M.Sc. Chemistry (PU-CET P.G) 2017 Panjab University Entrance Exam With Answers

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5.

An acetic acid was neutralized to 90% by adding NaOH. If the pH of the resulting solution is 4.70, the Ka of acetic acid is :

A: 1.795 x 10^-4
B: 1.795 x 10^-5
C: 6.712 x 10^-4
D: 6.712 x 10^-5

The answer is: A

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