The ionisation constant of acetic acid is 1.8 | vedclass

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

Question

(B) For a weak acid,the dissociation constant $K_a$ is related to the degree of dissociation $\alpha$ and concentration $C$ by the formula: $K_a = C\a

Vedclass · Accepted Answer

$0.045$

Vedclass · Answer

(B) For a weak acid,the dissociation constant $K_a$ is related to the degree of dissociation $\alpha$ and concentration $C$ by the formula: $K_a = C\alpha^2 / (1 - \alpha)$.Given $K_a = 1.8 	imes 10^{-5}$ and $\alpha = 2\% = 0.02$.Since $\alpha$ is very small,we can approximate $1 - \alpha \approx 1$.Thus,$K_a \approx C\alpha^2$.Substituting the values: $1.8 	imes 10^{-5} = C 	imes (0.02)^2$.$1.8 	imes 10^{-5} = C 	imes 0.0004$.$C = (1.8 	imes 10^{-5}) / (4 	imes 10^{-4}) = 0.45 	imes 10^{-1} = 0.045 \ M$.

The ionisation constant of acetic acid is $1.8 \times 10^{-5}$. The concentration at which it will be dissociated to $2\%$,is (in $M$)

Similar Questions

Derive the expression for the ionization constant $(K_b)$ of a weak base.

Sulphurous acid $(H_{2}SO_{3})$ has $Ka_{1} = 1.7 \times 10^{-2}$ and $Ka_{2} = 6.4 \times 10^{-8}$. The $pH$ of $0.588 \ M \ H_{2}SO_{3}$ is ..... . (Round off to the Nearest Integer)

$0.02 \, M$ monobasic acid dissociates $2 \%$,hence,$pH$ of the solution is

Select the $pK_a$ value of the strongest acid from the following.

$A$ weak monobasic acid is $2 \%$ dissociated in its $0.1 \ M$ solution. What is its dissociation constant?

Vedclass Test Series

Exam Paper Generator

Online Exam Module