IChO 1997 · Theory problem 4

Spectrophotometry of an acid-base indicator

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\(\ce{HIn}\) is a weakly acidic indicator.

Figure 1

also written as

Figure 2

At normal temperatures, the acid dissociation constant for this indicator is \(K_{\mathrm a}=2.93\times10^{-5}\).

The absorbance data (1.00 cm cells) for \(5.00\times10^{-4}\) M (\(\mathrm{mol\,dm^{-3}}\)) solutions of this indicator in strongly acidic and strongly alkaline solutions are given in the following table.

Absorbance Data (A)

\(\lambda\), nm \(pH=1.00\) \(pH=13.00\)
400 0.401 0.067
470 0.447 0.050
485 0.453 0.052
490 0.452 0.054
505 0.443 0.073
535 0.390 0.170
555 0.342 0.342
570 0.303 0.515
585 0.263 0.648
615 0.195 0.816
625 0.176 0.823
635 0.170 0.816
650 0.137 0.763
680 0.097 0.588
Figure 3

4.1 Predict the observed colour of the a) acidic and b) basic forms of the indicator.

Using a “50 nm wide bar”, shade the appropriate area of the wavelength scale on the answer sheet which would correspond to the colour of the indicator at the \(pH\) values given in the table.

For example, if observed colour is green, your answer would appear as:

Figure 4

4.2 A filter is located between the light source and the sample. What colour filter would be most suitable for the photometric analysis of the indicator in a strongly acidic medium?

4.3 What wavelength range would be most suitable for the photometric analysis of the indicator in a strongly basic medium?

4.4 What would be the absorbance of a \(1.00\times10^{-4}\) M (\(\mathrm{mol\,dm^{-3}}\)) solution of the indicator in alkaline form if measured at 545 nm in a 2.50 cm cell?

4.5 Solutions of the indicator were prepared in a strongly acidic solution (\(\ce{HCl}\), \(pH=1\)) and in a strongly basic solution (\(\ce{NaOH}\), \(pH=13\)). Perfectly linear relationships between absorbance and concentration were observed in both media at 490 nm and 625 nm, respectively.

The molar absorptivities at the two wavelengths are:

\(\lambda=490\ \mathrm{nm}\) \(\lambda=625\ \mathrm{nm}\)
\(\ce{HIn}\) (\(\ce{HCl}\)) \(9.04\times10^2\ \mathrm{M^{-1}\,cm^{-1}}\) \(3.52\times10^2\ \mathrm{M^{-1}\,cm^{-1}}\)
\(\ce{In-}\) (\(\ce{NaOH}\)) \(1.08\times10^2\ \mathrm{M^{-1}\,cm^{-1}}\) \(1.65\times10^3\ \mathrm{M^{-1}\,cm^{-1}}\)

\((\mathrm{M}=\mathrm{mol\,dm^{-3}})\)

Calculate the absorbance (1.00 cm cell) at the two wavelengths for an aqueous \(1.80\times10^{-3}\) M (\(\mathrm{mol\,dm^{-3}}\)) solution of the indicator \(\ce{HIn}\).

Official English text of the problem.

Answers

  • 4.1 a) Acidic form: orange; shade a 50 nm bar approximately over 600–650 nm. b) Basic form: blue; shade approximately 450–500 nm.
  • 4.2 A blue filter, transmitting near 485–490 nm.
  • 4.3 The absorption maximum around 625 nm, preferably approximately 615–635 nm.
  • 4.4 \(\boxed{A_{545}\approx0.128\approx0.13}\), using linear interpolation.
  • 4.5 \(\boxed{A_{490}=1.46,\qquad A_{625}=0.913}\).

Absorbance is dimensionless.

Solution

4.1

The indicator’s acidic form, \(\ce{HIn}\), and conjugate-base anion, \(\ce{In-}\), are generic labels: the problem supplies no molecular constitution from which a specific chemical name or structural SMILES could be determined.

Their equilibrium is

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

where \(\ce{H+}\) is the proton, SMILES [H+]. Thus,

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

At \(pH=1.00\),

\[ \frac{[\ce{In-}]}{[\ce{HIn}]} =\frac{2.93\times10^{-5}}{10^{-1}} =2.93\times10^{-4}, \]

so the spectrum is essentially that of the acidic form. At \(pH=13.00\), the ratio is \(2.93\times10^8\), so the spectrum is essentially that of the basic form.

The observed colour comes from the transmitted light and is approximately complementary to the principal absorbed colour.

  • a) Acidic form: The absorption maximum is near \(485\ \mathrm{nm}\), in the blue region. Absorption decreases toward the longer wavelengths, giving an orange appearance. On the answer sheet, shade a 50 nm interval in the orange region, approximately 600–650 nm.
  • b) Basic form: Absorption is strongest around \(625\ \mathrm{nm}\), in the orange–red region, while blue light is transmitted strongly. The solution therefore appears blue. Shade approximately 450–500 nm.

4.2

The Beer–Lambert law is

\[ A=\varepsilon bc, \]

where \(\varepsilon\) is molar absorptivity, \(b\) is optical path length, and \(c\) is concentration. Sensitivity to concentration is greatest where \(\varepsilon\) is largest.

The acidic form has its largest tabulated absorbance at \(485\ \mathrm{nm}\):

\[ A_{485}=0.453. \]

Therefore, use a blue filter, transmitting light near 485–490 nm.

4.3

For the basic form,

\[ A_{615}=0.816,\qquad A_{625}=0.823,\qquad A_{635}=0.816. \]

The region 615–635 nm, centred on 625 nm, combines high absorption with a relatively flat maximum. It therefore provides high concentration sensitivity and reduces sensitivity to small wavelength-setting errors.

4.4

First estimate the absorbance at \(545\ \mathrm{nm}\) for the reference solution, whose concentration and path length are

\[ c_{\mathrm{ref}}=5.00\times10^{-4}\ \mathrm{mol\,dm^{-3}}, \qquad b_{\mathrm{ref}}=1.00\ \mathrm{cm}. \]

Linear interpolation between \(535\) and \(555\ \mathrm{nm}\) gives

\[ A_{545,\mathrm{ref}} \approx 0.170+ \frac{545-535}{555-535}(0.342-0.170) =0.256. \]

At a fixed wavelength and chemical form, absorbance scales with \(bc\). Hence,

\[ \begin{aligned} A_{545} &=A_{545,\mathrm{ref}} \frac{bc}{b_{\mathrm{ref}}c_{\mathrm{ref}}}\\[2mm] &\approx 0.256 \left(\frac{2.50\ \mathrm{cm}}{1.00\ \mathrm{cm}}\right) \left( \frac{1.00\times10^{-4}\ \mathrm{mol\,dm^{-3}}} {5.00\times10^{-4}\ \mathrm{mol\,dm^{-3}}} \right)\\[2mm] &=\boxed{0.128\approx0.13}. \end{aligned} \]

4.5

The aqueous solution contains both indicator forms because the weak acid partially dissociates. Let

\[ C=1.80\times10^{-3}\ \mathrm{mol\,dm^{-3}} \]

be the analytical indicator concentration. Neglecting water’s contribution to the proton concentration, set

\[ [\ce{In-}]=[\ce{H+}]=x, \qquad [\ce{HIn}]=C-x. \]

Then

\[ K_{\mathrm a}=\frac{x^2}{C-x}, \]

so

\[ x^2+K_{\mathrm a}x-K_{\mathrm a}C=0. \]

The physically meaningful root is

\[ \begin{aligned} x &=\frac{-K_{\mathrm a}+ \sqrt{K_{\mathrm a}^{\,2}+4K_{\mathrm a}C}}{2}\\[1mm] &=\frac{-2.93\times10^{-5} +\sqrt{(2.93\times10^{-5})^2+ 4(2.93\times10^{-5})(1.80\times10^{-3})}}{2}\\[1mm] &=2.15469\times10^{-4}\ \mathrm{mol\,dm^{-3}}. \end{aligned} \]

Consequently,

\[ [\ce{In-}]=2.15469\times10^{-4}\ \mathrm{mol\,dm^{-3}}, \]

\[ [\ce{HIn}] =1.80\times10^{-3}-2.15469\times10^{-4} =1.58453\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \]

Approximately \(12.0\%\) of the indicator is dissociated, so retaining \(C-x\) in the equilibrium calculation is appropriate.

The two species’ absorbances add:

\[ A_\lambda =b\left( \varepsilon_{\ce{HIn},\lambda}[\ce{HIn}] + \varepsilon_{\ce{In-},\lambda}[\ce{In-}] \right). \]

Using \(b=1.00\ \mathrm{cm}\), the supplied molar absorptivities in \(\mathrm{M^{-1}\,cm^{-1}}\), and concentrations in \(\mathrm M\):

\[ \begin{aligned} A_{490} &=(1.00)\left[ (9.04\times10^2)(1.58453\times10^{-3}) +(1.08\times10^2)(2.15469\times10^{-4}) \right]\\ &=1.45569\\ &=\boxed{1.46}. \end{aligned} \]

Similarly,

\[ \begin{aligned} A_{625} &=(1.00)\left[ (3.52\times10^2)(1.58453\times10^{-3}) +(1.65\times10^3)(2.15469\times10^{-4}) \right]\\ &=0.913278\\ &=\boxed{0.913}. \end{aligned} \]

The solution was checked against the official solution.