Homework Answers
Answer: Option (E) 4.47
Calculation is shown below:
For HClO:
Given, Ka = 3.0 × 10-8, T = 25°C, CHClO = 0.0385
[H3O+] = √(Ka×CHClO) = √(3.0×10-8×0.0385) = 3.3 ×10-5
pH = -log [H3O+] = -log [3.3 ×10-5] = 4.48 ~ 4.47
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