Question

Deuterium oxide (\(\mathrm{D}_{2} \mathrm{O},\) 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

In pure (neutral) D2O at \(20^{\circ} \mathrm{C}\), the concentrations of D+ and OD- ions are equal and can be calculated using the given ion-product constant, \(K_w = 8.9 \times 10^{-16}\). By solving for their concentrations, we find that \([\mathrm{D}^+] = [\mathrm{OD}^-] = 2.98 \times 10^{-8} \ \mathrm{M}\) at this temperature.

Step-by-step solution

Write down the expression for the ion-product constant, \(K_w\), for D2O

The ion-product constant, \(K_w\), for water (H2O) is given by the product of the concentrations of H+ and OH- ions. Similarly, for deuterium oxide (D2O), the ion-product constant is the product of the concentrations of D+ and OD- ions: \[K_w = [\mathrm{D}^+][\mathrm{OD}^-]\]

Determine the conditions for pure (neutral) D2O

In pure (neutral) D2O, the concentrations of its ions, D+ and OD-, are equal. Let's denote the concentration of D+ and OD- as x: \[[\mathrm{D}^+] = [\mathrm{OD}^-] = x\]

Substitute the given value of \(K_w\) and conditions for pure D2O into the ion-product constant expression

We know that \(K_w = 8.9 \times 10^{-16}\) at \(20^{\circ} \mathrm{C}\). Substituting the given value of \(K_w\) and the conditions for pure D2O into the ion-product constant expression, we get: \[8.9 \times 10^{-16} = x^2\]

Solve for x, the concentration of D+ and OD- ions

To find the concentration of D+ and OD- ions (x), we need to solve the equation obtained in step 3: \[x^2 = 8.9 \times 10^{-16}\] Taking the square root of both sides: \[x = \sqrt{8.9 \times 10^{-16}}\] \[x = 2.98 \times 10^{-8}\]

Write down the final answer

We found that x = 2.98 x 10^{-8}. Since we denoted the concentrations of D+ and OD- ions as x, and they are equal in pure D2O, the final answer is: \[[\mathrm{D}^+] = [\mathrm{OD}^-] = 2.98 \times 10^{-8} \ \mathrm{M}\]

Key concepts

Ion-product constant

The ion-product constant, often symbolized as \(K_w\), represents a key component in understanding the behavior of water and similar compounds like deuterium oxide (\(\mathrm{D}_2\mathrm{O}\)). In any water-based solution, \(K_w\) is calculated as the product of the concentrations of the hydrogen ions (acidity) and hydroxide ions (basicity). For deuterium oxide, the equation parallels that of water:

• \(K_w = [\mathrm{D}^+][\mathrm{OD}^-]\)

This equation informs us about the internal balance of ions in \(\mathrm{D}_2\mathrm{O}\). The given \(K_w\) value is \(8.9 \times 10^{-16}\) at \(20^\circ\mathrm{C}\). This small number indicates a very slight ionization, showcasing the weak nature of the ion presence in pure deuterium oxide compared to more reactive solutions. Understanding this constant helps us predict the behavior and properties of deuterium oxide in various chemical contexts.

Neutral solutions

A neutral solution is one where the concentrations of hydrogen ions and hydroxide ions are equal. For deuterium oxide, instead of hydrogen ions, it’s deuterium ions \([\mathrm{D}^+]\) present. The idea of neutrality in \(\mathrm{D}_2\mathrm{O}\) is mirrored by the idea in water, which means that \([\mathrm{D}^+] = [\mathrm{OD}^-]\) in a neutral environment.

This equilibrium is deeply connected to the \(K_w\) value. In the exercise, we solved the equation \(x^2 = 8.9 \times 10^{-16}\) to find that the concentrations \([\mathrm{D}^+]\) and \([\mathrm{OD}^-]\) both equal \(2.98 \times 10^{-8} \ \mathrm{M}\).

• This reflects a pure, neutral deuterium oxide solution.
• Helps in understanding symmetries in water-like chemistry.

By equating the ion concentrations, it becomes simple to analyze changes to \(\mathrm{D}_2\mathrm{O}\) when mixed with other compounds or exposed to different conditions.

Deuterium chemistry

Deuterium is an isotope of hydrogen, denoted as \(\mathrm{D}\), distinguished by the presence of one proton and one neutron. This gives it a unique set of properties, subtly different from regular hydrogen (\(\mathrm{H}\)), which has only a proton.

When part of a compound such as deuterium oxide (\(\mathrm{D}_2\mathrm{O}\)), these differences matter. Here’s why deuterium oxide is of interest:

• It acts as a heavier "sibling" to regular water (\(\mathrm{H}_2\mathrm{O}\)).
• Its properties make it valuable in scientific research, especially in nuclear physics and spectroscopy.

In the context of ions, Deuterium does not significantly change the ionization constant, yet it provides insights into isotopic effects in chemistry.

The chemistry involving deuterium highlights the nuanced effects isotopes have on chemical reactions, solubility, and even boiling and freezing points of substances. Understanding deuterium chemistry enriches our grasp of basic principles, aiding advances in various scientific fields.