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

Explanation

According to Henderson.s equation
`pH=pK_a+"log" "[Salt]"/"[Acid]"`
pH=4.91 , `pK_a`=4.76
4.91=4.76+log `"[Salt]"/"[Acid]"`
log `"[NaAc]"/"[AcH]"`=4.91-4.76=0.15
`"[NaAc]"/"[AcH]"`=antilog (0.15) =1.41
if V is the volume of 0.1 M AcH required
`"[NaAc]"/"[AcH]"=((0.2xx50)/1000)/((0.1xxV)/1000)` =1.41
`(0.2xx50)/(0.1xxV)`=1.41
or V=`(0.2xx50)/(0.1xx1.41)`=70.92 mL
Volume of 0.1M acetic acid required = 70.92 mL