Two drops of indicator HIn (K a = 1.0 x10 -9 ), where HIn is yellow and In - is

Question

Two drops of indicator HIn (Ka = 1.0 x10-9), where HIn is yellow and In- is blue,are placed in 100.0mL of 0.10 M HCl. a) What color is the solution initially? b) The solution is titrated with 0.10 M NaOH. At what pH will the color change (yellow to greenish yellow)occur? c) What color will the solution be after 200.0 mL of NaOH hasbeen added? Two drops of indicator HIn (Ka = 1.0 x10-9), where HIn is yellow and In- is blue,are placed in 100.0mL of 0.10 M HCl. a) What color is the solution initially? b) The solution is titrated with 0.10 M NaOH. At what pH will the color change (yellow to greenish yellow)occur? c) What color will the solution be after 200.0 mL of NaOH hasbeen added?

Explanation / Answer

HIn acts like a weak acid: HIn  <-------> H+   + In- (a) In the presence of strong acid, the equilibrium will beshifted to the left, or the yellow color. (b) Greenish - yellow is a color in between the two sides, thepoint where both species are likely present in similarconcentraiton. The Henderson-Hasselbalch equation would tellus the pH = pKa = 9 (c)Here you have gone well past the endpoint and hydroxide ionpredominates. It reacts with the H+ in the above equation,causing the yellow species to further convert into the bluespecies.....this the solution becomes blue. (a) In the presence of strong acid, the equilibrium will beshifted to the left, or the yellow color. (b) Greenish - yellow is a color in between the two sides, thepoint where both species are likely present in similarconcentraiton. The Henderson-Hasselbalch equation would tellus the pH = pKa = 9 (c)Here you have gone well past the endpoint and hydroxide ionpredominates. It reacts with the H+ in the above equation,causing the yellow species to further convert into the bluespecies.....this the solution becomes blue.

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