Question

For a tribasic acid, \(\mathrm{H}_{3} \mathrm{~A}, K_{\mathrm{al}}=2 \times 10^{-5}\), \(K_{\mathrm{a} 2}=5 \times 10^{-9}\) and \(K_{\mathrm{a} 3}=4 \times 10^{-12}\). The value of \(\frac{\left[\mathrm{A}^{3-}\right]}{\left[\mathrm{H}_{3} \mathrm{~A}\right]}\) at equilibrium in an aqueous solution originally having \(0.2 \mathrm{M}-\mathrm{H}_{3} \mathrm{~A}\) is (a) \(5 \times 10^{-17}\) (b) \(5 \times 10^{-9}\) (c) \(1 \times 10^{-17}\) (d) \(2 \times 10^{-22}\)

Short answer

The value of \(\frac{[\mathrm{A}^{3-}]}{[\mathrm{H}_3\mathrm{A}]}\) is \(4 \times 10^{-27}\), but this value is not listed in the options provided, which suggests a mistake in the calculation or a typo within the choices given.

Step-by-step solution

Write the Dissociation Equations and Equilibrium Constants

Write the stepwise dissociation equations for the tribasic acid \(\mathrm{H}_3\mathrm{A}\).1. \(\mathrm{H}_3\mathrm{A} \rightleftharpoons \mathrm{H}^+ + \mathrm{H}_2\mathrm{A}^-\) with \(K_{\mathrm{a}1} = 2 \times 10^{-5}\)2. \(\mathrm{H}_2\mathrm{A}^- \rightleftharpoons \mathrm{H}^+ + \mathrm{HA}^{2-}\) with \(K_{\mathrm{a}2} = 5 \times 10^{-9}\)3. \(\mathrm{HA}^{2-} \rightleftharpoons \mathrm{H}^+ + \mathrm{A}^{3-}\) with \(K_{\mathrm{a}3} = 4 \times 10^{-12}\)

Calculate the Concentration of \(\mathrm{A}^{3-}\)

Find the concentration of \(\mathrm{A}^{3-}\) at equilibrium by multiplying the equilibrium constants: \[\frac{[\mathrm{A}^{3-}]}{[\mathrm{H}_3\mathrm{A}]} = K_{\mathrm{a}1} \cdot K_{\mathrm{a}2} \cdot K_{\mathrm{a}3}\] \[\frac{[\mathrm{A}^{3-}]}{[\mathrm{H}_3\mathrm{A}]} = (2 \times 10^{-5}) \cdot (5 \times 10^{-9}) \cdot (4 \times 10^{-12})\]

Simplify the Expression and Find the Ratio

Using the constants provided, simplify the expression: \[\frac{[\mathrm{A}^{3-}]}{[\mathrm{H}_3\mathrm{A}]} = 2 \times 5 \times 4 \times 10^{-5-9-12} = 40 \times 10^{-26}\] This simplifies to: \[\frac{[\mathrm{A}^{3-}]}{[\mathrm{H}_3\mathrm{A}]} = 4 \times 10^{-27}\], which we can compare with the given options to find the correct answer.

Key concepts

Acid Dissociation Constant

Understanding the acid dissociation constant, often denoted as Ka, is crucial when studying the behavior of acids in solution. The acid dissociation constant is a quantitative measure of the strength of an acid in solution. It illustrates the extent to which an acid can donate a proton to a base, which is the essence of the Bronsted-Lowry definition of acids and bases.

For the specific case of a tribasic acid such as \(\mathrm{H}_{3}\mathrm{A}\), this constant provides insights into the dissociation across the three steps involved. For example, a \(K_{\mathrm{a1}}\) value that is larger than \(K_{\mathrm{a2}}\) indicates that the first dissociation step is more favorable compared to the second. The magnitude of these constants, often expressed as powers of ten, dictates whether the dissociation is strong or weak. A lower \(\mathrm{p}K_{\mathrm{a}}\) suggests a stronger acid.

When we analyze \(K_{\mathrm{a}}\) values, like those given in the original exercise (\(K_{\mathrm{a1}}=2 \times 10^{-5}\), \(K_{\mathrm{a2}}=5 \times 10^{-9}\), and \(K_{\mathrm{a3}}=4 \times 10^{-12}\)), we understand the comparative strengths of the different protons being removed in each dissociation step.

Chemical Equilibrium

Chemical equilibrium refers to the state of a reaction in which the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time. This concept is paramount in understanding reactions such as the dissociation of tribasic acids. Even though the reactions may proceed in steps, each step will eventually reach its own equilibrium.

In our \(\mathrm{H}_{3}\mathrm{A}\) dissociation scenario, each dissociation step will have reached an equilibrium when the rate at which the acid donates protons equals the rate at which the conjugate base reaccepts them. This state does not imply that the reactants and products are in equal concentrations, but rather that their concentrations have stabilized in a ratio that will remain constant over time given constant conditions.

It's also important to note that equilibria are dynamic, and though the concentrations remain constant, particles continue to react. When interpreting equilibrium data, like in the exercise provided, the final equilibrium ratio gives insight into the extent of dissociation for the tribasic acid in question.

Equilibrium Constant Expression

The equilibrium constant expression for a reaction is written in terms of the concentrations of the reactants and products at equilibrium. Each species involved in the reaction is assigned either to the numerator or to the denominator based on whether it is a product or a reactant, respectively. The coefficients from the balanced chemical equation become exponents in the expression.

For acids, like the tribasic acid in our exercise, where multiple equilibria are at play, the overall dissociation can be understood by the product of the individual equilibrium constants. Mathematically, for our tribasic acid \(\mathrm{H}_{3}\mathrm{A}\), when we are looking for the ratio \(\frac{[\mathrm{A}^{3-}]}{[\mathrm{H}_{3}\mathrm{A}]}\) at equilibrium, we gauge this by multiplying the constants: \(K_{\mathrm{a1}} \cdot K_{\mathrm{a2}} \cdot K_{\mathrm{a3}}\). This approach underscores the cumulative favorability from one dissociation step to the next, ultimately leading us to the concentration ratio of fully dissociated species to the undissociated acid.

The calculation steps in the textbook example neatly incorporate this principle, simplifying our understanding of how the dissociation of an acid in a multi-step process leads us to determine the concentrations at equilibrium.