The enthalpies of neutralisation of a string acid `HA` and a weaker acid `HB` by `NaOH` are `-13.7` and `-12.7 kcal Eq^(-1)`, respectively. When one equivalent of `NaOH` is added to a mixture containing one equivalent of `HA` and `HB`, the enthalpy change was `-13.5kcal`. In what ratio is the base distributed between `HA` and `HB`?

Explanation

Let x equivalent of HA and y equivalent of HB are taken in the mixture
`x+y=1` . . . (i)
`x xx13.7+yxx12.7=13.5` . . . (ii)
Solving eqs. (i) and (ii), we get
`x=0.8,y=0.2`
`x:y=4:1`