The enthalpies of neutralization of a strong 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

To solve the problem, we need to determine the ratio in which the base (NaOH) is distributed between the strong acid (HA) and the weaker acid (HB) when one equivalent of NaOH is added to a mixture containing one equivalent of each acid.

### Step-by-Step Solution:

1. **Define Variables**:
Let \( x \) be the equivalent of NaOH that reacts with HA, and \( y \) be the equivalent of NaOH that reacts with HB. We know that:
\[
x + y = 1 \quad \text{(Equation 1)}
...