Question

The ionization constant Ka of an indicator HIn is 1.0x10^-6. The color of the nonionized form is red and that of the ionized form is yellow. What is the color of this indicator in a solution whose pH is 4.00?

Thanks!

Answer 1

K_HIn = (H^+)(In^-)/(HIn) = 1E-6

Solve for (In^-)/(HIn) = Ka/(H^+) =
1E-6/1E-4 = 1E-2 = 0.01
(HIn)(red) = 100(In^-)(yellow)
So what color is it?
Check my thinking.

Answer 2

would it be orange?

Answer 3

Is it because it is more acidic?

Answer 4

in my text book it says that if the calculated value is less than or equal to 0.1 then the color of the conjugate base (In- in this case) predominates. does this mean that the solution will be yellow then?

Answer 5

Covered under later listings.

Answer 6

the ratio of Hln:ln- is 100:1, much greater than 10, therefore it will be yellow

Answer 7

Disregard what I said yesterday, it was incorrect. The ratio of is correct, but that actually means it will be red, because the acid predominates (not enough ionization to change the color)