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A 0.61 m aqueous solution of a monoprotic acid (HA) freezes at -1.11°C. Find the pKa...

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.

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Answer #1

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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