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)

Question

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) (A) 4.76 × 10-5 (B) 4 × 10-5 (C) 8 × 10-5 (D) 2 × 10-5

Tardigrade Admin · Accepted Answer

Δ Tf =i × Kf × m ∴ i=(Δ Tf/Kf × m)=(0.19/1.86 × 0.1)=1.02 Again from, α =(i-1/n-1)=(1.02-1/2-1) =0.02=2.0 × 10-2 for CH 3 COOH longrightarrow CH 3 COO -+ H + K a =C α2 =0.1 ×(2 × 10-3)2 =4 × 10-5

Q. The measured freezing point depression for a $0.1 m$ aqueous $C H_{3} COO H$ solution is $0.1 9^{\circ} C$ . The acid dissociation constant $K_{a}$ at this concentration will be (Given $K_{f}$ , the molal cryoscopic constant $= 1.86 K k g m o l^{- 1}$ )

$4.76 \times 1 0^{- 5}$

$4 \times 1 0^{- 5}$

$8 \times 1 0^{- 5}$

$2 \times 1 0^{- 5}$

Q. The measured freezing point depression for a 0.1m aqueous CH3​COOH solution is 0.19∘C. The acid dissociation constant Ka​ at this concentration will be (Given Kf​, the molal cryoscopic constant =1.86Kkgmol−1)

4.76×10−5

4×10−5

8×10−5

2×10−5

ΔTf​=i×Kf​×m ∴i=Kf​×mΔTf​​=1.86×0.10.19​=1.02 Again from, α=n−1i−1​=2−11.02−1​ =0.02=2.0×10−2 for CH3​COOH⟶CH3​COO−+H+ Ka=Cα2 =0.1×(2×10−3)2 =4×10−5

Q. The measured freezing point depression for a $0.1 m$ aqueous $C H_{3} COO H$ solution is $0.1 9^{\circ} C$ . The acid dissociation constant $K_{a}$ at this concentration will be (Given $K_{f}$ , the molal cryoscopic constant $= 1.86 K k g m o l^{- 1}$ )

$4.76 \times 1 0^{- 5}$

$4 \times 1 0^{- 5}$

$8 \times 1 0^{- 5}$

$2 \times 1 0^{- 5}$