Question

Deuterium oxide \(\left(\mathrm{D}_{2} \mathrm{O}\right.\), where \(\mathrm{D}\) is deuterium, the hydrogen- 2 isotope) has an ion-product constant, \(K_{\mathrm{uu}}\) of \(8.9 \times 10^{-16}\) at \(20^{\circ} \mathrm{C}\) Calculate \(\left[\mathrm{D}^{+}\right]\) and \(\left[\mathrm{OD}^{-}\right]\) for pure(neutral) \(\mathrm{D}_{2} \mathrm{O}\) at this temperature.

Short answer

For a neutral deuterium oxide solution at \(20^{\circ} \mathrm{C}\), the concentrations of \(\mathrm{D}^{+}\) and \(\mathrm{OD}^{-}\) ions are approximately \(2.98 \times 10^{-8}\) mol/L each.

Step-by-step solution

Understand Ion-Product Constant

The ion-product constant, \(K_{\mathrm{w}}\), represents the equilibrium constant for the ionization of water (or in this case, deuterium oxide). The formula for the ion-product constant is: \[K_{\mathrm{w}} = [\mathrm{D}^{+}][\mathrm{OD}^{-}]\] We are given \(K_{\mathrm{w}} = 8.9 \times 10^{-16}\), and we need to determine the concentrations \([\mathrm{D}^{+}]\) and \([\mathrm{OD}^{-}]\) for a neutral solution.

Analyzing Neutral Solution

For a neutral solution of deuterium oxide, the concentrations of the cations and anions will be equal, that is, \[[\mathrm{D}^{+}] = [\mathrm{OD}^{-}]\] Let's denote their equal concentration as 'a': \[a = [\mathrm{D}^{+}] = [\mathrm{OD}^{-}]\]

Substitute Concentrations in Ion-Product Constant Formula

Now, let's substitute the equal concentrations (a) into the ion-product constant formula from Step 1: \[K_{\mathrm{w}} = [\mathrm{D}^{+}][\mathrm{OD}^{-}] = a \cdot a = a^2\]

Calculate Concentrations

We already have the value of \(K_{\mathrm{w}}\). Now we just need to solve for 'a': \[a^2 = K_{\mathrm{w}} = 8.9 \times 10^{-16}\] Now, to find 'a', we simply take the square root of both sides of the equation: \[a = \sqrt{8.9 \times 10^{-16}}\] \[a \approx 2.98\times10^{-8}\] Now that we have found the value of 'a', we can determine the concentrations of \(\mathrm{D}^{+}\) and \(\mathrm{OD}^{-}\): \[[\mathrm{D}^{+}] = a = 2.98\times10^{-8}\] \[[\mathrm{OD}^{-}] = a = 2.98\times10^{-8}\] So, the concentration of \(\mathrm{D}^{+}\) and \(\mathrm{OD}^{-}\) ions in the neutral deuterium oxide solution at this temperature is approximately \(2.98 \times 10^{-8}\) mol/L.

Key concepts

Deuterium Oxide

Deuterium oxide, commonly referred to as heavy water, is a form of water where the hydrogen atoms are replaced by deuterium, an isotope of hydrogen. While regular water has the formula H2O, deuterium oxide is denoted as D2O. The term isotopes indicate that while both hydrogen and deuterium hold the same number of protons, deuterium has an additional neutron. This slight difference leads to variations in physical and chemical properties between the two substances.

For instance, deuterium oxide has a higher freezing point and boiling point than regular water. Because of its unique properties, D2O is used in various applications, including nuclear reactors, where it serves as a neutron moderator. The isotopic difference also influences the ion-product constant of D2O compared to H2O, essentially affecting its chemical behavior in dissociation reactions.

Equilibrium Constant

The equilibrium constant, symbolized by K, is a crucial concept in chemical equilibrium, representing the constant ratio of the concentration of products to reactants for a reversible chemical reaction at equilibrium. The ion-product constant, Kw, is a specific type of equilibrium constant that applies to the self-ionization of water, where water molecules dissociate into hydrogen ions (H+) and hydroxide ions (OH-).

For deuterium oxide, the ion-product constant is symbolized by Kw and is specifically used to denote the equilibrium concentrations of deuterium ions [D+] and hydroxide variants [OD-]. The value of Kw varies with temperature and isotopic composition, and this variation must be accounted for when solving chemical problems related to deuterium oxide.

Isotopes

Isotopes are atoms of the same element that contain the same number of protons but a different number of neutrons, resulting in varying atomic masses. Hydrogen has three isotopes: protium (most common, no neutrons), deuterium (one neutron), and tritium (two neutrons).

Deuterium, being one of these isotopes, plays a key role in the properties and behavior of deuterium oxide. Isotopic differences can lead to divergent chemical and physical properties such as bond strength, reaction rates, and equilibrium constants. Understanding these isotopic effects is vital for accurately predicting the behavior of chemical species in various contexts, including isotopic substitution in reactions.

Chemical Concentration

Chemical concentration refers to the amount of a substance within a given volume of solution or mixture. It is usually expressed in moles per liter (mol/L) for solutions. In the context of deuterium oxide, understanding the concentration of deuterium ions [D+] and the hydroxide variants [OD-] is essential for solving problems related to its ion-product constant.

When deuterium oxide undergoes self-ionization, D2O dissociates into [D+] and [OD-]. In a neutral solution, the concentration of these ions will be equal, as the product of their concentrations equals the ion-product constant, Kw. Being skilled in calculating and understanding the equilibrium concentrations of these ions is important for students studying aqueous reactions and properties of acids and bases.