Question

The enthalpies of neutralization of a strong acid $HA$ and weaker acid $HB$ by NaOH are -13.7 and -12.7 kcal equivalent. When 1 equivalent of NaOH is added to a mixture containing 1 equivalent of $HA$ and $HB$, the enthalpy change was -13.5 kcal. The ratio of base distributed between $HA$ and $HB$ is:

• A. $4:1$
• B. $3:1$
• C. $2:1$
• D. $1:1$

Answer 1

The correct option is A $4:1$

Given
$NaOH+HA\to NaA+H_{2}O$ $\Delta H_N=-13.7kCaleq.$
$NaOH+HA\to NaA+H_{2}O$ $\Delta H_N=-12.7kCaleq.$
In these reactions, $1$ equivalent of $NaOH$ react with $1$ equivalent of $HA$ or $HB$ to release $\Delta H_N$
Now, we have a mixture containing $1$ equivalent of $HA$ and $HB$.
Let $HA=x$ equivalent ; $HB=(1-x)$equivalent
Now, $x$ equivalents of $HA$ will react with $x$ equivalents of $NaOH$ to release $E_1$ energy.
$E_1=-13.7x$
Similarly, $(1-x)$equivalents of $HB$ will react with $(1-x)$equivalents of $NaOH$ to release $E_2$ energy.
$E_2=-12.7(1-x)$
Now, enthalpy change in reaction of one equivalent
$NaOH$ with $1$ equivalent mixture $=-13.5KCal$
So, $E_1+E_2=-13.5$
$\Rightarrow -13.7x-12.7(1-x)=-13.5$
$\Rightarrow x=0.8$ or $1-x=0.2$
$\frac{x}{1-x}=$ (equivalents of $HA$ that used $NaOH$)\equivalents of $HB$ that used $NaOH$ $=\frac{0.8}{0.2}$
$\therefore$ $NaOH$ was distributed between $HA$ and $HB$ in $4:1$.