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Standard XII

Chemistry

Degree of Dissociation

Consider a we...

Question

Consider a weak electrolyte, $A B_{3}$ having a concentration of 0.1 M in solution. If the degree of dissociation is 0.2, Calculate the ionisation constant $K_{i}$ of $A B_{3}$ .

2.4 \times 10^{- 9}

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1.4 \times 10^{- 2}

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3.4 \times 10^{- 7}

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5.4 \times 10^{- 5}

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Solution

The correct option is D $5.4 \times 10^{- 5}$
$A B_{3}$ dissociates as:

$A B_{3} (a q .) ⇌ A^{3 +} (a q .) + 3 B^{-} (a q .) I n i t i a l : C 00 E q u i l i b r i u m : C (1 - α) C α 3 C α$

$K_{i} = \frac{[A^{3 +}] [B^{-}]^{3}}{[A B_{3}]} K_{i} = \frac{C α \times (3 C α)^{3}}{C (1 - α)} = \frac{(27 α) (C α)^{3}}{(1 - α)} K_{i} = \frac{27 \times 0.2 \times (0.1 \times 0.2)^{3}}{(1 - 0.2)} K_{i} = 5.4 \times 10^{- 5}$

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Degree of Dissociation

Standard XII Chemistry

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Consider a weak electrolyte, AB3 having a concentration of 0.1 M in solution. If the degree of dissociation is 0.2, Calculate the ionisation constant Ki of AB3.

The correct option is D 5.4×10−5 AB3 dissociates as: AB3 (aq.)⇌A3+ (aq.)+3 B− (aq.)Initial: C 0 0Equilibrium: C(1−α) Cα 3 Cα Ki=[A3+][B−]3[AB3]Ki=Cα×(3Cα)3C(1−α)=(27α)(Cα)3(1−α)Ki=27×0.2×(0.1×0.2)3(1−0.2)Ki=5.4×10−5

Consider a weak electrolyte, $A B_{3}$ having a concentration of 0.1 M in solution. If the degree of dissociation is 0.2, Calculate the ionisation constant $K_{i}$ of $A B_{3}$ .