Question

6gms of oil was saponified with 50ml of 0.5N alcoholic KOH solution. After refluxing for 2 hours the mixture was titrated with 25ml O.SN HCL Rnd the sponification value of Oil .

Answer 1

Given Data -  Weight of oil = 6 mgs     Blank titration reading = 50ml = V2
Back titration reading  = 25ml = V1
Solution - Volume of 0. 5N KOH required for saponification in terms of 0. 5N HCI
                                             = V2 - V1 = 50 - 25 = 25 ml
          `"sponification value of Oil" = ( "Volume of KOH xx Normality of KOH" xx 56 )/("weight og oil )"`

` = (25xxO.5 xx 56)/6 ` = 116.66 mg of KOH

Therefore the Saponification value of the oil is 116. 66 mg of oil.