Question

A volume of \(10 \mathrm{ml}\) of \(0.1 \mathrm{M}\) tribasic acid, \(\mathrm{H}_{3} \mathrm{~A}\) is titrated with \(0.1 \mathrm{M}-\mathrm{NaOH}\) solution. What is the ratio (approximate value) of \(\frac{\left[\mathrm{H}_{3} \mathrm{~A}\right]}{\left[\mathrm{A}^{3-}\right]}\) at the second equivalent point? Given: \(K_{1}=7.5 \times 10^{-4} ; K_{2}=10^{-8}\); \(K_{3}=10^{-12}\) (a) \(10^{-4}\) (b) \(10^{-3}\) (c) \(10^{-7}\) (d) \(10^{-6}\)

Short answer

Due to the mathematics involved, strictly speaking, the ratio \(\frac{[H_{3}A]}{[A^{3-}]}\) is zero at the second equivalent point. However, for the purpose of choosing the closest order of magnitude from the given options, the answer is (d) \(10^{-6}\).

Step-by-step solution

- Understanding the Equivalent Points in Titration

In the titration of a tribasic acid with a strong base, there are three equivalent points, corresponding to the neutralization of each acidic proton. The second equivalent point occurs when two protons have been neutralized, leaving the species \(H_{1}A^{2-}\) in equilibrium with \(HA^{3-}\).

- Identifying the Relevant Equilibrium and Constants

At the second equivalent point, the relevant equilibrium is the third ionization: \[ H_{1}A^{2-} \rightleftharpoons H^{+} + A^{3-} \] with the equilibrium constant \(K_{3} = 10^{-12}\).

- Writing the Expression for the Equilibrium Constant

Write the expression for \(K_{3}\): \[ K_{3} = \frac{[H^{+}][A^{3-}]}{[H_{1}A^{2-}]} \] At the second equivalent point, the concentration of \(H^{+}\) and \(A^{3-}\) is equal, because each \(H_{1}A^{2-}\) releases one \(H^{+}\) to form \(A^{3-}\).

- Simplifying the Expression for the Ratio

Because the concentrations of \(H^{+}\) and \(A^{3-}\) are equal at the second equivalent point, the expression simplifies to: \[ K_{3} = \frac{[A^{3-}]}{[H_{1}A^{2-}]} = [A^{3-}]^2 \]

- Calculating the Concentration Ratio

To find the ratio \(\frac{[H_{3}A]}{[A^{3-}]}\), we first find \(\sqrt{K_{3}}\) because \(K_{3} = [A^{3-}]^2\). Then, \[ [A^{3-}] = \sqrt{K_{3}} = \sqrt{10^{-12}} = 10^{-6} \mathrm{M} \] Since [H_{3}A] was initially 0.1 M and two protons have been neutralized, [H_{3}A] is effectively 0 at the second equivalent point. Therefore, the ratio approximates to the inverse of [A^{3-}], which is: \[ \frac{[H_{3}A]}{[A^{3-}]} \approx \frac{0}{10^{-6}} = 0 \] Even though mathematically the ratio is zero, for approximation purposes in the context of a multiple-choice question, we look to the options for the closest non-zero order of magnitude to our calculation of \(10^{-6}\).

Key concepts

Acid-Base Titration

Acid-base titration is a laboratory procedure used to determine the concentration of a known reactant in a solution. In the context of a tribasic acid being titrated with a strong base such as NaOH, the process involves the sequential neutralization of the acid's protons (hydrogen ions).

During the titration, there are specific points where the amount of base added equals the number of moles of protons available in the acid solution. These are known as equivalent points. In the case of tribasic acids, which have three ionizable protons, there will be three equivalent points. Understanding these points is crucial because they provide information about the titration progress and the acid's ionization stages.

To ensure that students fully comprehend the concept, it's essential to clarify that as the titration progresses, the nature of the solution changes from acidic to basic, and different ionic species predominate at each equivalent point. Through careful monitoring, usually with the aid of indicators or pH meters, chemists can accurately determine equivalent points and draw conclusions about the acid's strength and the titration's progress.

Equilibrium Constant

The equilibrium constant, designated as K, provides insight into the favorability of a chemical reaction at equilibrium. It is the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients. In the case of ionization reactions of acids or bases, separate equilibrium constants (like the given \(K_1\), \(K_2\), and \(K_3\)) are used to represent each ionization step.

For the third ionization step of the tribasic acid (\(H_2A^{-} \rightarrow H^+ + A^{3-}\)), the equilibrium constant is written as \(K_3 = \frac{[H^+][A^{3-}]}{[H_2A^-]}\). At the second equivalent point of titration, the amount of \(H^+\) generated is equal to the concentration of \(A^{3-}\) formed due to the neutralization reaction, leading to a simplification of the equilibrium expression. Hence, understanding the equilibrium constant in the context of our titration is pivotal, as it allows us to find the concentration ratio of different ions at the equivalent point, which illustrates the progression of the ionization process.

Ionization

Ionization refers to the process where a molecule or an atom gains or loses electrons and becomes charged, forming ions. In acid-base chemistry, ionization is the stepwise dissociation of hydrogen ions from an acid, which is crucial to understanding the behavior of acids and bases in solution. For tribasic acids, the ionization occurs in three distinct steps, with each step having a unique ionization constant (\(K_1\), \(K_2\), and \(K_3\)).

As seen with our tribasic acid example, at the second equivalent point, the third ionization reaction reaches equilibrium. Understanding ionization is key because it governs how acids and bases react in water and dictates the pH of the solution. When simplifying the chemical relationships, like ionization constants at equivalent points, it becomes possible to derive ratios like \(\frac{[H_3A]}{[A^{3-}]}\), which enlighten us about the dominant species in solution at any point during the titration. This understanding is not just academic; it's also practical, as it guides decisions in experimental design, such as choosing suitable indicators for titrations.