Question

HA + OH- --> H2O + A- + q1 kJ
H+ + OH- --> H2O + q2 kJ
The enthalpy of dissociation of HA is
1. q1 + q2
2.q1 - q2
3. q2 - q1
4. - q2 - q1

Answer 1

Answer:The correct answer is option 1.

Explanation:

$HA + OH^-\rightarrow H_2O + A^-\Delta H_1= q_1 kJ$...(1)

$H^+ + OH^-\rightarrow H_2O, \Delta H_2= q_2 kJ$...(2)

The enthalpy of the dissociation of HA.

$HA \rightarrow H^++A^-,\Delta H_d=?$...(3)

(1)+(2)=(3)

$\Delta H_d=q_1+q_2$

Hence , the correct answer is option 1.

Answer 2

3. q2 - q1

Explanation:

HA + OH− ⟶ H2​O +  A− +q1 ​kJ -------------(1)

H+ + OH− ⟶ H2​O + q2​ kJ

H2​O ⟶  H+ + OH− − q2 ​kJ -------------(2)

Adding 1 and 2, we get:

HA + OH− + H2​O ⟶ H2​O + A− + H+ + OH− + (q1​−q2​) kJ

HA ⟶ H+ + OH− ; ΔH = −(q1​−q2​) kJ

Therefore, the enthalpy of dissociation of HA = (q2​−q1​) kJ.

Option 3 is the answer.