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(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

$2.3$

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(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 $= 2\,\%$)

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