Question

A certain acid-base indicator has acid form red and blue is basic form 75% of the indicator is present in the solution in its blue form at pH = 5. ${K_a}$ of the indicator is:
A. ${K_a} = 2 \times {10^{ - 5}}$
B. ${K_a} = 9 \times {10^{ - 5}}$
C. ${K_a} = 3 \times {10^{ - 6}}$
D. ${K_a} = 3 \times {10^{ - 5}}$

Answer 1

Hint: To solve this type of question, first we need to understand what is ${K_a}$ given here. ${K_a}$ is the acid dissociation constant which measures the strength of an acid in a particular solution. The acid dissociation constant is also known as an equilibrium constant for an acid base chemical reaction.

Complete step by step answer:
Let’s consider the acid as ‘A’ and base as ‘B’ to solve this question.
Then, the acid base equilibrium chemical reaction formed will be:
$AB \rightleftharpoons {A^ + } + {B^ - }$
So, from the above acid base equation the acid dissociation constant will be
${K_a} = \dfrac{{[{A^ + }][{B^ - }]}}{{[AB]}}$ (Formula used in this question)
From this formula, we will derive the value of the acidic form
$[{A^ + }] = {10^{ - pH}}$
Here, the pH value is equal to 5 so, we will substitute the pH value to find the acidic form.
Thus, Acidic form $[{A^ + }] = {10^{ - 5}}$
Now, we need to find the basic form which is given 75% of the indicator.
$[{B^ - }] = 75\% $
So, the concentration of the whole solution [AB] will be
⇒(100-75) % = 25%
So, let’s substitute all the values to find out the value of ${K_a}$ of the indicator
${K_a} = \dfrac{{{{10}^{ - 5}} \times 75 \times 100}}{{25 \times 100}}$
So, ${K_a} = 3 \times {10^{ - 5}}$
So, the answer will be option D.

Note:
-As we know that ${K_a}$ is the equilibrium constant for dissociation reaction of an acid in a solution. So, we need to understand the basic difference between equilibrium and dissociation for basic understanding.
-Equilibrium is the state of a reaction in which the forward and reverse reactions' speed is equal.
-Dissociation is the method by which a mixture breaks into its integral ions in an aqueous solution.