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Depression in freezing point of 0 .01 molal aqueous solution of HA is 0 .01924 K . The p K a value of HA is [Given = molarity = molality, K f H 2 O = 1 .85 Kkg mol − 1 , log 2 = 0 .3

Correct answer

4.80

Step-by-step solution

Δ T f = i × m × K f 0 .01924 = i × 0 .01 × 1 .85 i = 1 .04 i = 1 + α(n − 1) 1 .04 = 1 + α(2 − 1) α = 0.04 K a = α 2 c = 0 .04 2 × 0 .01 = 16 × 10 − 6 pK a = 4 .8

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