Problem 83 - General Chemistry: Principles and Modern Applications

Question

Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2},\) is a somewhat stronger acid than water. Its ionization is represented by the equation \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HO}_{2}^{-}\) In \(1912,\) the following experiments were performed to obtain an approximate value of \(\mathrm{p} K_{\mathrm{a}}\) for this ionization at \(0^{\circ} \mathrm{C} .\) A sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was shaken together with a mixture of water and 1 -pentanol. The mixture settled into two layers. At equilibrium, the hydrogen peroxide had distributed itself between the two layers such that the water layer contained 6.78 times as much \(\mathrm{H}_{2} \mathrm{O}_{2}\) as the 1 -pentanol layer. In a second experiment, a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was shaken together with 0.250 M NaOH(aq) and 1-pentanol. At equilibrium, the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was \(0.00357 \mathrm{M}\) in the 1-pentanol layer and 0.259 M in the aqueous layer. In a third experiment, a sample of \(\mathrm{H}_{2} \mathrm{O}_{2}\) was brought to equilibrium with a mixture of 1 -pentanol and \(0.125 \mathrm{M}\) \(\mathrm{NaOH}(\mathrm{aq}) ;\) the concentrations of the hydrogen peroxide were \(0.00198 \mathrm{M}\) in the 1 -pentanol and \(0.123 \mathrm{M}\) in the aqueous layer. For water at \(0^{\circ} \mathrm{C}, \mathrm{p} K_{\mathrm{w}}=14.94\) Find an approximate value of \(\mathrm{pK}_{\mathrm{a}}\) for \(\mathrm{H}_{2} \mathrm{O}_{2}\) at \(0^{\circ} \mathrm{C}\) [Hint: The hydrogen peroxide concentration in the aqueous layers is the total concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) and \(\mathrm{HO}_{2}^{-}\). Assume that the 1 -pentanol solutions contain no ionic species.

Step-by-Step Solution

Answer

The approximate value of \(\mathrm{pK}_{\mathrm{a}}\) for \(\mathrm{H}_{2}\mathrm{O}_{2}\) at \(0^{\circ} C\) is evaluated in Step 3.

Step 1: Calculate the concentration of ionized species

First examine the equilibrium reaction given, \(\mathrm{H}_{2}\mathrm{O}_{2}+\mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}+\mathrm{HO}_{2}^{-}\). From this, understand that the hydrogen peroxide \(\mathrm{H}_{2}\mathrm{O}_{2}\) in the aqueous layer is ionized to provide \(\mathrm{H}_{3}\mathrm{O}^{+}\) and \(\mathrm{HO}_{2}^{-}\). The total concentration of \(\mathrm{H}_{2}\mathrm{O}_{2}\) in the aqueous layer, thus, equals the concentrations of \(\mathrm{H}_{3}\mathrm{O}^{+}\) and \(\mathrm{HO}_{2}^{-}\). Hence, calculate these concentrations using the given data from the third experiment - concentration of \(\mathrm{H}_{2}\mathrm{O}_{2}\) in the aqueous layer was \(0.123 M\). Therefore, \([H_3O^+] = [HO_2^-] = 0.123 M\).

Step 2: Use the ionization constant expression

The ionization constant \(\mathrm{K}_{\mathrm{a}}\) for the ionization of \(\mathrm{H}_{2}\mathrm{O}_{2}\) can be written as \(\mathrm{K}_{\mathrm{a}} = [\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{HO}_{2}^{-}]/[\mathrm{H}_{2}\mathrm{O}_{2}]\). Substitute the concentrations of the species into this equation. Here, the concentration of \(\mathrm{H}_{2}\mathrm{O}_{2}\) is the concentration before ionization, which is the concentration of \(\mathrm{H}_{2}\mathrm{O}_{2}\) found in the pentanol layer. From the third experiment, this is \(0.00198 M\). So, \(\mathrm{K}_{\mathrm{a}} = (0.123)^2 / 0.00198\).

Step 3: Calculate \(\mathrm{pK}_{\mathrm{a}}\)

\(\mathrm{pK}_{\mathrm{a}}\) can be calculated from \(\mathrm{K}_{\mathrm{a}}\) using the relation \(\mathrm{pK}_{\mathrm{a}} = -\log(\mathrm{K}_{\mathrm{a}})\). Evaluate \(\mathrm{K}_{\mathrm{a}}\) from Step 2 and substitute into this equation to get the \(\mathrm{pK}_{\mathrm{a}}\) value for \(\mathrm{H}_{2}\mathrm{O}_{2}\) at \(0^{\circ} C\).

Key Concepts

Acid Ionization Constant

Understanding the acid ionization constant, denoted as \( K_a \), is crucial for students studying chemistry. It quantifies the strength of an acid in an aqueous solution by representing the equilibrium concentration of the acid and its ionized form. The acid ionization constant can be written as \( K_a = \frac{[H_3O^+][A^-]}{[HA]} \), where \( [HA] \) represents the concentration of the non-ionized acid, and \( [H_3O^+] \) and \( [A^-] \) are the concentrations of the hydronium and the corresponding anion, respectively.

To further elucidate this concept, let's take the ionization of hydrogen peroxide (\( H_2O_2 \)) as an example. When \( H_2O_2 \) is added to water, it partially dissociates to produce hydronium ions (\( H_3O^+ \) and hydroperoxyl ions (\( HO_2^- \)). By calculating the concentrations of these ions in the aqueous solution, we can determine \( K_a \) for hydrogen peroxide. A higher \( K_a \) value indicates a stronger acid, meaning it donates protons to water more readily, achieving greater ionization. To find the pKa, which often helps in better understanding the acid strength, we simply take the negative logarithm of the \( K_a \) value.

Chemical Equilibrium

The state of chemical equilibrium is central to numerous concepts in chemistry, including the ionization of acids. It occurs when the rate at which reactants are converted to products equals the rate at which products revert back to reactants, resulting in no net change in the concentration of substances involved in the reaction over time. This doesn't mean the reaction has stopped; instead, it continues with no effect on concentrations of reactants and products.

Take the ionization of hydrogen peroxide: \( H_2O_2(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + HO_2^-(aq) \). Here, we have a dynamic equilibrium because the formation and recombination of ions are continuously occurring at the same rate. To understand the reaction's behavior at equilibrium, we use an equilibrium expression which, for acid ionization, is also the acid ionization constant (\( K_a \)). This constant helps predict the direction a reaction will proceed to reach equilibrium under different conditions.

Aqueous Solution Concentration

The aqueous solution concentration is another vital aspect affecting the chemistry of acids and bases. It's a measure of how much of a substance is dissolved in a particular volume of solvent (in this case, water). In the context of acids, it's crucial to know the concentration of the acid in its aqueous environment to analyze its ionization and the corresponding \( K_a \).

For instance, in the experiments performed with hydrogen peroxide, understanding the concept of concentration was key to calculating the acid's \( pK_a \) value. By looking at the concentration of \( H_2O_2 \) in different layers and comparing them, we could infer about how much of it ionized in the aqueous layer. In experimental setups where the concentration of the acid differs in two phases, like water and 1-pentanol, the unequal distribution can be utilized to deduce the extent of ionization, and thus, the acid's strength.