Question

Deuterium oxide, \(\mathrm{D}_{2} \mathrm{O},\) ionizes like water. At \(20^{\circ} \mathrm{C}\) its \(K_{\mathrm{w}}\) or ion product constant, analogous to that of water, is \(8.9 \times 10^{-16} .\) Calculate \(\left[\mathrm{D}^{+}\right]\) and \(\left[\mathrm{OD}^{-}\right]\) in deuterium oxide at \(20^{\circ} \mathrm{C}\). Calculate also the \(\mathrm{pD}\) and the \(\mathrm{pOD}\). What would be the neutral \(\mathrm{pD}\) ?

Short answer

The concentrations are [D+] = [OD-] = sqrt(8.9 x 10^-16) = 9.43 x 10^-8 M. The pD and pOD are both -log(9.43 x 10^-8) = 7.025. Neutral pD is half of pKw, which is 7.025.

Step-by-step solution

Understanding the Ion Product of Water

The ion product of water, Kw, is the product of the molar concentrations of H+ and OH- ions in water. For Deuterium oxide (D2O) at 20°C, the ion product constant is given as 8.9 x 10^-16. This can be represented as Kw = [D+][OD-] where [D+] is the concentration of deuterium ions and [OD-] is the concentration of deuterium hydroxide ions.

Determine the Concentration of Deuterium Ions and Deuterium Hydroxide Ions

Because D2O ionizes similar to H2O, and assuming it dissociates completely into D+ and OD- ions, the concentrations of these ions would be equal at equilibrium. Thus, [D+] = [OD-]. We can express this mathematically using the ion product constant: Kw = [D+]^2. Solving for [D+] or [OD-] yields [D+] = [OD-] = sqrt(Kw).

Calculate the Concentration of D+ and OD-

Substitute the given Kw value into the equation from the previous step: [D+] = [OD-] = sqrt(8.9 x 10^-16).

Determine pD and pOD

pD and pOD are calculated similar to pH and pOH in standard water. The formulas are pD = -log[D+] and pOD = -log[OD-].

Calculate the Neutral pD

Neutral pD can be found when [D+] equals [OD-]. Under neutral conditions pD = pOD, therefore neutral pD can be calculated as half of the pKw, pKw being the negative logarithm of the ion product constant of deuterium oxide. pKw = -log(Kw).

Key concepts

Deuterium Oxide D2O

Deuterium oxide, also known as heavy water, has a chemical formula \( \mathrm{D}_{2} \mathrm{O} \). It is similar to water (\mathrm{H}_{2}\mathrm{O}) but has the hydrogen isotope deuterium, denoted by \( \mathrm{D} \), which contains one neutron and one proton in its nucleus, making it heavier than the proton-only isotope of hydrogen.

Like regular water, deuterium oxide also undergoes self-ionization, albeit at a different ion product constant due to the unique properties of deuterium. Even with this difference, \( \mathrm{D}_{2}\mathrm{O} \) ionizes to form \( \mathrm{D}^{+} \) and \( \mathrm{OD}^{-} \) ions. These ions are essential for different chemical processes and have significant implications in nuclear chemistry, particularly in nuclear reactors, where \( \mathrm{D}_{2}\mathrm{O} \) is often used as a moderator to slow down neutrons.

pD Calculation

The pD of a solution with deuterium oxide is analogous to the pH in regular water, representing the negative logarithm of the \( \mathrm{D}^{+} \) ion concentration. Calculating pD involves finding the concentration of \( \mathrm{D}^{+} \) ions and then taking the negative logarithm of that value. Since \( \mathrm{D}_{2} \mathrm{O} \) ionizes producing equal amounts of \( \mathrm{D}^{+} \) and \( \mathrm{OD}^{-} \) ions, for a solution in equilibrium, this can be simplified.

The steps begin with knowing the \( K_{\mathrm{w}} \) of \( \mathrm{D}_{2}\mathrm{O} \) and then calculating the square root to give equal concentrations for \( \mathrm{D}^{+} \) and \( \mathrm{OD}^{-} \) due to the dissociation equilibrium. Subsequently, \( \mathrm{pD} \) and \( \mathrm{pOD} \) are calculated similar to \( \mathrm{pH} \) and \( \mathrm{pOH} \) with the use of logarithms, catering for the properties of deuterium ions.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate, leading to an unchanging system where the reactants and products have constant concentrations over time. When deuterium oxide ionizes to form \( \mathrm{D}^{+} \) and \( \mathrm{OD}^{-} \) ions, it reaches a chemical equilibrium, characterized by a constant called the ion product constant (\( K_{\mathrm{w}} \)).

Understanding the behavior of \( \mathrm{D}_{2}\mathrm{O} \) at equilibrium is crucial for calculating pD, as it involves applying the principles of equilibrium to the dissociation process. The calculated concentrations of \( \mathrm{D}^{+} \) and \( \mathrm{OD}^{-} \) are those present when the forward and reverse reactions are balanced, demonstrating the core role equilibrium principles play in determining the ion concentration, pD, and pOD of deuterium oxide solutions.