Question

. The enthalpies of neutralization of strong acid $HA$and weaker acid $HB$by $NaOH$are $- 13.7{\text{ }}and{\text{ }} - 12.7{\text{ }}kcal$equivalent. When 1 equivalent of $NaOH$is added to the mixture containing 1 equivalent of$HA{\text{ }}and{\text{ }}HB$, the enthalpy change was$- 13.5{\text{ }}kcal$. The ratio of base distributed between HA and HB is: -
A.$4:1$
B.$3:1$
C.$\;2:1$
D.$1:1$

Answer 1

Consider $HA$as $x$and $HB$as $y$and form the two equations from the provided data and solve these equations. These will give you the ratio of $x{\text{ }}and{\text{ }}y{\text{ }}i.e.{\text{ }}HA{\text{ }}and{\text{ }}HB.$

Complete step by step answer:
In the given question, the reaction is acid base reaction i.e. when an acid reacts with base it becomes neutralization reaction. In the given question, the neutralization reaction that is occurring is as follows: -
HA{\text{ }} + {\text{ }}NaOH{\text{ }} \to {\text{ }}NaA{\text{ }} + {\text{ }}{H_2}O$$$$\Delta H{\text{ }} = {\text{ }} - 13.7{\text{ }}kcal
HB{\text{ }} + {\text{ }}NaOH{\text{ }} \to {\text{ }}NaB{\text{ }} + {\text{ }}{H_2}O$$$$\Delta H{\text{ }} = {\text{ }} - 12.7{\text{ }}kcal
This is how the reactions are occurring and now we will come to the solving part in which HA and HB mixture is provided and their total is 1 equivalent.
So, let us consider that x equivalent of HA and y equivalent of HB are mixed and then according to the question.
$X{\text{ }} + {\text{ }}Y{\text{ }} = {\text{ }}1$ (because total 1 equivalent mixture is there) --------(1)
Now from the above two reactions provided, we see that x equivalent of HA will combine with x equivalent of NaOH and will give -13.7 kcal of energy and similarly y equivalent of HB will combine with y equivalent of NaOH and will release -12.7 kcal of energy and when both are mixed then they will give -13.5 kcal of energy.
So, the change in enthalpy in the mixture: -
$- 13.7x{\text{ }} + {\text{ }}\left( { - 12.7} \right){\text{ }}y{\text{ }} = {\text{ }} - 13.5$ ----------(2)
So, we see that we have got 2 linear equations now which could be solved by multiplying 1 by 13.7 and then adding both the equations.
We will get the final value of x and y
Therefore, $x{\text{ }} = {\text{ }}0.8{\text{ }}and{\text{ }}y{\text{ }} = {\text{ }}0.2$
$\dfrac{{HA}}{{HB}} = \dfrac{x}{y} = \dfrac{{0.8}}{{0.2}} = \dfrac{4}{1}$
So, the ratio in which HA and HB are present is $4:1.$

Hence the correct answer among all the options is option A i.e. $4:1$

Note:
These types of similar questions can be solved when you have knowledge of basic reactions and enthalpies like enthalpy of formation, dissociation, neutralization etc. and after that it is just simple algebra which would lead you directly to the answer. So, concept clarity of these topics is required to solve and along with basic equation forming and solving in algebra.