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Indicators

Indicators. This is the basis of indicators. Many chemicals, such as the pigments of pelargonia, have different colours depending on whether they are in the acid form or its conjugate base. Demonstration. 1) Pipette out 25ml of acid into a conical flask.

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Indicators

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  1. Indicators

  2. This is the basis of indicators. Many chemicals, such as the pigments of pelargonia, have different colours depending on whether they are in the acid form or its conjugate base.

  3. Demonstration • 1) Pipette out 25ml of acid into a conical flask. • 2) Fill a burette with the base. • 3) Add a few drops of universal indicator. • 4) Titrate until the colour change indicates pH 7 has been achieved. Note the volume. • 5) Repeat using phenolphthalein. • 6) Repeat using methyl orange.

  4. Indicators as weak acids. Indicators are weak acids. Acid HIn ⇌ H+ + In- Conjugate base In acid the eqm is to the LHS In alkali equm is to the RHS. HIn ⇌ H+ + In- HIn ⇌ H+ + In- Solution has the colour of the undissociated acid form. Solution has the colour of the conjugate base.

  5. Phenolphthalein pH < 8.2 pH > 10.0

  6. Eg; Phenolphthalein • Acid form – colourless, Conjugate base - pink

  7. Methyl orange pH < 3.2  pH > 4.4

  8. Methyl orange • Acid form – red, conjugate base – yellow.

  9. Bromothymol blue • Under acid conditions (LHS) bromothymol blue is yellow. • Under basic conditions (RHS) it is blue.

  10. End points • At the end point; • [In-] = [HIn] • So the colour will be half way between the two. • The end point should be sharp, such that the addition of a small amount of acid or base causes a complete colour change. • NB The contribution of the indicator to the pH of a solution is minute as only a few drops are used.

  11. pH and endpoint • HIn⇌ H+ + In- • Kin = [H+] [In-] / [HIn] • pKin = pH + log ([HIn] / [In-]) • pH = pKin - log ([HIn] / [In-]) • At the end point [HIn]= [In-] • pH = pKin – log1 • log 1 = 0 • So pH = pKin

  12. It is assumed that the “acid” colour is totally obscured by the “base” colour when [In-] = 10 x [HIn] • Substituting; • pH = pKin - log ([HIn] / [In-]) • pH = pKin - log([HIn] / 10[HIn]) • log0.1 = -1 • pH = pKin – (-1) • pH = pKin + 1

  13. Similarly the “base” colour will be obscured when; • [HIn] = 10 [In-] • Subsituting; • pH = pKin - log([HIn] / [In-]) • pH = pKin - log(10[In-] / [In-]) • log10 = 1 • pH = pKin – (+1) • pH = pKin - 1

  14. Effective range of an indicator • Generally the effective range of an indicator is; • pKin +/- 1 • ie 2 pH units. • For a titration the end point of the indicator must correspond with the equivalence point. • The point where neutralisation has taken place and the reaction has finished.

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