Question

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

Answer 1

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.