Question

Which one of the following is true for any diprotic acid, \(\mathrm{H}_{2} \mathrm{X} ?\) (a) \(\mathrm{K}_{\mathrm{a}_{2}}=\mathrm{K}_{\mathrm{a}_{1}}\) (b) \(\mathrm{K}_{\mathrm{a}_{2}}>\mathrm{K}_{\mathrm{a}}\) (c) \(\mathrm{K}_{\mathrm{a}_{\mathrm{l}}}>\mathrm{K}_{\mathrm{a}_{2}}\) (d) \(\mathrm{K}_{\mathrm{a}_{2}}=1 / \mathrm{K}_{\mathrm{a}_{1}}\)

Short answer

(c) \(K_{a_1} > K_{a_2}\) is true for any diprotic acid.

Step-by-step solution

Understanding Diprotic Acids

A diprotic acid releases two protons (\(Hx\) and \(H_{2}x\)). The second proton release is generally harder than the first.

Comparing Acid Dissociation Constants

For any diprotic acid, the first dissociation constant (\(K_{a_1}\)) relates to the release of the first proton, while the second dissociation constant (\(K_{a_2}\)) relates to the release of the second proton.

Analyzing the Given Options

Option (a) suggests \(K_{a_2} = K_{a_1}\), which cannot be true as \(K_{a_2}\) is weaker. Option (b) incorrectly states \(K_{a_2} > K_{a_1}\). Option (c) states \(K_{a_1} > K_{a_2}\), which aligns with the general behavior of diprotic acids. Option (d) makes no sense as \(K_{a_2} = 1 / K_{a_1}\).

Choose the Correct Answer

Based on the understanding that the first dissociation is stronger (having a higher \(K_{a}\)), option (c) \(K_{a_1} > K_{a_2}\) is correct for any diprotic acid.

Key concepts

Acid Dissociation Constant

The acid dissociation constant, often denoted as \(K_a\), is a crucial concept in understanding how acids behave in solution. It provides a measure of how easily an acid can donate protons (H⁺ ions) to the solution. Higher \(K_a\) values indicate stronger acids that dissociate more completely, releasing more protons. This key parameter helps chemists predict the behavior of acids in various chemical reactions. Understanding \(K_a\) is fundamental when dealing with different types of acids, including diprotic acids, which can donate two protons sequentially. In solutions, the dissociation of acids to form their respective bases is an equilibrium process, represented as follows: \[ HA \rightleftharpoons H^+ + A^- \] The equilibrium constant expression for this reaction, \(K_a\), is given by \[ K_a = \frac{[H^+][A^-]}{[HA]} \] where \([HA]\) is the concentration of the undissociated acid, and \([H^+]\) and \([A^-]\) are the concentrations of the resulting ions.

First Dissociation Constant

Diprotic acids, such as \(\mathrm{H}_2X\), have the ability to donate two protons. The first dissociation constant, \(K_{a_1}\), describes the equilibrium process of releasing the first proton. This process can be represented as: \[ \mathrm{H}_2X \rightleftharpoons \mathrm{H}^+ + \mathrm{HX}^- \] The expression for \(K_{a_1}\) is given by \[ K_{a_1} = \frac{[\mathrm{H}^+][\mathrm{HX}^-]}{[\mathrm{H}_2X]} \] The first dissociation step is generally more favorable than the second, meaning \(K_{a_1}\) is typically larger than the second dissociation constant, \(K_{a_2}\). This reflects the tendency of diprotic acids to more readily release the first proton compared to the second. A higher \(K_{a_1}\) indicates a higher efficiency in donating the initial proton.

Second Dissociation Constant

In a diprotic acid, the second dissociation constant, \(K_{a_2}\), addresses the release of a second proton from the anion formed after the first dissociation. The reaction is expressed as: \[ \mathrm{HX}^- \rightleftharpoons \mathrm{H}^+ + \mathrm{X}^{2-} \] The expression for \(K_{a_2}\) is: \[ K_{a_2} = \frac{[\mathrm{H}^+][\mathrm{X}^{2-}]}{[\mathrm{HX}^-]} \] Typically, \(K_{a_2}\) is significantly smaller than \(K_{a_1}\). This means that the second proton is released much less readily. The smaller \(K_{a_2}\) value also indicates that once the first proton is removed, the remaining species, \(\mathrm{HX}^-\), is a weaker acid than the original \(\mathrm{H}_2X\). Analyzing these constants allows us to understand how the strength of an acid changes upon losing a second proton.