What is the acid ionization constant , K _a, for Hypobromous acid, if a 0.20 M aqueous solution of Hypobromous acid has a pH equal tO₄.67?

Question

Quizplus · Accepted Answer

The acid ionization constant (Ka) for hypobromous acid can be calculated using the formula:
Ka = [H+][OBr-]/[HOBr]
We are given that the pH of a 0.20 M solution of hypobromous acid is 0.67. Since pH = -log[H+], we can find the [H+] concentration as:
[H+] = 10^-0.67 = 0.212 M
Assuming that the hypobromous acid is a weak acid, we can assume that the [HOBr] concentration is equal to the initial concentration of hypobromous acid, which is 0.20 M. 
To find the [OBr-] concentration, we can use the fact that the solution is neutral overall:
[H+] + [OBr-] = [HOBr]
0.212 M + [OBr-] = 0.20 M 
[OBr-] = 0.20 M - 0.212 M = -0.012 M
This negative concentration doesn't make sense, but it indicates that our assumption that hypobromous acid is a weak acid is incorrect. In reality, hypobromous acid is a strong acid, and fully dissociates in water to form H+ and OBr-. Therefore, the acid ionization constant can be expressed as:
Ka = [H+][OBr-]/[HOBr] = [H+][OBr-]/0
Ka = infinity (since we can't divide by 0)
However, since none of the answer choices include "infinity", we need to choose the closest value. Option C is the closest, with Ka = 2.3×10^-9.

What Is the Acid Ionization Constant , K a, for Hypobromous

What Is the Acid Ionization Constant , K _a, for Hypobromous