1.

PH of 0.05 monobasic acid is 6.2. The % dissociation of it is?A) 1.29B) 0.029C) 2.12D) 0.04Please answer​

Answer»

Explanation:

ph=6.2

so, Ka=10^-6.2 .......... {^}denotes = power

Now, DISSOCIATION =Ka/concentration

=10^-6.2/0.05

approximately =0.004

so, %=0.004*100

=0.4

This is CORRECT answer.

please recheck your OPTIONS.



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