For a concentrated solution of a weak electro | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The dissociation reaction is: $A_2B_3 (aq.) \rightleftharpoons 2 A^{3+} (aq.) + 3 B^{2-} (aq.)$ At equilibrium,the concentrations are: $[A_2B_3] =

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The dissociation reaction is: $A_2B_3 (aq.) \rightleftharpoons 2 A^{3+} (aq.) + 3 B^{2-} (aq.)$ At equilibrium,the concentrations are: $[A_2B_3] = c(1 - \alpha)$,$[A^{3+}] = 2c\alpha$,and $[B^{2-}] = 3c\alpha$. The equilibrium constant expression is: $K_{eq} = \frac{[A^{3+}]^2 [B^{2-}]^3}{[A_2B_3]} = \frac{(2c\alpha)^2 (3c\alpha)^3}{c(1 - \alpha)}$. For a weak electrolyte,$\alpha \ll 1$,so $(1 - \alpha) \approx 1$. Thus,$K_{eq} = \frac{(4c^2\alpha^2)(27c^3\alpha^3)}{c} = 108c^4\alpha^5$. Solving for $\alpha$: $\alpha^5 = \frac{K_{eq}}{108c^4} \implies \alpha = \left(\frac{K_{eq}}{108 c^4}\right)^{\frac{1}{5}}$.

For a concentrated solution of a weak electrolyte ($K_{eq} = $ equilibrium constant) $A_2B_3$ of concentration '$c$',the degree of dissociation '$\alpha$' is

Similar Questions

What is the ratio of the number of $H^{+}$ ions to the number of $H_2O$ molecules in $1 \ L$ of water?

At $25\, ^oC$,the dissociation constant of $CH_3COOH$ and $NH_4OH$ in aqueous solution are almost the same. The $pH$ of a $0.01\, N$ $CH_3COOH$ solution is $4.0$ at $25\, ^oC$. The $pH$ of a $0.01\, N$ $NH_4OH$ solution at the same temperature would be:

Calculate the ionic product of the solution formed by mixing equal volumes of $0.004 \, M \, CaSO_4$ and $0.002 \, M \, H_2SO_4$.

$A$ solution contains $8 \, g$ of $NaOH$ and $4.9 \, g$ of $H_2SO_4$ in $1 \, L$ of solution. What is its $pH$?

$5\%$ ionization occurs in $0.01 \ M$ $CH_3COOH$ solution. Calculate its dissociation constant.

Vedclass Test Series

Exam Paper Generator

Online Exam Module