1. Tardigrade
  2. Question
  3. Chemistry
  4. The measured freezing point depression for a 0.1 m aqueous CH3COOH solution is 0.19° C. The acid dissociation constant Ka at this concentration will be (Given Kf, the molal cryoscopic constant = 1.86 K kg mol-1)

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Q. The measured freezing point depression for a $0.1 \,m$ aqueous $CH_3COOH$ solution is $0.19^{\circ}\,C$. The acid dissociation constant $K_a$ at this concentration will be (Given $K_f$, the molal cryoscopic constant $= 1.86\, K \,kg \,mol^{-1}$)

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Solution:

$ \Delta T_{f} =i \times K_{f} \times m $
$ \therefore i=\frac{\Delta T_{f}}{K_{f} \times m}=\frac{0.19}{1.86 \times 0.1}=1.02$
Again from, $\alpha =\frac{i-1}{n-1}=\frac{1.02-1}{2-1} $
$=0.02=2.0 \times 10^{-2} $
for $CH _{3} COOH \longrightarrow CH _{3} COO ^{-}+ H ^{+}$
$ K a =C \alpha^{2} $
$=0.1 \times\left(2 \times 10^{-3}\right)^{2} $
$=4 \times 10^{-5} $