The pH of a monoacidic weak base is 10.9. Cal | vedclass

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Question

(B) For a monoacidic weak base $BOH$: $BOH_{(aq)} \rightleftharpoons B_{(aq)}^{+} + OH_{(aq)}^{-}$ Given $pH = 10.9$,we calculate $pOH$: $pOH = 14 - p

Vedclass · Accepted Answer

$3.95$

Vedclass · Answer

(B) For a monoacidic weak base $BOH$: $BOH_{(aq)} ightleftharpoons B_{(aq)}^{+} + OH_{(aq)}^{-}$ Given $pH = 10.9$,we calculate $pOH$: $pOH = 14 - pH = 14 - 10.9 = 3.1$ The concentration of hydroxide ions is: $[OH^{-}] = 10^{-pOH} = 10^{-3.1} = 10^{-4} 	imes 10^{0.9} \approx 7.94 	imes 10^{-4} \ M$ For a weak base,$[OH^{-}] = C 	imes \alpha$,where $C = 0.02 \ M$: $\alpha = \frac{[OH^{-}]}{C} = \frac{7.94 	imes 10^{-4}}{0.02} = 3.97 	imes 10^{-2}$ The percent dissociation is $\alpha 	imes 100$: $\% \alpha = 3.97 	imes 10^{-2} 	imes 100 = 3.97 \% \approx 3.95 \%$ Thus,the correct option is $B$.

The $pH$ of a monoacidic weak base is $10.9$. Calculate the percent dissociation in a $0.02 \ M$ solution. (in $\%$)

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