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? (The question mentions that you can use the ratio of HIn/In).

I NEED this step-by-step. Please don't just give me the answer, please explain how you got there and what you did. I have to find the concentrations of HIn, In and H, correct? I can get the concentration of H from pH, but that's all I can do.

Answer 1

pH = pKa + log (In^-)/(HIn)
Using pKa and pH of 4.00, you can solve for (In/HIn)
Remember the ratio must be at least 10:1 to change color (or 1:10 depending upon how you look at it).