Asked by Carlos
A 0.649 m aqueous solution of a monoprotic acid (HA) freezes at -2.34°C. Find the pKa of this monoprotic acid. Kf of water = 1.86 °C/m
Answers
Answered by
Scott
Ka = [H+][A-] / [HA]
2.34 / 1.86 = 1.26 M
2 x + (.649 - x) = 1.26
Ka = x^2 / (.649 - x)
pKa = -log(Ka)
2.34 / 1.86 = 1.26 M
2 x + (.649 - x) = 1.26
Ka = x^2 / (.649 - x)
pKa = -log(Ka)
Answered by
DrBob222
dT = i*Kf*m
i = 2.34/1.86*0.649
i = 1.938 so it's not quite 2.0 which means it's 93.8% ionized.
........HA ==> H^+ + A^-
I....0.649.....0.....0
C......-x......x.....x
E....0.649-x...x.....x
You know x = 0.938*0.649 = 0.609 = (H^+). Plug those numbers into Ka expression and solve for Ka, then convert to pKa.
i = 2.34/1.86*0.649
i = 1.938 so it's not quite 2.0 which means it's 93.8% ionized.
........HA ==> H^+ + A^-
I....0.649.....0.....0
C......-x......x.....x
E....0.649-x...x.....x
You know x = 0.938*0.649 = 0.609 = (H^+). Plug those numbers into Ka expression and solve for Ka, then convert to pKa.
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