A 0.61 m aqueous solution of a monoprotic acid (HA) freezes at -1.11°C. Find the pKa of this monoprotic acid. Kf of water = 1.86 °C/m Enter to 3 decimal places.
Solution:
Depression in freezing point (ΔT) is related to molality(m) as,
ΔTf = - i kf m
Where,
i = Vant Hoff factor
Kf = molar depression constant
Thus,
i = -ΔT / kf m
= - (-1.11) °C / 1.86 •C/m x 0.61 m
= 0.98
Thus,
Degree of dissociation (α) = 1 - i = 1 - 0.98 = 0.02
Therefore,
Ka = c α2 / 1 - α
Ka = 0.61 x (0.02)^2 / 1 - 0.02
Ka = 2.49 x 10^-4
pka = - log ka
pka = - log 2.49 x 10^-4 = 4 - 0.396
= 3.604
Thus,
pka = 3.604
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