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?


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  1. 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.

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  2. would it be orange?

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  3. Is it because it is more acidic?

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  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?

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  5. Covered under later listings.

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