[scott_wright7]

Hello, I'm doing my exams next week and i need help on one past exam paper question, i have the solutions but it doesnt give me a proper answer.
Bromothymol blue (HIn) is an indicator and a weak acid with a Ka of 1.0 x 10^7. It is yellow in its acidic form and blue in its basic form. The equilibrium reaction is
                                HIn(aq) ? H+ (aq) + In- (aq)
If a few drops of Bromothylmol blue are added to an acidic solution whose PH is 1.0 the position of the equilibrium in the reaction shown above:
A. shifts to the right and the solution turns blue
B. shifts left and the solution turns yellow
C. shifts right and the solution turns yellow
D. shifts left and the solution turns blue

I've come to the conclusion that is is pH of 1.0 making is acidic, therefore knocking out A and D, and i also thought an adition of a reactant (HIn) would cause a net foward reaction? therefore making the reaction go right. however the solutions say B, left shift and yellow. why?
Thanks Scott

[Borek]

pKa is 7 (you made a mistake omitting minus in the Ka constant exponent).

To calculate ratio of HIn to In^- you may either use Ka definition of HH equation.

Whichever you will use you will came to the conlusion that only about 10^-6 indicator is in In^- form. Reaction can't go right from there, as pH defines this ratio and you were not told that pH was changed. Besides, inidcators are usually added in such minute amounts that they can't change anything.