Question

1) Phenol red is an indicator that changes color over a range from pH=6.4 to pH=8.0. What is Ka of phenol red? Express to 1 sig fig

(I put 0.2 but it's wrong)

2) Alizarin yellow R, Ka=7.9x10^-12, is yellow in its protonated form (HX) and red in its ionized form (X-). At what pH will alizarin yellow R be a perfect orange color?

Express to 1 dec place

Answer 1

Ans: 1. The pH obtained as:

pH = pK_a + log[X^-]/[HX]

X- = deprotonated form of the indicator

HX = protonated form of the indicator

When a single indicator is used, pH range of the indicator is taken as (pK_a ± 1).

So, for phenol red we can take the middle point f the pH range as the value of pK_a. Middle point of 6.4-8 is 7.2.

So pK_a = 7.2

K_a = 10^-7.2 = 6.3*10^-8

2. When [X-]=[HX] the color of the solution will be a perfect orange.

pH = pK_a + log[X^-]/[HX]

= pK_a + log[HX]/[HX]

= pK_a

= - log(7.9x10^-12)

= 11.1