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 is

Options

A. `240mL`
B. `150mL`
C. `100mL`
D. None

Answer

A

Explanation

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`