posted by George on .
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?
KHIn = (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.
would it be orange?
Is it because it is more acidic?
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?
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