Dissociation constant for a monobasic acid is | vedclass

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

Question

(C) For a monobasic acid,the dissociation is given by $HA \rightleftharpoons H^{+} + A^{-}$.Given,degree of dissociation $\alpha = 2\,\% = 0.02$.Disso

Vedclass · Accepted Answer

$2.3$

Vedclass · Answer

(C) For a monobasic acid,the dissociation is given by $HA ightleftharpoons H^{+} + A^{-}$.Given,degree of dissociation $\alpha = 2\,\% = 0.02$.Dissociation constant $K_{a} = C \alpha^{2} = 10^{-4}$.$C = \frac{K_{a}}{\alpha^{2}} = \frac{10^{-4}}{(0.02)^{2}} = \frac{10^{-4}}{4 	imes 10^{-4}} = 0.25 \ M$.Concentration of $H^{+}$ ions is $[H^{+}] = C \alpha = 0.25 	imes 0.02 = 0.005 \ M = 5 	imes 10^{-3} \ M$.$pH = -\log[H^{+}] = -\log(5 	imes 10^{-3}) = 3 - \log 5$.Since $\log 5 \approx 0.7$,$pH = 3 - 0.7 = 2.3$.

Dissociation constant for a monobasic acid is $10^{-4}$. What is the $pH$ of the monobasic acid? (If $\%$ dissociation $= 2\,\%$)

Similar Questions

The $pH$ of a $0.1 \ M$ aqueous solution of a very weak acid $(HA)$ is $3$. What is its degree of dissociation in percentage?

Calculate $[H_3O^{+}]$ of a monobasic acid if it is $0.04 \%$ dissociated in $0.05 \ M$ solution.

If the $pH$ of $0.10 \ M$ monobasic acid at $298 \ K$ is $5.0$,the value of $p K_a$ at the same temperature is

Dissociation constant of $0.01 \ M$ weak acid is $10^{-4}$. What is percent dissociation of acid (in $\%$)?

For a weak base $BOH$ at $25\,^oC$,the dissociation constant $K_b$ is $1.0 \times 10^{-12}$. The concentration of hydroxyl ions in its $0.01\,M$ aqueous solution will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module