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 for pure \(\mathrm{D}_{2} \mathrm{O}\) at \(20^{\circ} \mathrm{C}\) are approximately \(2.98 \times 10^{-8}\,\text{M}\).

Step-by-step solution

Write down the ion-product constant expression for \(\mathrm{D}_{2} \mathrm{O}\)

The ion-product constant expression is given by: \[K_{w} = [\mathrm{D}^+][\mathrm{OD}^-]\] As \(\mathrm{D}_{2} \mathrm{O}\) is neutral, the concentration of \(\mathrm{D}^+\) ions is equal to \(\mathrm{OD}^-\) ions. Therefore, we can simplify the equation by replacing both concentrations with a single variable, say \(x\). \[K_{w} = x^2\] Now, we are given the value of \(K_{w} = 8.9 \times 10^{-16}\) and we can plug this value into the equation and solve for \(x\).

Solve for \(x\)

We have the equation: \[8.9 \times 10^{-16} = x^2\] To solve for \(x\), we take the square root of both sides of the equation: \[x = \sqrt{8.9 \times 10^{-16}}\] Calculating this value, we get: \[x \approx 2.98 \times 10^{-8}\]

Determine the concentrations of \(\mathrm{D}^+\) and \(\mathrm{OD}^-\)

Now that we have found the value of \(x\), we can determine the concentration of \(\mathrm{D}^+\) and \(\mathrm{OD}^-\) ions. As we noted earlier, the concentration of \(\mathrm{D}^+\) ions is equal to the concentration of \(\mathrm{OD}^-\) ions. \[[\mathrm{D}^+] = [\mathrm{OD}^-] = x \] Therefore, the concentrations of \(\mathrm{D}^+\) and \(\mathrm{OD}^-\) ions are approximately \(2.98 \times 10^{-8}\,\text{M}\).

Key concepts

Ion-Product Constant

In chemistry, the ion-product constant is essential for understanding the behavior of water-based systems. This concept, often represented as \(K_w\), defines the equilibrium concentrations of hydrogen ions \([ ext{H}^+]\) and hydroxide ions \([ ext{OH}^-]\) in pure water. This constant is crucial because it helps us determine the acidity or basicity of a solution. For pure water at 25°C, \(K_w\) is typically \(1.0 \times 10^{-14}\). However, when discussing deuterium oxide or heavy water \(( ext{D}_2 ext{O})\), the \(K_w\) changes due to the presence of deuterium. For \( ext{D}_2 ext{O}\) at 20°C, \(K_w\) is given as \(8.9 \times 10^{-16}\). Understanding this value helps us calculate the concentration of deuterium ions \([ ext{D}^+]\) and the deuterium equivalent of hydroxide ions \([ ext{OD}^-]\) in a neutral solution, which are essential for quantifying chemical equilibrium in such settings.

Deuterium

Deuterium is an isotope of hydrogen, distinctive in having one neutron in addition to the single proton found in the nucleus of a hydrogen atom. This extra neutron doubles the mass of deuterium compared to protium, the most common hydrogen isotope. Deuterium is symbolized as \( ext{D}\), and forms compounds like deuterium oxide \( ext{D}_2 ext{O}\), commonly known as heavy water. Due to its additional mass, deuterium has slightly different physical and chemical properties compared to ordinary hydrogen. These differences become significant in nuclear reactors and in various fields of chemistry where isotopic considerations are essential. Deuterium's role in heavy water, for instance, makes it crucial in calculating ion-product constants and understanding the subtle changes in chemical equilibrium.

Neutral D2O

Neutral deuterium oxide \((\text{D}_2\text{O})\), commonly known as heavy water, behaves similarly to normal water in terms of neutrality. In its pure form, \( ext{D}_2\text{O}\) maintains equal concentrations of \([ ext{D}^+]\) ions and \([ ext{OD}^-]\) ions. This balance ensures that the \(pD\) scale, similar to the \(pH\) scale in normal water, indicates neutrality, meaning neither acidic nor basic. The neutrality in \( ext{D}_2\text{O}\) ensures that chemical reactions occurring in heavy water do not favor either \( ext{D}^+\) or \( ext{OD}^-\), maintaining a stable chemical equilibrium. This aspect of \( ext{D}_2\text{O}\) is particularly useful in scientific research where the behavior and reactivity of substances in heavy water are studied without the interference of extra acidity or basicity.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse proceed at the same rate, leading to stable concentrations of the reactants and products. This dynamic process ensures that there are no net changes in the concentration of the substances involved. In the context of deuterium oxide \((\text{D}_2\text{O})\), chemical equilibrium is reflected in the constant concentrations of \([ ext{D}^+]\) and \([ ext{OD}^-]\) ions. The ion-product constant \(K_w\) provides a quantitative measure of this equilibrium.

• When disturbed, such as by temperature changes, chemical equilibrium can shift, causing a temporary imbalance that eventually stabilizes again.
• For \( ext{D}_2\text{O}\) at 20°C, the specific \(K_w\) value of \(8.9 \times 10^{-16}\) dictates the equilibrium concentrations of ions, impacting the chemical properties of the solution.
• This understanding is vital in fields such as analytical chemistry and pharmaceuticals, where precise control over reaction conditions is essential.

Understanding these dynamics helps in predicting the behavior of reactions within deuterium environments effectively.