How much volume of `0.1M Hac` should be added to `50mL` of `0.2M NaAc` solution to have a `pH 4.91`?

Explanation

It is an acidic buffer
`pH = pK_(a) + "log" (["Salt"])/(["Acid"])`
`4.91 = 4.76 + log [("Salt")/("Acid")]`
`:. log [("salt")/("Acid")] = (4.91 - 4.76 ) = 0.15`
`(["Salt"])/(["Acid"]) = "Antilog" (0.15) = 1.41`
mmoles of `NaAc = 50 xx 0.2 = 10`
mmoles of HAc required `= 0.1 M xxV`
`[("Salt")/("Acid")] = 1.41 = (10)/(0.1xxV)`
`rArr V = (10)/(0.1 xx 1.41) = 70.92mL`
`:.` Volume of 0.1M HAc required `= 70.92mL`