Calculate the pH of 0.02 M monobasic acid ha | vedclass

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

Question

(A) For a monobasic acid $HA$,the dissociation is given by: $HA \rightleftharpoons H_{(aq)}^{+} + A_{(aq)}^{-}$Concentration of $H^{+}$ ions is calcul

Vedclass · Accepted Answer

$3.4$

Vedclass · Answer

(A) For a monobasic acid $HA$,the dissociation is given by: $HA ightleftharpoons H_{(aq)}^{+} + A_{(aq)}^{-}$Concentration of $H^{+}$ ions is calculated as $[H^{+}] = \alpha 	imes C$,where $\alpha$ is the degree of dissociation and $C$ is the molar concentration.Given $\alpha = 2 \% = 0.02$ and $C = 0.02 \ M$.$[H^{+}] = 0.02 	imes 0.02 = 0.0004 \ M = 4 	imes 10^{-4} \ M$.$pH = -\log[H^{+}] = -\log(4 	imes 10^{-4}) = 4 - \log(4) = 4 - 0.602 = 3.398 \approx 3.4$.

Calculate the $pH$ of $0.02 \ M$ monobasic acid having $2 \%$ dissociation.

Similar Questions

What is the percent ionization $(\alpha)$ of a $0.01 \ M \ HA$ solution? $......\%$ $(K_a = 10^{-6})$

$A$ $0.1 \ M$ solution of $HF$ is $1 \%$ ionized. What is the $K_a$?

For a $0.1 \ M$ aqueous pyridine solution,if the pyridinium ion concentration is produced such that the degree of dissociation is $0.013 \%$,calculate the concentration of the pyridinium ion.

$A$ weak monobasic acid dissociates to $0.001 \%$ in its $0.01 \ M$ solution. What is its dissociation constant?

$A$ weak base is $1.42 \%$ dissociated in its $0.05 \ M$ solution. Calculate its dissociation constant.

Vedclass Test Series

Exam Paper Generator

Online Exam Module