Question: 59. Phosphoric acid ionizes in three steps with their ionization constant values \mathrm{K}_{\mathrm{a}_{1}}, \mathrm{~K}_{\mathrm{a}_{2}} and \mathrm{K}_{\mathrm{a}_{3}}, respectively, while K is the overall ionization constant. Which of the following statements are true?
A. \quad \log \mathrm{K}=\log \mathrm{K}_{\mathrm{a}_{1}}+\log \mathrm{K}_{\mathrm{a}_{2}}+\log \mathrm{K}_{\mathrm{a}_{3}}
B. \mathrm{H}_{3} \mathrm{PO}_{4} is stronger acid than \mathrm{H}_{2} \mathrm{PO}_{4}^{-} and \mathrm{HPO}_{4}^{2-}.
C. \mathrm{K}_{\mathrm{a}_{1}}>\mathrm{K}_{\mathrm{a}_{2}}>\mathrm{K}_{\mathrm{a}_{3}}
D. \mathrm{K}_{\mathrm{a}_{1}}=\frac{\mathrm{K}_{\mathrm{a}_{3}}+\mathrm{K}_{\mathrm{a}_{2}}}{2}
Choose the correct answer from the options given below:
(1) A and B only
(2) A and C only
(3) B, C and D only
(4) A, B and C only
Answer: Option (4)
Explanation:
Phosphoric acid is a tribasic acid and ionizes in three successive steps.
The overall ionization constant K is the product of the stepwise ionization constants.
Thus, K = K_{a_1} \times K_{a_2} \times K_{a_3}.
Taking logarithm on both sides:
\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}, so statement A is correct.
In successive ionization, removal of each proton becomes more difficult due to increasing negative charge on the species. Therefore, the undissociated acid is always stronger than its conjugate bases.
Hence, \mathrm{H}_{3}\mathrm{PO}_{4} is a stronger acid than \mathrm{H}_{2}\mathrm{PO}_{4}^{-} and \mathrm{HPO}_{4}^{2-}, so statement B is correct.
For any polybasic acid, the magnitude of ionization constants decreases with each successive dissociation because of increased electrostatic attraction between the remaining proton and the negatively charged ion.
Therefore, \mathrm{K}_{a_1} > \mathrm{K}_{a_2} > \mathrm{K}_{a_3}, making statement C correct.
Statement D has no theoretical or mathematical basis and is incorrect.
Thus, the correct statements are A, B and C.