InterviewSolution
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InterviewSolution
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The dissociation constant of a monobasic acid which is `3.5%` dissociated in `(N)/(20)` solution at `20^(@)C` isA. `3.5xx10^(-2)`B. `5xx10^(-3)`C. `6.34xx10^(-5)`D. `6.75xx10^(-2)` |
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Answer» Correct Answer - C Concentration of acid `=(N)/(20)=0.05N` Out of `100` molecules, `3.5` molecules has been dissociated `:.` Out of `1` molecule the no. of disscoiated molecules `=(35)/(100)=0.035=alpha` `K_(a)=(c alpha^(2))/((1-alpha))=(0.05xx(0.035)^(2))/(1-0.035)` `=6.34xx10^(-5)` |
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