At what concentration of 0.2 M CH_3COOH will | vedclass

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

Question

(A) For a weak acid,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$.Let the initial concentration be $C_1 = 0.2 \ M$ a

Vedclass · Accepted Answer

$0.05$

Vedclass · Answer

(A) For a weak acid,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$.Let the initial concentration be $C_1 = 0.2 \ M$ and the initial degree of dissociation be $\alpha_1$.Then $\alpha_1 = \sqrt{\frac{K_a}{C_1}}$.We want the new degree of dissociation $\alpha_2 = 2\alpha_1$.So,$\alpha_2 = \sqrt{\frac{K_a}{C_2}} = 2 \sqrt{\frac{K_a}{C_1}}$.Squaring both sides,we get $\frac{K_a}{C_2} = 4 \frac{K_a}{C_1}$.This simplifies to $C_2 = \frac{C_1}{4}$.Substituting $C_1 = 0.2 \ M$,we get $C_2 = \frac{0.2}{4} = 0.05 \ M$.

At what concentration of $0.2 \ M$ $CH_3COOH$ will its degree of dissociation become double (in $M$)? (Given: $K_a = 1.8 \times 10^{-5}$ for $CH_3COOH$)

Similar Questions

Nicotinic acid $(K_a = 10^{-5})$ is represented by the formula $HNiC$. Calculate the percentage of dissociation in a solution containing $0.1 \ mol$ of nicotinic acid in $2 \ L$ of solution.

The percentage of pyridine $(C_5H_5N)$ that forms pyridinium ion $(C_5H_5N^{+}H)$ in a $0.10 \ M$ aqueous pyridine solution ($K_b$ for $C_5H_5N = 1.7 \times 10^{-9}$) is (in $\%$)

Ionisation constant of $CH_3COOH$ is $1.7 \times 10^{-5}$ and concentration of $H^{+}$ ions is $3.4 \times 10^{-4} \ M$. Find the initial concentration of $CH_3COOH$ molecules.

$pH$ of $0.05 \ M$ aqueous solution of diethyl amine is $10$. The value of equilibrium constant $(K_b)$ is:

The dissociation constant of a weak monobasic acid is $2.7 \times 10^{-5}$. If the degree of dissociation of the acid is $3 \times 10^{-2}$,what is the concentration of the acid (in $M$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module