Question

Deuterium oxide \(\left(\mathrm{D}_{2} \mathrm{O},\right.\) where \(\mathrm{D}\) is deuterium, the hydrogen- 2 isotope) has an ion-product constant, \(K_{w^{+}}\) 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

The concentrations of \(\mathrm{D}^+\) and \(\mathrm{OD}^-\) ions in pure deuterium oxide at \(20^{\circ}\mathrm{C}\) are both \(2.98 \times 10^{-8} \mathrm{M}\).

Step-by-step solution

Write the equilibrium expression for \(K_w\)

For the ionization of \(\mathrm{D}_2\mathrm{O}\), the equilibrium constant is given by: $$K_w = [\mathrm{D}^+][\mathrm{OD}^-]$$ We are given the value of \(K_w = 8.9 \times 10^{-16}\) at \(20^{\circ}\mathrm{C}\).

Set up the relationship between the ion concentrations

Since the concentration of \(\mathrm{D}^+\) and \(\mathrm{OD}^-\) ions are equal at neutrality, we can set up the following relationship: $$[\mathrm{D}^+] = [\mathrm{OD}^-]$$ Let \(x\) represent the concentration of both \(\mathrm{D}^+\) and \(\mathrm{OD}^-\) ions.

Substitute the values into the equilibrium expression

We can now substitute the values of \(K_w\) and the relationship between the ion concentrations into the equilibrium expression from Step 1: $$8.9 \times 10^{-16} = x^2$$

Solve for x

Now, we need to solve the equation for \(x\) to find the concentration of both \(\mathrm{D}^+\) and \(\mathrm{OD}^-\) ions: $$x = \sqrt{8.9 \times 10^{-16}}$$ $$x \approx 2.98 \times 10^{-8} \mathrm{M}$$

Calculate the concentrations of ions

Since \(x\) represents the concentration of both \(\mathrm{D}^+\) and \(\mathrm{OD}^-\) ions in the pure deuterium oxide, we have: $$[\mathrm{D}^+] = [\mathrm{OD}^-] = 2.98 \times 10^{-8} \mathrm{M}$$ Thus, the concentrations of \(\mathrm{D}^+\) and \(\mathrm{OD}^-\) ions in pure deuterium oxide at \(20^{\circ}\mathrm{C}\) are both \(2.98 \times 10^{-8} \mathrm{M}\).

Key concepts

Equilibrium Constant

Understanding the equilibrium constant is crucial for solving problems related to chemical equilibrium. When a reversible reaction reaches equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of the reactants and products remain constant.

For any reversible reaction, the equilibrium constant (\(K\text{eq}\) or simply, \(K\)) is a fixed value that quantifies the ratio of the concentrations of products to reactants, each raised to the power of their respective coefficients in the balanced equation.For example, in the ionization of deuterium oxide (\(\mathrm{D}_2\mathrm{O}\)), the equilibrium constant (\(K_w\)) is given by the expression:\[K_w = [\mathrm{D}^+][\mathrm{OD}^-]\]where \( [\mathrm{D}^+] \) and \( [\mathrm{OD}^-] \) are the concentrations of the ions at equilibrium. The provided value of \(8.9 \times 10^{-16}\) at \(20^\circ\text{C}\) represents the specific ion-product constant for deuterium oxide, different from that of regular water due to isotopic effects.

Isotopes in Chemistry

Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons. This causes them to have the same atomic number but different mass numbers. For example, hydrogen has three isotopes: protium (\(\text{H}\)), deuterium (\(\text{D}\)), and tritium (\(\text{T}\)). Deuterium, where the hydrogen atom has one neutron, is what gives us deuterium oxide (\(\text{D}_2\text{O}\)), commonly referred to as 'heavy water' due to its higher mass compared to ordinary water (\(\text{H}_2\text{O}\)).

Isotopes can have different physical properties, such as melting points, boiling points, and densities, but they typically behave identically in chemical reactions. However, the slight differences can affect reaction rates and equilibrium constants, leading to variations such as the unique ion-product constant of deuterium oxide when compared to regular water.

Acid-Base Neutral Solutions

In acid-base chemistry, a neutral solution is one in which the concentrations of hydrogen ions (\(\text{H}^+\)) and hydroxide ions (\(\text{OH}^-\)) are equal, resulting in a pH of 7 at 25°C. However, when dealing with solutions of deuterium oxide, the deuterium ions (\(\text{D}^+\)) and the deutoxide ions (\(\text{OD}^-\)) play analogous roles as \(\text{H}^+\) and \(\text{OH}^-\) do in water.For pure \(\text{D}_2\text{O}\), a neutral solution means that the concentrations of \(\text{D}^+\) and \(\text{OD}^-\) are equal. As highlighted by the exercise, this condition allows us to simplify the equilibrium expression to find the individual ion concentrations that characterize a neutral deuterium oxide solution.