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Question: Heat of neutralization of \[CsOH\] with all strong acid is \( 13.4Kcalmo{l^{ - 1}} \). The heat rele...

Heat of neutralization of CsOHCsOH with all strong acid is 13.4Kcalmol−113.4Kcalmo{l^{ - 1}}. The heat released on neutralization of CsOHCsOH with HFHF (weak acid) is 16.4Kcalmol−116.4Kcalmo{l^{ - 1}}. ΔH∘\Delta {H^\circ } of ionization of HFHF is
A. 3.0Kcal3.0Kcal
B. −3.0Kcal- 3.0Kcal
C. 6.0Kcal6.0Kcal
D. 0.3Kcal0.3Kcal

Explanation

Solution

As we all are familiar with, the heat of neutralization is the change in enthalpy that takes place when 1g1g equivalent of strong acid is completely neutralized by 1g1g equivalent of strong base.

Complete answer:
As we know that the heat of neutralization is the change in enthalpy that takes place when 1g1g equivalent of strong acid is completely neutralized by 1g1g equivalent of strong base or we can say that it is the amount of energy released when 11 mole of water is formed from its ions is its heat of neutralization. It is an exothermic process and the enthalpy change is always negative and the amount of energy released is equal to 13.7Kcal13.7Kcal per gram equivalent.
So, from the given question, the heat of neutralization of CsOHCsOH with strong acids is given as: CsOH+H+→Cs++H2O;  ΔH1=−13.4KcalCsOH + {H^ + } \to C{s^ + } + {H_2}O;\;\Delta {H_1} = - 13.4Kcal

Similarly, heat of neutralization of CsOHCsOH with weak acid is given as:
CsOH+HF→F−+H2O;  ΔH2=−16.4KcalCsOH + HF \to {F^ - } + {H_2}O;\;\Delta {H_2} = - 16.4Kcal
So, the heat of ionization of HFHF will be given as:
⇒ΔH3=ΔH2−ΔH1 ⇒ΔH3=−16.4−(−13.4) ⇒ΔH3=−3.0Kcal \Rightarrow \Delta {H_3} = \Delta {H_2} - \Delta {H_1} \\\ \Rightarrow \Delta {H_3} = - 16.4 - ( - 13.4) \\\ \Rightarrow \Delta {H_3} = - 3.0Kcal
**Therefore, the correct answer is (B).

Note:**
Some important points to remember:
1. When either acid, base or both are weak then the heat of neutralization depends on pH and is less than 13.7Kcal13.7Kcal because some amount of heat is used in complete dissociation of weak acid or weak base and this amount of energy is called heat of dissociation.
2. Heat of neutralization = Heat of dissociation −13.7Kcal- 13.7Kcal.
3. Heat of neutralization thus is affected by the quantity of acid and base, basicity of acid and base and the strength of acid and base.
4. The heat of neutralization is always constant for any strong and strong base as they completely ionize in dilute solution and change in enthalpy is always negative.
5. Change in enthalpy takes place when one mole of a compound is formed from its element which is present in gaseous state.