Question

Under the same condition how many $ mL $ of $ 0.1MNaOH $ and $ 0.05M $ $ {H_2}A $ (strong diprotic acid) solution should be mixed for a total volume of $ 100mL $ produce the highest rise in temperature.
(A) $ 25:75 $
(B) $ 50:50 $
(C) $ 75:25 $
(D) $ 66.66:33.33 $

Answer 1

Hint : During neutralisation reaction, when acid and base react to form water and salt large amounts of energy also release which is the reason for increase in temperature. Thus, maximum energy release when acid and base completely react with each other.

Complete Step By Step Answer:
The enthalpy change during the neutralisation of acid and base is known as enthalpy of neutralisation. For one gram equivalent of strong acid and base it is constant. But for weak acid or weak base it is slightly less due to some energy required to completely neutralise weak acid or base.
For acid and base we always use normality for expressing concentration as they can have different acidity or basicity. For complete neutralisation-
 $ {\left( {NV} \right)_1} - {\left( {NV} \right)_2} = 0 $ [for complete neutralisation]
Relation Normality and Morality is-
Normality = Molarity x n-factor
n-factor for $ NaOH $ is $ 1 $ and for $ {H_2}A $ is $ 2 $
Hence normality of $ NaOH $ $ \Rightarrow 0.1 \times 1 = 0.1N $
And normality of $ {H_2}A $ $ \Rightarrow 0.05 \times 2 = 0.1 $
As given volume of complete solution is $ 100mL $
Let volume of sodium hydroxide $ = x $
And volume of $ {H_2}A $ $ = 100 - x $
Hence the equation is given as-
 $ {\left( {NV} \right)_1} - {\left( {NV} \right)_2} = 0 $
 $ 0.1 \times x - 0.1 \times \left( {100 - x} \right) = 0 $
 $ x = 50 $
Hence the volume of sodium hydroxide required, $ x = 50 $
And thus the volume of $ {H_2}A $ $ = 100 - x $ $ \Rightarrow 100 - 50 = 50 $
So the ratio of volume so that the solution on neutralization produce maximum rise in temperature is-
 $ 50:50 $
Hence the correct answer is (b).

Note :
The enthalpy of neutralisation for gram equivalent of strong acid and base is constant that is equal to $ - 57.3KJe{q^{ - 1}} $ or $ - 13.7kcal.e{q^{ - 1}} $ . For weak base and weak acid enthalpy of neutralisation is slightly lower as some energy is required to completely dissociate weak base and acid. On increasing or decreasing the amount of acid and base equally, change in temperature of the reaction mixture will remain constant but unequally always decrease in temperature.