InterviewSolution
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InterviewSolution
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`HCOOH` and `CH_(3)COOH` solutions have equal `pH.` If `K_(1)//K_(2)` is `4`, the ratio of their molar concentration will beA. `0.25`B. `0.5`C. `2`D. `4` |
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Answer» Correct Answer - A Since `pH` is same for two acids. `:. [H^(o+)]_(1) = [H^(o+)]_(2)` Thus, `C_(1)alpha_(1) = C_(2)alpha_(2)` `(K_(a_(1)))/(K_(a_(2))) = (C_(1)alpha_(1)^(2))/(C_(1)alpha_(2)^(2)) = (alpha_(1))/(alpha_(2)) …(i)` `alpha_(1) = sqrt((K_(a_(1)))/(C_(1))), alpha_(2) = sqrt((K_(a_(2)))/(C_(2)))` `:. (alpha_(1))/(alpha_(2)) = sqrt((K_(a_(1))C_(2))/(K_(a_(2))C_(1)))` Substituting from equation (i) the value of `alpha_(1)//alpha_(2)` `(K_(a_(1)))/(K_(a_(2))) = sqrt((K_(a_(1))C_(2))/(K_(a_(2))C_(1)))` `:. (K_(a_(1)))/(K_(a_(2))) = (C_(2))/(C_(1))` `4 = (C_(2))/(C_(1))` `:. C_(1) : C_(2) = (1)/(4) = 0.25` |
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