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

Explanation

Let V mL of Hac on mixing with NaAc gives a pH of `4.91`. Thus total volume after mixing becomes `(V+50)mL`.
m mole of `[Hac]= 0.1xxV`
`:. [Hac]= (0.1xxV)/((V+50))`
m mole of `NaAc= 50xx0.2`
`:. [NaAc]= (10)/((V+50))`
Also pH of acid buffer mixiture is given by:
`pH= pK_(a)+log((["Salt"])/(["Acid"]))`
`:. 4.91=4.76+log ((10//(V+50))/((0.1xxV)//(V+50)))`
`log((10)/((0.1xxV)))=0.15`
`:. V= 70.80 mL`