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)`
Ethalpy change `=x xx13.7 +y xx12.7 = 13.5 …(ii)`
Solving equation (i) and (ii), we get
`x = 0.8, y = 0.2`
`x:y = 4:1`