The Ka of a certain indicator is 4 ×10−7. The color of HIn is green and that of In− is red. A few drops of the indicator are added to an HCl solution, which is then titrated against an NaOH solution. Over which pH range will the indicator change color?

7.4 - 5.4.

Explanation:

The relationship between HIn that is the non-ionized form and the In^-; the ionized form is an equilibrium Reaction which is given below;

HIn(aq) + H2O(l) <> H3O^+(aq). + In^-(aq).

From the question, we are given that the ka is 4 ×10^−7, therefore, the mathematical relationship between HIn and In^- is given below;

Ka = [H3O^+] [In^-] / [HIn] = 4 ×10^−7.

The formula we are going to be using to solve this question is given below;

pH = pKa (+ or -) 1.

Recall that the relationship between the ka and pKa;

pKa = - log (ka).

pKa = - log ( 4 ×10^−7).

pKa = 6.4

Therefore, for it to have a distinct color change, the pH range will be;

pH = pKa (+ or -) 1.

pH= 6.4 + 1 = 7.4.

pH= 6.4 - 1 = 5.4.

Hence, the pH range = 7.4 - 5.4.

the answer is option a.