Question

1) calculate the [OH-] of a solution having a [H3O+]=2.85x10-6M.

2) A solution is tested with a series of indicatores. the solution turns blue when bromcresol green is added, it turns purple when bromcresol purple is added, and its colorless when phenolphthalein is added. what is the estimated pH of this solution?

3) how does a pH indicator work? provide an equilibrium equation to illustrate your answer.

Answer 1

1) [H3O+] = 2.85 x 10^-6 M

So, [OH-] = Kw/[H3O+] = 1 x 10^-14/2.85 x 10^-6 = 3.51 x 10^-9 M

2) The estimated pH of solution would be 6.5 which is close to high end for bromocreasol purple givign purple color to solution.

3) An indicator measures of [H+] concentration change in a solution and show color change accordingly

Let indicator be HIn, then

HIn <==> H+ + In-

Ka = [H+][In-]/[HIn]

Ka/[H+] = [In-][/[HIn]

So, as the [H+] goes up, [HIn] will also go up. the solution will retain the color of non-dissociated indicator.

When [H+] goes down, so more basic solution, [In-] concentration goes up and we would see the color change at high pH end for indicator.

Say, for Phenolphthalein, we have an acidic solution, the pH would be less than 8.3, the solution would remain colorless. As the pH goes up, the solution becomes more basic, Phenolphthalein goes to the conjugate anion form and we see a pink color around pH 10.