Question

The ion product of \(\mathrm{D}_{2} \mathrm{O}\) is \(1.35 \times 10^{-15}\) at \(25^{\circ} \mathrm{C}\). (b) For (a) Calculate \(\mathrm{pD}\) where \(\mathrm{pD}=-\log \left[\mathrm{D}_{3} \mathrm{O}^{+}\right]\). what values of \(\mathrm{pD}\) will a solution be acidic in \(\mathrm{D}_{2} \mathrm{O} ?\) (c) Derive a relation between \(\mathrm{pD}\) and \(\mathrm{pOD}\).

Short answer

(a) pD = 6.93. (b) Acidic if pD < 6.93. (c) pD + pOD = 15.13.

Step-by-step solution

Calculate the pD

The ion product of \(\mathrm{D}_{2}\mathrm{O}\) at \(25^\circ \mathrm{C}\) is given as \(K_w = 1.35 \times 10^{-15}\). To find the pD, we use the relation: \[ \mathrm{pD} = -\log [\mathrm{D}_3\mathrm{O}^+] \] and the fact that \( [\mathrm{D}_3\mathrm{O}^+][\mathrm{OD}^-] = 1.35 \times 10^{-15} \). At neutral conditions, where \([\mathrm{D}_3\mathrm{O}^+] = [\mathrm{OD}^-]\), the concentration of \([\mathrm{D}_3\mathrm{O}^+]\) can be found using: \[ [\mathrm{D}_3\mathrm{O}^+] = \sqrt{K_w} = \sqrt{1.35 \times 10^{-15}} \approx 1.16 \times 10^{-7} \] Therefore, \(\mathrm{pD} = -\log(1.16 \times 10^{-7}) \approx 6.93\).

Determine acidic conditions

A solution is considered acidic if \(\mathrm{pD} < 6.93\). This is because a lower \(\mathrm{pD}\) indicates a higher concentration of \([\mathrm{D}_3\mathrm{O}^+]\), similar to how \(\mathrm{pH} < 7\) indicates acidity in water.

Derive relation between pD and pOD

The relation between \(\mathrm{pD}\) and \(\mathrm{pOD}\) can be derived by taking the negative logarithm of both sides of the ion product equation \([\mathrm{D}_3\mathrm{O}^+][\mathrm{OD}^-] = 1.35 \times 10^{-15}\), giving us: \[ \log([\mathrm{D}_3\mathrm{O}^+]) + \log([\mathrm{OD}^-]) = \log(1.35 \times 10^{-15}) \] Substituting the definitions of \(\mathrm{pD}\) and \(\mathrm{pOD}\), we have: \[ -\mathrm{pD} - \mathrm{pOD} = -\log(1.35 \times 10^{-15}) \] \[ \mathrm{pD} + \mathrm{pOD} = 15 - \log(1.35) \approx 15.13 \] Thus, the relation is \(\mathrm{pD} + \mathrm{pOD} = 15.13\).

Key concepts

pD Calculation

Calculating the \(\mathrm{pD}\) of heavy water, denoted as \(\mathrm{D}_2\mathrm{O}\), involves understanding the ion product constant given by \(K_w\). This constant at \(25^{\circ} \mathrm{C}\) is \(1.35 \times 10^{-15}\). To determine \(\mathrm{pD}\), you use the formula: \[\mathrm{pD} = -\log ([\mathrm{D}_3\mathrm{O}^+])\]. The \(\mathrm{D}_3\mathrm{O}^+\) ions represent the deuterated form of hydronium ions found in heavy water.
Find the concentration of deuteron ions in neutral conditions using: \[ [\mathrm{D}_3\mathrm{O}^+] = \sqrt{K_w} = \sqrt{1.35 \times 10^{-15}} \approx 1.16 \times 10^{-7} \]
This approximation leads to a \(\mathrm{pD}\) value at neutrality:

• \(\mathrm{pD} = -\log(1.16 \times 10^{-7}) \approx 6.93\)

By determining the \(\mathrm{pD}\), you gain insight into the strength and concentration of deuterium ions in \(\mathrm{D}_2\mathrm{O}\).

Acidic Conditions

Understanding when a solution is acidic in heavy water involves comparing \(\mathrm{pD}\) values to the neutrality standard. A \(\mathrm{pD}\) less than 6.93 indicates an acidic environment. This mirrors the concept of pH in regular water systems, where lower values mean higher acidity due to increased hydronium ions.
Recognize key points:

• Lower \(\mathrm{pD}\) translates to a higher concentration of \(\mathrm{D}_3\mathrm{O}^+\).
• Being acidic parallels the idea that pH less than 7 represents acidity in H extsubscript{2}O.

Conceptually, grasping acidity in heavy water enables a deeper comprehension of how properties like ion concentration reflect the solution’s nature.

Relation between pD and pOD

Establishing a relationship between \(\mathrm{pD}\) and \(\mathrm{pOD}\) involves assessing their collective interaction within heavy water’s ion balance. Both represent logarithmic scales of ion concentrations, similar to \(pH\) and \(pOH\) in regular water. The basic relationship originates from the ion product equation: \[ [\mathrm{D}_3\mathrm{O}^+][\mathrm{OD}^-] = 1.35 \times 10^{-15} \]
This transforms into logarithmic form:

• \(\log([\mathrm{D}_3\mathrm{O}^+]) + \log([\mathrm{OD}^-]) = \log(1.35 \times 10^{-15})\)

By using definitions:

• \(-\mathrm{pD} - \mathrm{pOD} = -\log(1.35 \times 10^{-15})\)

Ultimately creating:

• \(\mathrm{pD} + \mathrm{pOD} = 15 - \log(1.35) \approx 15.13\)

This crucial formula illustrates that the combined acidity and basicity measures in heavy water sum to a fixed constant, defining the delicate balance of ion concentrations.