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Enthalpy of neutralisation of a weak dibasic acid by NaOH is - 24.0 kcal. Thus, enthalpy of ionisation of dibasic acid is
- a)10.3 kcal mol-1
- b)3.4 kcal mol-1
- c)13.7 kcal mol-1
- d)-13.7 kcal mol-1
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Enthalpy of neutralisation of a weak dibasic acid by NaOH is - 24.0 kc...
Enthalpy of neutralisation = ∆HH+ + OH- → H2O + ∆Hionisation
-24.0 = -27.4 + ∆Hionisation (since we have 2 moles of H+ and OH-, so enthalpy will also double)
∆Hionisation = 3.4 kcal mol-1
-24.0 = -27.4 + ∆Hionisation (since we have 2 moles of H+ and OH-, so enthalpy will also double)
∆Hionisation = 3.4 kcal mol-1
Most Upvoted Answer
Enthalpy of neutralisation of a weak dibasic acid by NaOH is - 24.0 kc...
Enthalpy of Neutralization:
Enthalpy of neutralization of a weak dibasic acid by NaOH is - 24.0 kcal.
Enthalpy of Ionisation:
Enthalpy of ionization of dibasic acid can be calculated using the following equation:
ΔHionization = ΔHneutralization + ΔHdissociation
Here, ΔHdissociation is the enthalpy of dissociation of water which is - 13.7 kcal/mol.
Substituting the given values, we get:
ΔHionization = - 24.0 kcal/mol - (-13.7 kcal/mol)
ΔHionization = -10.3 kcal/mol
Therefore, the enthalpy of ionization of dibasic acid is 3.4 kcal/mol (Option B).
Enthalpy of neutralization of a weak dibasic acid by NaOH is - 24.0 kcal.
Enthalpy of Ionisation:
Enthalpy of ionization of dibasic acid can be calculated using the following equation:
ΔHionization = ΔHneutralization + ΔHdissociation
Here, ΔHdissociation is the enthalpy of dissociation of water which is - 13.7 kcal/mol.
Substituting the given values, we get:
ΔHionization = - 24.0 kcal/mol - (-13.7 kcal/mol)
ΔHionization = -10.3 kcal/mol
Therefore, the enthalpy of ionization of dibasic acid is 3.4 kcal/mol (Option B).
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Community Answer
Enthalpy of neutralisation of a weak dibasic acid by NaOH is - 24.0 kc...
Enthalpy of dissociation of dibasic acid = Enthalpy of neutralisation of strong acid with a strong base - enthalpy of neutralisation of weak dibasic acid with a strong base.... So.. = - 24-(-2*13.7)=3.4.. Hence option b is the right choice
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Enthalpy of neutralisation of a weak dibasic acid by NaOH is - 24.0 kcal. Thus, enthalpy of ionisation of dibasic acid isa)10.3 kcal mol-1b)3.4kcal mol-1c)13.7kcal mol-1d)-13.7kcal mol-1Correct answer is option 'B'. Can you explain this answer?
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