For a concentrated solution of a weak electrolyte ( K eq = equilibrium constant) A 2 B 3 , of concentration ' c ' , the degree of dissociation is [2023]

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Q.
For a concentrated solution of a weak electrolyte ( $K_{eq} =$ equilibrium constant) $A_{2} B_{3}$ , of concentration $' c'$ , the degree of dissociation $' α'$ is [2023]

1 ${(\frac{K_{eq}}{6 c^{5}})}^{1 / 5}$

2 ${(\frac{K_{eq}}{25 c^{2}})}^{1 / 5}$

3 ${(\frac{K_{eq}}{5 c^{4}})}^{1 / 5}$

4 ${(\frac{K_{eq}}{108 c^{4}})}^{1 / 5}$

Ans.
(4)

$\begin{matrix} A_{2} B_{3} & ⇌ & 2 A & + & 3 B \\ Initial conc. & C & 0 & 0 \\ C - C α & 2 C α & 3 C α \end{matrix}$

$K = \frac{{[A]}^{2} {[B]}^{3}}{[A_{2} B_{3}]}$

$K = \frac{108 C^{5} α^{5}}{C} = 108 C^{4} α^{5}$

$α = {[\frac{K}{108 C^{4}}]}^{1 / 5}$

Solution

Q.
For a concentrated solution of a weak electrolyte ( $K_{eq} =$ equilibrium constant) $A_{2} B_{3}$ , of concentration $' c'$ , the degree of dissociation $' α'$ is [2023]

1 ${(\frac{K_{eq}}{6 c^{5}})}^{1 / 5}$

2 ${(\frac{K_{eq}}{25 c^{2}})}^{1 / 5}$

3 ${(\frac{K_{eq}}{5 c^{4}})}^{1 / 5}$

4 ${(\frac{K_{eq}}{108 c^{4}})}^{1 / 5}$

Ans.
(4)

$\begin{matrix} A_{2} B_{3} & ⇌ & 2 A & + & 3 B \\ Initial conc. & C & 0 & 0 \\ C - C α & 2 C α & 3 C α \end{matrix}$

$K = \frac{{[A]}^{2} {[B]}^{3}}{[A_{2} B_{3}]}$

$K = \frac{108 C^{5} α^{5}}{C} = 108 C^{4} α^{5}$

$α = {[\frac{K}{108 C^{4}}]}^{1 / 5}$

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Solution

Q. For a concentrated solution of a weak electrolyte (Keq= equilibrium constant) A2B3, of concentration 'c', the degree of dissociation 'α' is [2023]

1 (Keq6c5)1/5

2 (Keq25c2)1/5

3 (Keq5c4)1/5

4 (Keq108c4)1/5

Ans. (4) A2B3⇌2A+3BInitial conc.C00C-Cα2Cα3Cα K=[A]2[B]3[A2B3] K=108 C5α5C=108 C4α5 α=[K108 C4]1/5

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Q.
For a concentrated solution of a weak electrolyte ( $K_{eq} =$ equilibrium constant) $A_{2} B_{3}$ , of concentration $' c'$ , the degree of dissociation $' α'$ is [2023]

1 ${(\frac{K_{eq}}{6 c^{5}})}^{1 / 5}$

2 ${(\frac{K_{eq}}{25 c^{2}})}^{1 / 5}$

3 ${(\frac{K_{eq}}{5 c^{4}})}^{1 / 5}$

4 ${(\frac{K_{eq}}{108 c^{4}})}^{1 / 5}$

Ans.
(4)

$\begin{matrix} A_{2} B_{3} & ⇌ & 2 A & + & 3 B \\ Initial conc. & C & 0 & 0 \\ C - C α & 2 C α & 3 C α \end{matrix}$

$K = \frac{{[A]}^{2} {[B]}^{3}}{[A_{2} B_{3}]}$

$K = \frac{108 C^{5} α^{5}}{C} = 108 C^{4} α^{5}$

$α = {[\frac{K}{108 C^{4}}]}^{1 / 5}$