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Bromocresol blue (BCB) [VISUAL] is an organic dye, an acid-base indicator, a weak diprotic acid (H2AOrganic Chemistry Chemistry Question

Determination of the acidity constant of bromocresol blue (3′,3′′,5′,5′′-tetrabromo-m-cresolsulfonephthalein, bcb)

Bromocresol blue (BCB) [VISUAL] is an organic dye, an acid-base indicator, a weak diprotic acid (H2A). In aqueous solutions in the pH range of 3-6 BCB changes its color from yellow to blue due to dissociation of the second proton:

HA– (yellow) ⇄ A2– (blue) + H+

On the base of the absorbance of BCB solution measured as a function of the pH one can calculate the second acidity constant of BCB, pKa2.

Reagents and solutions required
* Bromocresol blue, 0.25% solution in 50% aqueous ethanol,
* Mixture of acids for preparation of buffer solutions: an aqueous solution containing H3PO4, and H3BO3 with a concentration 0.04 mol dm–3 each,
* NaOH, solutions with concentrations of 0.2 mol dm–3 and 2 mol dm–3,
* HCl, solution (c = 2 mol dm–3).

1. Choice of the wavelength for the Ka2 determination
1.1 Into each of two 50.0 cm3 volumetric flasks place 1.00 cm3 of the BCB solution and 10.00 cm3 of the mixture of acids (see reagent list). Then add 1.00 cm3 of NaOH solution (c = 0.2 mol dm–3) into the first and 6.00 cm3 of NaOH solution (c = 2 mol dm–3) into the second flask. Dilute the solutions to the mark with water and mix.
1.2 Measure the pH of the solutions prepared. The first one must have the pH in the range of 2 – 3, the second one in the range within 7 – 8. Under such conditions all BCB is in the form of either HA– or A2– respectively. If either of the pH is different from the required, adjust it by adding few drops of HCl solution (2 mol dm–3) or NaOH solution (2 mol dm–3).
1.3 Measure the absorption spectra of the solutions in the range of 400 – 700 nm; 5 – 10 data points would be sufficient.
1.4 Choose the wavelength at which the absorbances of the solutions differ most greatly. Usually that wavelength corresponds to the maximum of absorbance of one of the species or close to it. Further carry out all the measurements at that wavelength.

2. Preparation of series of BCB solutions, measuring their absorbance and the pH
2.1 Into each of twelve 50 cm3 volumetric flasks place 1.00 cm3 of BCB solution and 10.00 cm3 of the mixture of acids. Then add NaOH solution (c = 0.2 mol dm–3) to each flask in the amount indicated in Table below:

| Flask number | Solution NaOH, cm3 |
|---|---|
| 1 | 0.75 |
| 2 | 1.50 |
| 3 | 2.50 |
| 4 | 2.75 |
| 5 | 3.00 |
| 6 | 3.25 |
| 7 | 3.50 |
| 8 | 3.75 |
| 9 | 4.00 |
| 10 | 4.25 |
| 11 | 5.25 |
| 12 | 6.25 |

Dilute the solutions to the mark with water and mix.

*Note. It is of essential importance that the concentrations of BCB be strictly the same in all the solutions. When preparing the solutions pay especial attention to that requirement!*

2.2 For each solution measure the pH and the absorbance at the chosen wavelength.
2.3 Using the data obtained calculate logKa2 for each of the solutions unless fraction of either of the species involved in the acid-base equilibrium is negligible.
2.4 Calculate the average logKa2 value.

Questions
Denote as:
[HA–], [A2–], c – equilibrium concentrations of the corresponding BCB forms and its total concentration, respectively;
l – cuvette length;
Ka2 – acidity constant of HA–;
εHA, εA – extinction coefficients of the corresponding forms at the chosen wavelength;
AHA, AA, A – absorbances of BCB solution containing only HA–, only A2– and their mixture, respectively.

32.1.

Write down the equations for AHA, AA and A as functions of [HA–], [A2–] and c.

Model Answer

AHA=εHAlcA_{HA} = \varepsilon_{HA} l c
AA=εAlcA_A = \varepsilon_A l c
A=(εHA[HA]+εA[A2])lA = (\varepsilon_{HA} [HA^-] + \varepsilon_A [A^{2-}]) l

32.2.

Express A as a function of AHA, AA and [H+].

Model Answer

Since [HA]=[H+][H+]+Ka2c[HA^-] = \frac{[H^+]}{[H^+] + K_{a2}} c and [A2]=Ka2[H+]+Ka2c[A^{2-}] = \frac{K_{a2}}{[H^+] + K_{a2}} c:
A=AHA[H+]+AAKa2[H+]+Ka2A = \frac{A_{HA} [H^+] + A_A K_{a2}}{[H^+] + K_{a2}}

32.3.

Write down the equation for calculation of Ka2 from AHA, AA, A and [H+].

Model Answer

Ka2=[H+]AAHAAAAK_{a2} = [H^+] \frac{A - A_{HA}}{A_A - A}

32.4.

Consider the wavelength at which εHA = εA. It is called the isosbestic point.
a) Is it possible to determine Ka of a dye by measuring the absorbance at the isosbestic point?
b) What analytical information can be obtained from such measurement?

Model Answer

a) No. If εHA=εA=ε\varepsilon_{HA} = \varepsilon_A = \varepsilon, then at any pH: A=(ε[HA]+ε[A2])l=εlc=AHA=AAA = (\varepsilon [HA^-] + \varepsilon [A^{2-}]) l = \varepsilon l c = A_{HA} = A_A. Therefore, calculation of Ka2K_{a2} is not possible.
b) The total concentration (cc) of the dye.

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