K_{a_1}, K_{a_2} and K_{a_3} are the respecti | vedclass

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

Question

(D) The ionization constant for reaction $(a)$ is $K_{a_1} = \frac{[H^{+}][HC_2O_4^-]}{[H_2C_2O_4]}$.The ionization constant for reaction $(b)$ is $K_

Vedclass · Accepted Answer

$K_{a_3} = K_{a_1} 	imes K_{a_2}$

Vedclass · Answer

(D) The ionization constant for reaction $(a)$ is $K_{a_1} = \frac{[H^{+}][HC_2O_4^-]}{[H_2C_2O_4]}$.The ionization constant for reaction $(b)$ is $K_{a_2} = \frac{[H^{+}][C_2O_4^{2-}]}{[HC_2O_4^-]}$.The reaction $(c)$ is the sum of reactions $(a)$ and $(b)$.For reaction $(c)$,the equilibrium constant $K_{a_3} = \frac{[H^{+}]^2[C_2O_4^{2-}]}{[H_2C_2O_4]}$.Multiplying $K_{a_1}$ and $K_{a_2}$ gives: $K_{a_1} 	imes K_{a_2} = \frac{[H^{+}][HC_2O_4^-]}{[H_2C_2O_4]} 	imes \frac{[H^{+}][C_2O_4^{2-}]}{[HC_2O_4^-]} = \frac{[H^{+}]^2[C_2O_4^{2-}]}{[H_2C_2O_4]} = K_{a_3}$.Thus,$K_{a_3} = K_{a_1} 	imes K_{a_2}$.

Similar Questions

State the law of chemical equilibrium for an equilibrium process.

In the reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $B$ was $1.5$ times that of $[A]$. At equilibrium,the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is:

For the system $A_{(g)} + 2B_{(g)} \rightleftharpoons C_{(g)}$,the equilibrium concentrations are $[A] = 0.06 \ mol/L$,$[B] = 0.12 \ mol/L$,and $[C] = 0.216 \ mol/L$. The $K_{eq}$ for the reaction is:

For the reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$,the equilibrium mixture is given. If $1 \ mol$ of $H_2S$,$0.2 \ mol$ of $H_2$,and $0.8 \ mol$ of $S_2$ are taken in a $2 \ L$ vessel,find the value of $K_c$.

If the equilibrium constant for the reaction $A + 3B \rightleftharpoons 2C$ is $K_1$,then the equilibrium constant $K_2$ for the reaction $2C \rightleftharpoons A + 3B$ should be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module