Chemistry, asked by samyrabudhwani8368, 9 months ago

Ph of a weak monobasic acid is 3.2 in its 0.02 m solution. calculate its dissociation constant

Answers

Answered by sanapalassb

46

Answer:

pH=- log[H+]=2.52. Therefore [H+]=10^-2.52= 3.02 x10^-3 , [H+]=CX where C= concentration and X=degree of dissociation. =3.02 x 10^-3/0.02= 0.151.

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