A solution is prepared by mixing 500.mL of 0.10 M NaOCl and 500.mL of 0.20 M HOCl.What is the pH of this solution? [K_a(HOCl)= 3.2 × 10^-8]

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA]), where pKa is the dissociation constant of HOCl.

pKa for HOCl is given: pKa(HOCl) = 7.5.

To use the Henderson-Hasselbalch equation, we need to find [A-] and [HA].

[A-] = [NaOCl] = 0.10 M
[HA] = [HOCl] = 0.20 M

Substituting these values into the Henderson-Hasselbalch equation:

pH = 7.5 + log(0.10/0.20) = 7.19

Therefore, the pH of the solution is 7.19. The best choice is (C) with this explanation.

A Solution Is Prepared by Mixing 500