Question

The enthalpies of neutralisation of a strong acid $HA$ and weak acid $HB$ by NaOH are -13.68 kcal/equivalent and -2.9 kcal/equivalent respectively. If one equivalent of $HA$ and one equivalent of $HB$ are mixed with one equivalent of NaOH, the enthalpy change was found to be - 6.9 kcal/equivalent calculate the distribution of NaOH to neutralise $HA$ and $HB$.

Answer 1

Let $a$ Eq. of NaOH be used by $HA$, and $b$ Eq. of NaOH be used by $HB$.
$AlsoheatofdissociationofHB=+13.68-2.9=10.78kcal/eq$.
$\therefore a\times (-13.68)+b\times (-2.9)+b\times \left(10.78\right)=-6.9$
$\therefore -13.68a+7.88b=-6.9$
or $-13.68a+7.88(1-a)=-6.9$ $(a+b=1)$
$\therefore -21.56a=-14.78$
$\therefore a=0.685$
Hence, 0.685 Eq. of NaOH be used by $HA$, and $1-0.685=0.315$ Eq. of NaOH be used by $HB$.