NEET Questions Solved

NEET - 2007

A weak add. HA, has a Ka of 1.00 x 10-5 . If 0.100 mole of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to :

(a) 99.0%

(b) 1.00%

(c) 99.9%

(d) 0.100%

HA H+ + A-

Ka = [H+][A+]/[HA] = [H+]2/[HA]

[H+] = Ka[HA] =1x10-5x0.1=1x10-6 = 1x10-3

α = actual ionization / molar concentration = 10-3 / 0.1 = 10-2

% of acid dissociated = 10-2 x 100 = 1%

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