A 0.595 m aqueous solution of a monoprotic acid (HA) freezes at -2.21°C. Find the pKa of this monoprotic acid. Kf of water = 1.86 °C/m
freezing point depression=delta T
here: delta Tf = 2.21 oC
usw:
delta Tf = i kf molality
so, i=frezing point/(kf * molality)
i=2.21/(1.86*0.595)
i=1.99
Now use:
for dissociation,
degree of dissociation=x=i-1
x=1.99-1
x=0.99
HA --> H+ + A-
c 0 0 (initial)
c-cx cx cx (at equlibrium)
use:
ka=(c*x)2 / (c - c*x)
Ka= (0.595*0.99)^2 / (0.595 - 0.595*0.99)
Ka = 58.3
Now use:
pKa = -log Ka
= -log (58.3)
= -1.76
Answer: -1.76
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