Question

A dye that is an acid and appears as different colors in its protonated and deprotonated forms can be used as a pH indicator. Suppose that you have a 0.001 M solution of a dye with a p K a of 7.2. From the color, the concentration of the protonated form is found to be 0.0002 M. Assume that the remainder of the dye is in the deprotonated form. What is the pH of the solution?

Answer 1

Answer:

pH of the solution = 7.8

Explanation:

Acid-base indicators show different colours in their protonated (acid form) and deprotonated from (Basic form or salt).

An acid-base indicator dissociates like weak acid as:  

HIn <=> ln- + H+

pH  is calculated using Henderson-Hasselbalch equation,

$pH=p_{Ka} + \frac{[Salt]}{[Acid]}$

Hln = Protonated form (Acid)

ln- = Deprotonated form (Salt)

Given,

Initial concentration of acid = 0.001 M

Protonated form [Hln] = 0.0002 M

Deprotonated form [ln-] = 0.001 - 0.0002 = 0.0008 M

pKa = 7.2

Now, put the values in Henderson-Hasselbalch equation

$pH=7.2 + \frac{[0.0008]}{[0.0002]}$

pH = 7.2 + log4

pH = 7.2 + 0.2020

pH = 7.8