InterviewSolution
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InterviewSolution
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`K_(a)` for `HCN` is `5 xx 10^^(-10)` at `25^(@)C`. For maintaining a constant `pH` of `9.0`, the volume of `5M KCN` solution required to be added to `10mL` of `2M HCN` solution isA. `9.3 mL`B. `7.95 mL`C. `4mL`D. `2mL` |
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Answer» Correct Answer - D `pK_(a) =- log(5 xx 10^(-10))` `=- 0.7 + 10 =9.3` Acidic buffer will be formed. `pH = pK_(a) + "log" ([KCN])/([HCN])` Let `V mL` of `KCN` is added Total volume `= (V +10)mL` mmol of `KCN = 5 xx V` `:. [KCN] = (5 xx V "mmol")/((V + 10)mL)` mmol of `HCN = 2 xx 10 = 20` `:. [HCN] = (20 "mmol")/((V +10)mL)` Thus, from equation (i), `9 = 9.3 + log[(5V//(V+10))/(20//(V+10))]` `- 0.3 = log ((5V)(20))` `log (V)/(4) = - 0.3` `(V)/(4) = "Antilog" (-0.3 + 1-1)` `= "Antilog" (bar(1).7) = 5 xx 10^(-1)` `V = 20 xx 10^(-1) = 2 mL` |
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