\); bromphenol blue and Congo red. The usual simplified ampholyte approximation gives pH \(4.07\).
  • Second equivalence point: \(\boxed{\mathrm{pH}\approx9.37}\); thymol blue and phenolphthalein.

SMILES identifying these indicators are supplied in Solution 2.4.

Solution

2.1 — Dissociation constants

The acid and its conjugate bases are:

Symbol Name SMILES
\(\ce{H2A}\) Maleic acid, \((Z)\)-but-2-enedioic acid O=C(O)/C=C\C(=O)O
\(\ce{HA^-}\) Hydrogen maleate ion O=C(O)/C=C\C(=O)[O-]
\(\ce{A^{2-}}\) Maleate ion O=C([O-])/C=C\C(=O)[O-]

These SMILES encode the \(Z\) alkene geometry; they do not specify a particular conformation of the carboxyl groups. Other species used below are water O, hydronium [OH3+], hydroxide [OH-], and sodium hydroxide [Na+].[OH-], whose sodium ion [Na+] is a spectator.

The successive dissociations are

\[ \ce{H2A + H2O <=> H3O+ + HA^-}, \]

\[ \ce{HA^- + H2O <=> H3O+ + A^{2-}}. \]

Write \(h=[\ce{H3O+}]\). Equilibrium expressions below use molar concentrations, with the \(1\ \mathrm{mol\,L^{-1}}\) standard state implicit. Then

\[ K_1=\frac{h[\ce{HA^-}]}{[\ce{H2A}]} =h\frac{\alpha_1}{\alpha_0}, \qquad K_2=\frac{h[\ce{A^{2-}}]}{[\ce{HA^-}]} =h\frac{\alpha_2}{\alpha_1}. \]

When \(\alpha_0=\alpha_1\), \(K_1=h\). Consequently,

\[ pK_1=1.92, \qquad \boxed{K_1=10^{-1.92}=1.2023\times10^{-2}\approx1.2\times10^{-2}}. \]

Similarly, \(\alpha_1=\alpha_2\) implies \(K_2=h\), giving

\[ pK_2=6.22, \qquad \boxed{K_2=10^{-6.22}=6.0256\times10^{-7}\approx6.0\times10^{-7}}. \]

2.2 — Distribution fractions

The analytical concentration is

\[ c=[\ce{H2A}]+[\ce{HA^-}]+[\ce{A^{2-}}], \qquad \alpha_0+\alpha_1+\alpha_2=1. \]

The equilibrium ratios give

\[ [\ce{HA^-}]=\frac{K_1}{h}[\ce{H2A}], \qquad [\ce{A^{2-}}]=\frac{K_1K_2}{h^2}[\ce{H2A}]. \]

Therefore, defining \(D=h^2+K_1h+K_1K_2\),

\[ \boxed{ \alpha_0=\frac{h^2}{D},\qquad \alpha_1=\frac{K_1h}{D},\qquad \alpha_2=\frac{K_1K_2}{D}. } \]

Let

\[ r=\frac{K_2}{K_1}=10^{-4.30}=5.0119\times10^{-5}. \]

At pH \(1.92\), \(h=K_1\), so

\[ (\alpha_0,\alpha_1,\alpha_2) =\frac{(1,1,r)}{2+r}. \]

Thus

\[ \alpha_0=\alpha_1=0.49998747, \qquad \alpha_2=2.50587\times10^{-5}. \]

At pH \(6.22\), \(h=K_2\), so

\[ (\alpha_0,\alpha_1,\alpha_2) =\frac{(r,1,1)}{2+r}. \]

Thus

\[ \alpha_0=2.50587\times10^{-5}, \qquad \alpha_1=\alpha_2=0.49998747. \]

2.3 — Maximum hydrogen maleate fraction

Rewrite the expression for \(\alpha_1\) as

\[ \alpha_1=\frac{1}{1+h/K_1+K_2/h}. \]

For positive \(h\), the arithmetic–geometric mean inequality gives

\[ \frac{h}{K_1}+\frac{K_2}{h} \geq 2\sqrt{\frac{K_2}{K_1}}, \]

with equality precisely when

\[ \frac{h}{K_1}=\frac{K_2}{h}, \qquad h=\sqrt{K_1K_2}. \]

This minimizes the denominator and hence gives the global maximum:

\[ \boxed{ \mathrm{pH}_{\max} =\frac{pK_1+pK_2}{2} =\frac{1.92+6.22}{2} =4.07. } \]

The maximum fraction is

\[ \boxed{ \alpha_{1,\max} =\frac{1}{1+2\sqrt{K_2/K_1}} =\frac{1}{1+2(10^{-2.15})} =0.98603876\approx0.986. } \]

At this pH, the remaining fractions are equal:

\[ \alpha_0=\alpha_2 =\frac{1-\alpha_{1,\max}}{2} \approx0.00698. \]

2.4 — Equivalence points and indicators

The two neutralization stages are

\[ \ce{H2A + OH^- -> HA^- + H2O}, \]

\[ \ce{HA^- + OH^- -> A^{2-} + H2O}. \]

Let the initial acid volume be \(V_0\). Since acid and titrant both have concentration \(0.1\ \mathrm{mol\,L^{-1}}\), the first and second equivalence points require titrant volumes \(V_0\) and \(2V_0\), respectively.

Use the conventional value \(K_w=1.0\times10^{-14}\) at \(25^\circ\mathrm C\).

First equivalence point.

Dilution gives

\[ C_1=\frac{0.1V_0}{2V_0} =0.050\ \mathrm{mol\,L^{-1}}. \]

The solution contains predominantly the amphiprotic hydrogen maleate ion. Its acid and base reactions are

\[ \ce{HA^- + H2O <=> H3O+ + A^{2-}}, \]

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

The familiar ampholyte approximation is

\[ h\approx\sqrt{K_1K_2}, \qquad \mathrm{pH}\approx\frac{pK_1+pK_2}{2}=4.07. \]

For the stated concentration, a small correction is warranted because \(K_1/C_1\approx0.24\). At this equivalence point, \([\ce{Na+}]=C_1\), so charge balance requires

\[ h+C_1=\frac{K_w}{h}+C_1\alpha_1+2C_1\alpha_2. \]

Using the distribution fractions,

\[ h-\frac{K_w}{h} =C_1(\alpha_2-\alpha_0) =C_1\frac{K_1K_2-h^2}{h^2+K_1h+K_1K_2}. \]

Solving this equation gives

\[ \boxed{\mathrm{pH}_{\mathrm{eq},1}=4.1174\approx4.12.} \]

Both \(4.07\) and \(4.12\) fall within the listed intervals of bromphenol blue and Congo red.

Second equivalence point.

The analytical maleate concentration is now

\[ C_2=\frac{0.1V_0}{3V_0} =\frac{0.1}{3} =0.0333\ \mathrm{mol\,L^{-1}}. \]

Maleate hydrolyses:

\[ \ce{A^{2-} + H2O <=> HA^- + OH^-}. \]

Its base dissociation constant is

\[ K_b=\frac{K_w}{K_2} =\frac{10^{-14}}{6.0256\times10^{-7}} =1.6596\times10^{-8}. \]

Writing \(x=[\ce{OH^-}]\), the weak-base approximation gives

\[ K_b=\frac{x^2}{C_2-x}\approx\frac{x^2}{C_2}, \]

\[ x\approx\sqrt{K_bC_2} =\sqrt{(1.6596\times10^{-8})(0.03333)} =2.3520\times10^{-5}\ \mathrm{mol\,L^{-1}}. \]

Here \(x/C_2\approx7.1\times10^{-4}\), confirming the approximation. Hence

\[ \mathrm{pOH}=4.6286, \qquad \boxed{\mathrm{pH}_{\mathrm{eq},2}=14.00-4.6286\approx9.37.} \]

This lies within the intervals of thymol blue and phenolphthalein.

The indicator entries in Table 1 are therefore:

Equivalence point pH Indicator Listed transition interval
First \(4.12\), or \(4.07\) in the simplified approximation Bromphenol blue 3.0–4.6
First Same Congo red 3.0–5.2
Second 9.37 Thymol blue 8.0–9.6
Second Same Phenolphthalein 8.2–10.0

These occupy two indicator lines in each box; the remaining lines can be left blank.

The following SMILES identify the selected compounds. For the three phthalein/sulfonephthalein indicators, neutral closed-ring reference forms are shown; Congo red is shown as its usual disodium salt. Their solution forms change with pH.

Indicator SMILES
Bromphenol blue (bromophenol blue) O=S1(=O)OC(c2cc(Br)c(O)c(Br)c2)(c2cc(Br)c(O)c(Br)c2)c2ccccc21
Congo red [Na+].[Na+].Nc1c(/N=N/c2ccc(-c3ccc(/N=N/c4cc(S(=O)(=O)[O-])c5ccccc5c4N)cc3)cc2)cc(S(=O)(=O)[O-])c2ccccc12
Thymol blue O=S1(=O)OC(c2cc(C(C)C)c(O)cc2C)(c2cc(C(C)C)c(O)cc2C)c2ccccc21
Phenolphthalein O=C1OC(c2ccc(O)cc2)(c2ccc(O)cc2)c2ccccc21

The solution was checked against the official solution.