Answer to Question #115939 in Physical Chemistry for Kimberly

Question

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.

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?

Answer 1

From conditions:

K(HIn) = [H+]*[In-]/[HIn]

pK(HIn) = pH + p([In-]/[HIn])

4.95 = 4.55 + p([In-]/[HIn])

p([In-]/[HIn]) = 0.40

[In-]/[HIn] = 10^(-0.40) = 0.398

[In-] = 0.398*[HIn]

Percentage of HIn is

[HIn]/([HIn] + [In-])*100% =

= [HIn]/([HIn] + 0.398*[HIn])*100% =

= 1/(1 + 0.398)*100% = 71.5 %

Answer: percentage of HIn is 71.5%