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Equilibrium
Ionization of Polybasic Acids
1 question from 1 paper, each with its answer.
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Concepts tested
- Polyprotic Acid Equilibrium
All Questions
2025 1 question
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Phosphoric acid ionizes in three steps with their ionization constant values $K_{a_1}, K_{a_2}$ and $K_{a_3}$ respectively, while $K$ is the overall ionization constant. Which of the following statements are true?
A. $\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$
B. $\text{H}_3\text{PO}_4$ is stronger acid than $\text{H}_2\text{PO}_4^-$ and $\text{HPO}_4^{2-}$
C. $K_{a_1} > K_{a_2} > K_{a_3}$
D. $K = \frac{K_{a_1} + K_{a_2} + K_{a_3}}{2}$
Choose the correct answer from the options given below:Overall constant is $K = K_{a_1} \cdot K_{a_2} \cdot K_{a_3}$, so $\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$. Removing successive protons from negatively charged species becomes increasingly difficult, so $K_{a_1} > K_{a_2} > K_{a_3}$ and $\text{H}_3\text{PO}_4$ is the strongest acid. Thus A, B, and C are true.
For polyprotic acid $\text{H}_3\text{PO}_4$: 1. Stepwise ionizations add to give overall reaction $\text{H}_3\text{PO}_4 \rightleftharpoons 3\text{H}^+ + \text{PO}_4^{3-}$, so overall $K = K_{a_1} \times K_{a_2} \times K_{a_3} \implies \log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$ (Statement A is true). 2. Due to electrostatic attraction, losing a proton from a neutral molecule is much easier than from negatively charged ions $\text{H}_2\text{PO}_4^-$ and $\text{HPO}_4^{2-}$, so $K_{a_1} > K_{a_2} > K_{a_3}$ and $\text{H}_3\text{PO}_4$ is the strongest acid among them (Statements B and C are true). Statement D is mathematically incorrect. Thus, A, B, and C are true.