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Standard XII

Chemistry

Stoichiometric Calculations

The boiling p...

Question

The boiling point of a 0.5 molal aqueous solution of $N a H S O_{4}$ is $100.64^{\circ} C$ . What is the dissociation constant for the following reaction? $[K_{b} of H_{2} O = 0.512 K k g m o l^{- 1}]$
$H S O_{4}^{-} ⇌ H^{+} + S O_{4}^{2 -}$

0.04

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0.2

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0.125

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0.25

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Solution

The correct option is D 0.25
$N a H S O_{4}$ will dissociate as,
$N a H S O_{4} \to N a^{+} + H S O_{4}^{-}$
$0.5 0.5 0.5$
Again $H S O_{4}^{-}$ will dissociate as,
$\begin{matrix} H S O_{4}^{-} & ⇌ & H^{+} + & S O_{4}^{2 -} 0.5 (1 - α) & 0.5 α & 0.5 α \end{matrix}$
$m = 0.5 + 0.5 (1 + α)$
$Δ T_{b} = K_{b} i m$
$\Rightarrow 0.64 = 0.512 \times 0.5 (2 + α)$
$\Rightarrow α = 0.5$
$Again K_{a} = \frac{C α^{2}}{1 - α}$
$= \frac{0.5 \times (0.5)^{2}}{1 - 0.5}$
$= 0.25$

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Similar questions

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 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

molal solution of

N a H S O

- {0.372}^{0}

C. Calculate the dissociation constant for the reaction.

H S O_{4} ⇌ H^{\oplus} + S O_{4}^{2 -}

K_{f}

for water =

1.86 K m^{- 1}

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)

A B_{2}

10 %

dissociated in water to

A^{2 +}

and

B^{-}

. The boiling point of a

10.0

molal aqueous solution of

A B_{2}

^{\circ} C .

(Round off to the nearest Integer).
[Given: Molal elevation constant of water

K_{b} = 0.5 {K kg mol}^{- 1}

, boiling point of pure water

= 100^{\circ} C

]

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The boiling point of a 0.5 molal aqueous solution of NaHSO4 is 100.64 ∘C. What is the dissociation constant for the following reaction? [Kb of H2O=0.512 Kkgmol−1] HSO−4⇌H++SO2−4

The correct option is D 0.25NaHSO4 will dissociate as, NaHSO4→Na++HSO−4 0.5 0.5 0.5 Again HSO−4 will dissociate as, HSO−4⇌H++SO2−40.5(1−α)0.5α0.5α m=0.5+0.5(1+α) ΔTb=Kbim ⇒0.64=0.512×0.5(2+α) ⇒α=0.5 Again Ka=Cα21−α =0.5×(0.5)21−0.5 =0.25

The boiling point of a 0.5 molal aqueous solution of $N a H S O_{4}$ is $100.64^{\circ} C$ . What is the dissociation constant for the following reaction? $[K_{b} of H_{2} O = 0.512 K k g m o l^{- 1}]$
$H S O_{4}^{-} ⇌ H^{+} + S O_{4}^{2 -}$