. HF H+ + OH-
Initial 1.2 M. 0 0
Change. -x +x +x
Change. (1.2-x)M xM x M
Given pH = 1.551
Since pH = - log [H+]
H+ = 0.0281 M
Percentage ionisation of the acid = (acid amount ionized/ initial acid )X100= (0.0281 M/1.2 M)X100= 2.34 Answer
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