Please can you help,sample paper question - diprotic indicators/thymol blue.

[swlondon]

Guys,

I have been trying to get my head around this question without success. I know the answer is there but I am just not 'seeing it'.

It is question 36 from Unit 12 in the Acer Sample Questions booklet.

It mentions that 2 acid/base indicators are added to separate samples of an uknown colourless solution of unknown pH and gives the resulting colours of each one, which indicates whether their acid or base form is predominantly present. It then proceeds to ask what the colour of thymol blue would be if it was also added to a sample of the unknown solution. Thymol blue is a diprotic indicator with 2 pKa values, one is 2.0 and the other is 8.8.

From the resulting colours of either of the first two indicators I am able to work out the pH of the unknown solution which is 9. This corresponds to the solution provided in the back of the booklet.

But the solution then says that because thymol blue has a pKa value, 8.8, that is close in value to the pH of the unknown solution, 9, then this means that it is present in both acid and base forms of almost equal quantity, resulting in its colour being green.

But my problem with this is why is the other pKa value of thymol blue, 2.0, not taken into account here? What dictates that we use the pKa value of 8.8 instead?

The way I understand it, we COULD, if we wanted to, use the pKa value of 2.0 instead. Plugging this into the equation:

pKA = pH - log In-/HIn (this equation may be slightly incorrect, it may need a + sign instead of a - sign before the log), means that we do not have equal amounts of the 2 forms of thymol blue, but one will be 10 times more than the other meaning the colour will NOT be green.

I am guessing that because the pH of the unknown solution is 9, and this value is extremely close to the pKA value of 8.8, that we use that value of pKa and not 2.0???? But why??

What dictates what here please?