The indicator methyl red has a pKHIn = 4.95. It changes from red to yellow over the pH range from 4.4 to 6.2.

Question

If the indicator is placed in a colourless buffer solution of pH = 4.55, what percent of the indicator will be present in the acid form, HIn?

What colour would you expect the solution in question 13 above to most look like?
Select one:
a. red
b. colourless
c. magenta
d. yellow

Answer 1

pH = pKa + log (In^-/HIn)
4.55 = 4.95 + log (In^-/HIn)
-0.4 = log (In^-/HIn)
(In^-/HIn) = 0.398 or
(In^-) = 0.398(HIn) so when (HIn) = 1, (In*-) = 0.398
%HIn = [(HIn)/(HIn) + (In^-)]*100 = [1/(1+0.398)]*100 = ?
What color is it at this pH? I don't know because I've never done a titration like that however, it is largely red. So red (a) is mostly right. It will not be b or d. You would expect it to be red to yellow or orange at about 5 or 5.1 So the almost all red will have some yellow or yellow orange mixed in with it. I don't think magenta has orange in it but I don't know. If I had to answer I would choose a as the answer since I don't have a clue what magenta is. :-)

Answer 2

Disregard this answer.

Answer 3

I believe this is an easier way to explain.
pH = pKa + log [(In^-)/(HIn)]
4.55 = 4.95 + log[(In^-)/(HIn)]
-0.4 = log [(In^-)/(HIn)]
In^-/HIn = 0.398 and
In^- = 0.398*HIn is equation 1
equation 2 is In^- + HIn = 1.0
solve the two simultaneously to obtain
0.398 HIn + HIn = 1
HIn = 1/1.398 = 0.715 and
In^- = 1 - 0.715 = 0.285
So HIn percent = 71.5% and In^- = 28.5%
For the color:
When pH = 4.95 we have base/acid = 1 so acid and base are @ 50%.
At the pKa value of 4.95 we have a 50-50 mixture of red and yellow which makes orange. I am told that mixtures of red and yellow will not give magenta so I'm going with a red color @ pH 4.55 which is about 70% red and about 30% yellow