Question

A weak acid, HA, has a pK, of 4.971. If a solution of this acid has a pH of 4.437, what percentage of the acid is not ionized? Assume all H+ in the solution came from the ionization of HA. percentage not ionized:

Answer 1

Solution: - The aud HA will ionize as, A HA cag) - Ht lag & A laq), ka at t=0 C 0 0 at tit (1-4) ca ea We have Pkg = 4.971, So kg = 1.061155 and PH = 4.437 , so [ht] = 3: 65x105 le (&= 3,65x105m. wherre (= concentration, a = degree of dissociation now Ka= [HA] [A] [HAD

. Ka= ca ² 11.06810 5 = (3.65X105). 0:29 (1-4)=& So, a = 0.22 le then 22%. Of the auld get dissovated 78% does not dissociated.
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