Calculate the pH of 0.02 M monobasic acid ha | vedclass

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(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.

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