Question

pH of a weak monobasic acid is 3.2
in its 0.02 M solution. Calculate its
dissociation constant.​

Answer 1

Explanation:

HX⇌H++X−

Given,

pH=2.52

⇒[H+]=10−2.52=3×10−3

α=C[H+]=0.023×10−3=0.151

Kc=1−αCα2

⇒Kc=1−0.1510.02×0.1512=5.37×10−4

Answer 2

Answer in attachment!

Explanation:

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