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Standard XII

Chemistry

Abnormal Colligative Properties

The freezing ...

Question

The freezing point of a 0.08 molar aqueous solution of $N a H S O_{4}$ is $- {0.372}^{\circ} C$ . The dissociation constant for the following reaction is:
( $K_{f}$ for water= 1.86 $K k g m o l^{- 1}$ )
$H S O_{4}^{-} \to H^{+} + S O_{4}^{- 2}$ .
(Assume molarity and molality to be same)

0.04

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0.02

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0.01

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0.2

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Solution

The correct option is A 0.04

$0.08 M N a H S O_{4} = 0.08 M N a^{+} + 0.08 M H S O_{4}^{-}$

$H S O_{4}^{-} \to H^{+} S O_{4}^{- 2} 0.08 M (1 - α) 0.08 M α 0.08 M α$

Effective molality = $[N a^{+}] + i [H S O_{4}^{-}] = 0.08 m + 0.08 m (1 + α)$

$Δ T_{f} = K_{f} i m \Rightarrow 0.372 = 1.86 \times 0.08 (2 + α)$

$α = 0.5$
So, $K_{a} = \frac{c α^{2}}{1 - α} = \frac{0.08 \times (0.5)^{2}}{1 - 0.5} = \frac{0.02}{0.5} = 0.04$

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Similar questions

Q. The freezing point of a 0.08 molar aqueous solution of

N a H S O_{4}

- {0.372}^{\circ} C

. The dissociation constant for the following reaction is:
(

K_{f}

for water= 1.86

K k g m o l^{- 1}

)

H S O_{4}^{-} \to H^{+} + S O_{4}^{- 2}

.
(Assume molarity and molality to be same)

Q. The freezing point of a 0.08 molar aqueous solution of

N a H S O_{4}

- {0.372}^{\circ} C

. The dissociation constant for the following reaction is:
(

K_{f}

for water= 1.86

K k g m o l^{- 1}

)

H S O_{4}^{-} \to H^{+} + S O_{4}^{- 2}

.
(Assume molarity and molality to be same)

Q. The freezing point of an aqueous solution of

N a H S O_{4}

- 0.372^{o} C

. The dissociation constant for the reaction

H S O_{4}^{-} \to H^{+} + S O_{4}^{- 2}

is 0.4. Find the concentration of

N a H S O_{4}

in mol/kg.

[

K_{f} f o r H_{2} O = 1.86 K k g m o l^{- 1}

]
(Assume dilute solution and take molality = molarity)

Q. The freezing point of a

0.08

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N a H S O

- {0.372}^{0}

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K_{f}

for water =

1.86 K m^{- 1}

Q. A

0.025 M

aqueous solution of a monobasic acid has a freezing point of

- {0.06}^{\circ} C

.
Calculate acid dissociation constant

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(K_{f})_{H_{2} O} = 1.86 K k g m o l^{- 1}

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The freezing point of a 0.08 molar aqueous solution of NaHSO4 is −0.372∘C. The dissociation constant for the following reaction is: (Kf for water= 1.86Kkgmol−1) HSO−4→H++SO−24. (Assume molarity and molality to be same)

The correct option is A 0.04 0.08 M NaHSO4=0.08 M Na++0.08 M HSO−4 HSO−4 →H+ SO−240.08 M(1−α) 0.08Mα 0.08Mα Effective molality = [Na+]+i[HSO−4]=0.08m+0.08m(1+α) ΔTf=Kfim⇒0.372=1.86×0.08(2+α) α=0.5 So, Ka=cα21−α=0.08×(0.5)21−0.5=0.020.5=0.04

The freezing point of a 0.08 molar aqueous solution of $N a H S O_{4}$ is $- {0.372}^{\circ} C$ . The dissociation constant for the following reaction is:
( $K_{f}$ for water= 1.86 $K k g m o l^{- 1}$ )
$H S O_{4}^{-} \to H^{+} + S O_{4}^{- 2}$ .
(Assume molarity and molality to be same)