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Fixed volume of `0.1M` benzoic acid `(pK_(a) = 4.2)` solution is added into `0.2M` sodium benzote solution and formed a `300mL`, resulting acidic buffer solution. If `pH` of the resulting solution is `3.9`, then added volume of banzoic acid isA. `240mL`B. `150mL`C. `100mL`D. None |
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Answer» Correct Answer - A Let `VmL` of acid solution is taken. mmol of acid `= 0.1 xx V` mmol of salt `= 0.2M xx (300 - V) mL` `pH = pK_(a) + "log" (["Salt"])/(["Acid"])` `3.9 = 4.2 +"log" (["Salt"])/(["Acid"])` `-0.3 = "log"(["Salt"])/(["Acid"])` `"log" (["Acid"])/(["Salt"]) = 0.3` `(["Acid"])/(["Salt"]) = "Antilog" (0.3) = 2` `:. ["Acid"] =2 ["Salt"]` `(0.1 xx V)/(cancel 300) = 2 xx [(0.2xx(300-V))/(cancel300)]` Solve for `V:` `V = 240 mL` |
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