`0.1` mol each of ethyl alcohol and acetic acid are allowed to react and at equilibrium, the acid was exactly neutralised by `100 mL` of `0.85 N NaOH`. If no hydrolysis of ester is supposed to have undergo, finf `K_(c)`.

Answer

1

Explanation

`{:(,CH_(3)COOH,+,C_(2)H_(5)OH,hArr,CH_(3)COC_(2)H_(5),+,H_(2)O),(Att=0,0.1,,0.1,,0,,0),("At equilibrium",0.1-x,,0.1-x,,x,,x):}`
Meq of acetic acid left`="Meq. Of" NaOH` used.
`=100xx0.75`
`=75`
Milomoles of acetic acid left`=75` (`because` monobasic)
Moles of acetic acid left`=0.075`
`0.1-x=0.075`
`x=0.025`
`K_(c)=(X^(2))/((0.1-x)^(2))=((0.025)^(2))/((0.075)^(2))=0.111=1.11xx10^(-2)`