Question

When \(20 \mathrm{ml}\) of \(0.2 \mathrm{M}-\mathrm{DCl}\) solution is mixed with \(80 \mathrm{ml}\) of \(0.1 \mathrm{M}-\mathrm{NaOD}\) solution, \(\mathrm{pD}\) of the resulting solution becomes \(13.6 .\) The ionic product of heavy water, \(\mathrm{D}_{2} \mathrm{O}\), is (a) \(10^{-15}\) (b) \(10^{-16}\) (c) \(4 \times 10^{-15}\) (d) \(4 \times 10^{-16}\)

Short answer

The ionic product of heavy water, \(D_{2}O\), is \(4 \times 10^{-15}\).

Step-by-step solution

Determine the amount of substance in each solution

Calculate the moles of DCl and NaOD using the volume and concentration. For DCl, the moles are given by volume in liters multiplied by molarity, which is 0.02 L * 0.2 M = 0.004 moles. Similarly, for NaOD, 0.08 L * 0.1 M = 0.008 moles.

Calculate the changes in moles after reaction

The reaction between DCl and NaOD is a neutralization reaction where D+ from DCl will react with OD- from NaOD forming D2O and NaCl. As there is twice as much NaOD as DCl, all DCl will react and 0.004 moles of NaOD will remain unreacted.

Calculate the concentration of OD- in the final solution

The final volume of the solution is the sum of the volumes of the two solutions mixed, which is 20 ml + 80 ml = 100 ml or 0.1 liters. The concentration of OD- will be the moles of OD- that didn't react divided by the total volume, which is 0.004 moles/0.1 L = 0.04 M.

Convert pD to [D+]

Using the pD value provided (13.6), we can calculate the concentration of D+ as follows: pD = -log[D+], so [D+] = 10^{-pD}. Substituting the pD value, [D+] = 10^{-13.6}.

Calculate the ionic product of D2O (\(K_{D2O}\))

The ionic product is given by [D+][OD-] = 10^{-13.6} * 0.04 M. This calculation gives us the value of the ionic product of heavy water.

Find the correct answer based on the calculated ionic product

After performing the calculation from the previous step we get the ionic product of heavy water as 4 x 10^{-15}. This matches answer choice (c).

Key concepts

Chemical Equilibrium

Understanding the concept of chemical equilibrium is crucial when analyzing reactions like the one between DCl and NaOD. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, and the concentrations of reactants and products remain constant over time.

However, when DCl reacts with NaOD in the given problem, we assume the reaction goes to completion rather than reaching a dynamic equilibrium. This is typical for neutralization reactions, which are generally considered irreversible under standard conditions. In this specific case, DCl and NaOD fully react to form D₂O and NaCl.

To really grasp chemical equilibrium, it is key to recognize that not all reactions reach this state. In the context of the problem, understanding that the reaction goes to completion allows us to simplify the calculations by not considering the reverse reaction when calculating the ionic product of heavy water.

Neutralization Reaction

A neutralization reaction is a type of chemical reaction where an acid and a base react to form a salt and water. In our problem, the reactants are deuterated hydrochloric acid (DCl) and sodium deuteroxide (NaOD), which are deuterium analogs of HCl and NaOH, respectively.

The key to a neutralization reaction is the transfer of a proton—though, in the case of heavy water, it's a deuteron (D+). Here's the reaction:

Neutralization Reaction: DCl + NaOD → NaCl + D₂O

It's essential to note the stoichiometry; one mole of acid reacts with one mole of base. Since there's an excess of NaOD, after all DCl has been neutralized, we're left with unreacted NaOD, which directly affects the pD calculation and the ionic product of D₂O.

pD Calculation

The pD is analogous to pH but for deuterium ions (D+). It quantifies the deuterium ion concentration in a solution and is calculated using a negative logarithm, similar to pH calculations. In the provided problem:

pD and [D+]: pD = -log[D+]

Given a pD of 13.6, we can deduce the [D+] using the formula:
[D+] = 10^-pD or [D+] = 10^-13.6. This formula provides the concentration of D+ ions in the solution after the neutralization reaction.

Once the [D+] is known, combined with the concentration of OD- ions that were not consumed in the reaction, we can calculate the ionic product for heavy water (D₂O), which is the product of concentrations of the deuterium ion (D+) and the deuteroxide ion (OD-). Accurate pD calculation is fundamental in determining the ionic product, which is a unique property in heavy water just as the ionic product of water (Kw) is for regular water.