The enthalpy of neutralization of `HNO_(3)` by `NaOH=-13680 cal//eqt.` When one equivalent of `NaOH` is added to a dilute solution containing one equivalent of `HNO_(3)` and one equivalent of a certain monoprotic weak acid `13960` cals are evolved. Assume that the base is distributed between `HNO_(3)` and the weak acid in the ratio `3:1` and the weak acid is practically nonionized. Calculate the enthalpy of ionization of the weak acid.

Options

A. `-1120` cals
B. `-2110` cals
C. `1210` cals
D. `+1210 cals`

Answer

3

Explanation

`13960=-13680xx(3)/(4)-13680xx(1)/(4)+(1)/(4)x`
where x is enthalpy of dissociation/ionisation of weak acid.
so `x=1120` calories.