0.01(M) solution an acid HA freezes at-0.0205°C If Kf for water is 1.86 K kg mol-1, the ionization constant of the conjugate base of the acid will be (consider molarity ≃ molality)

Question

0.01(M) solution an acid HA freezes at-0.0205°C If Kf for water is 1.86 K kg mol-1, the ionization constant of the conjugate base of the acid will be (consider molarity ≃ molality) (A) 1.1 × 10-4 (B) 1.1 × 10-3 (C) 9× 10-11 (D) 9× 10-12

Tardigrade Admin · Accepted Answer

Δ Tf =kf × m =1.86×0.01=0.0186 i=(0.0205/0.0186)=1.10+1+α or α=0.1 Ka=(Cα2/1-α) =(0.01×(0.1)2/1-0.1)=(1/9)×10-3 or Kb=9×10-11

Q. $0.01 (M)$ solution an acid HA freezes at- $0.020 5^{\circ} C$ If $K_{f}$ for water is $1.86 K k g m o l^{- 1}$ , the ionization constant of the conjugate base of the acid will be (consider molarity $≃$ molality)

$1.1 \times 1 0^{- 4}$

$1.1 \times 1 0^{- 3}$

$9 \times 1 0^{- 11}$

$9 \times 1 0^{- 12}$

$Δ T_{f} = k_{f} \times m = 1.86 \times 0.01 = 0.0186$
$i = \frac{0.0205}{0.0186} = 1.10 + 1 + α or α = 0.1$
$K_{a} = \frac{C α ^{2}}{1 - α}$
$= \frac{0.01 \times ( 0.1 ) ^{2}}{1 - 0.1} = \frac{1}{9} \times 1 0^{- 3}$
or $K_{b} = 9 \times 1 0^{- 11}$

Q. 0.01(M) solution an acid HA freezes at-0.0205∘C If Kf​ for water is 1.86Kkgmol−1, the ionization constant of the conjugate base of the acid will be (consider molarity ≃ molality)

1.1×10−4

1.1×10−3

9×10−11

9×10−12