Heat of neutralization of a strong acid HA and a weaker acid $\mathrm{HB}$ with $\mathrm{KOH}$ are $-13.7$…
Heat of neutralization of a strong acid HA and a weaker acid $\mathrm{HB}$ with $\mathrm{KOH}$ are $-13.7$ and $-12.7 \mathrm{k}$ cal $\mathrm{mol}^{-1}$. When 1 mole of $\mathrm{KOH}$ was added to a mixture containing 1 mole each of HA and $\mathrm{HB}$, the heat change was $-13.5 \mathrm{kcal}$. In what ratio is the base distributed between HA and HB.
$3: 1$
$1: 3$
$4: 1$
$1: 4$
Solution
Let $x$ mole of $\mathrm{KOH}$ be neutralized by the strong acid $H A$. Then, moles neutralized by $H B=1-x$
Hence, $-13.7 \times x+(-12.7) \times(1-x)=-13.5$
$\Rightarrow \quad x=0.8 ; \quad \frac{x}{1-x}=\frac{0.8}{0.2}=\frac{4}{1} =4: 1$