Problem 112 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

A solution contains a mixture of \(\mathrm{Ag}^{+}\) \((0.10 \mathrm{M})\) and \(\mathrm{Hg}_{2}^{2+}(0.10 \mathrm{M})\), which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What per cent of that metal ion is precipitated, before the start of precipitation of second metal ion? \(K_{\mathrm{sp}}(\mathrm{AgI})=8.5 \times 10^{-17}\) and \(K_{\mathrm{sp}}\left(\mathrm{Hg}_{2} \mathrm{I}_{2}\right)=2.5 \times 10^{-26}\) (a) \(5 \times 10^{-13} \mathrm{M}, 99.83 \%\) (b) \(8.5 \times 10^{-16} \mathrm{M}, 99.83 \%\) (c) \(2.5 \times 10^{-25} \mathrm{M}, 100 \%\) (d) \(5 \times 10^{-13} \mathrm{M}, 98.3 \%\)

Step-by-Step Solution

Answer

\(5 \times 10^{-13} \mathrm{M}, 100 \%\)

Step 1 - Write Down the Equilibria

For the precipitation reaction, we have two equilibria to consider: 1) For AgI: \[\mathrm{Ag}^+ + \mathrm{I}^- \rightleftharpoons \mathrm{AgI}(s)\] with \(K_{\mathrm{sp}}(\mathrm{AgI}) = 8.5 \times 10^{-17}\).2) For \(\mathrm{Hg}_2\mathrm{I}_2\): \[\mathrm{Hg}_2^{2+} + 2\mathrm{I}^- \rightleftharpoons \mathrm{Hg}_2\mathrm{I}_2(s)\] with \(K_{\mathrm{sp}}(\mathrm{Hg}_2\mathrm{I}_2) = 2.5 \times 10^{-26}\).

Step 2 - Calculate the Saturation Concentration of I- for Ag+ Precipitation

Set the expression for the solubility product of AgI: \[K_{\mathrm{sp}}(\mathrm{AgI}) = [\mathrm{Ag}^+][\mathrm{I}^-]\] Substitute the known \(K_{sp}\) of AgI and the concentration of \(\mathrm{Ag}^+\) to find \([\mathrm{I}^-]\): \[8.5 \times 10^{-17} = (0.10\,\mathrm{M}) \times [\mathrm{I}^-]\] Solve for \([\mathrm{I}^-]\): \[\begin{align*}[\mathrm{I}^-] &= \frac{8.5 \times 10^{-17}}{0.10\,\mathrm{M}}\&= 8.5 \times 10^{-16}\,\mathrm{M}\end{align*}\]

Step 3 - Calculate the Saturation Concentration of I- for Hg22+ Precipitation

Set up the expression for the solubility product of \(\mathrm{Hg}_2\mathrm{I}_2\): \[K_{\mathrm{sp}}(\mathrm{Hg}_2\mathrm{I}_2) = [\mathrm{Hg}_2^{2+}][\mathrm{I}^-]^2\] Substitute the \(K_{sp}\) of \(\mathrm{Hg}_2\mathrm{I}_2\) and the concentration of \(\mathrm{Hg}_2^{2+}\) to find \([\mathrm{I}^-]\): \[2.5 \times 10^{-26} = (0.10\,\mathrm{M}) \times [\mathrm{I}^-]^2\] Solve for \([\mathrm{I}^-]\): \[\begin{align*}[\mathrm{I}^-]^2 &= \frac{2.5 \times 10^{-26}}{0.10\,\mathrm{M}}\ &= 2.5 \times 10^{-25}\end{align*}\] Take the square root: \[\begin{align*}[\mathrm{I}^-] &= \sqrt{2.5 \times 10^{-25}}\&= 5 \times 10^{-13}\,\mathrm{M}\end{align*}\]

Step 4 - Compare the Saturation Concentration of I-

The lower concentration of iodide ions that would result in precipitation is the one that will precipitate first. So, compared the saturation concentrations calculated from Step 2 and Step 3, we choose the smaller value. Since \(5 \times 10^{-13}\,\mathrm{M} < 8.5 \times 10^{-16}\,\mathrm{M}\), the \(\mathrm{Hg}_2\mathrm{I}_2\) precipitation will occur first.

Step 5 - Calculate the Percentage Precipitation of \(\mathrm{Hg}_2^{2+}\)

At the moment \(\mathrm{Hg}_2^{2+}\) begins to precipitate, there is no \(\mathrm{I}^-\) left to precipitate \(\mathrm{Ag}^{+}\) because all iodide is consumed by \(\mathrm{Hg}_2^{2+}\). Thus, essentially, 100% of the \(\mathrm{Hg}_2^{2+}\) is precipitated before the start of precipitation of \(\mathrm{Ag}^{+}\).

Key Concepts

Solubility Product Constant (Ksp)

Understanding the solubility product constant, often abbreviated as Ksp, is crucial when diving into the realm of selective precipitation in chemistry. This constant is a special type of equilibrium constant that measures the solubility of a sparingly soluble compound. In simple terms, it represents the maximum amount of the compound that can dissolve in a solution to form a saturated solution before excess solid precipitates out.

The value of Ksp is unique for every sparingly soluble compound and is influenced by temperature. For a general ionic compound \textsf{A}a\textsf{B}b that dissociates into 'a' cations of \textsf{A} and 'b' anions of \textsf{B}, the Ksp expression would be:
\[ K_{\mathrm{sp}} = [\mathrm{A}^{+}]^a [\mathrm{B}^{-}]^b \]
It's essential to note that the concentrations are those at equilibrium, where the solid is in a dynamic balance with its ions. In the exercise, the Ksp values for AgI and \textsf{Hg}\(_2\)I\(_2\) provided the necessary information to calculate the saturation concentrations of iodide ions required to begin precipitation of each respective solid.

Saturation Concentration Calculation

When dealing with selective precipitation, the concept of saturation concentration is a key player. It refers to the concentration of a dissolved substance in a solution at which no more of that substance can dissolve; it’s the tipping point before precipitation occurs. Calculating this concentration is a matter of algebraic manipulation of the Ksp value of the compound in question.

Using the solubility product constant, you set up an equation that relates the molarities of the ions in solution at saturation:
\[ K_{\mathrm{sp}} = [\mathrm{Cation}]^n [\mathrm{Anion}]^m \]
Where 'n' and 'm' correspond to the stoichiometry of the ions in the precipitation equation. For instance, in the exercise, we saw the calculation for the maximum concentration of iodide ions for both AgI and \textsf{Hg}\(_2\)I\(_2\) prior to precipitation. Knowing how to calculate the saturation concentration is vital for predicting when a compound will precipitate from solution.

Percent Precipitation Calculation

Percent precipitation is a measure of how much of a particular ion is removed from solution in the form of a precipitate. It's an insightful way to understand the efficiency of a selective precipitation process. This percentage is calculated by comparing the amount of the ion before and after the precipitation reaction completes, typically expressed as:
\[ \% \, \mathrm{Precipitation} = \left(\frac{\text{Initial Concentration - Final Concentration}}{\text{Initial Concentration}}\right) \times 100 \% \]
In the context of our exercise, we looked at the moment at which \textsf{Hg}\(_2^{2+}\) begins to precipitate, upon reaching the saturation concentration of iodide ions. Since the saturation concentration for \textsf{Hg}\(_2\)^\textsf{2+} is lower than that for Ag\(^+\), \textsf{Hg}\(_2\)^\textsf{2+} precipitates first, and we can consider that virtually 100% of \textsf{Hg}\(_2\)^\textsf{2+} is precipitated before Ag\(^{+}\) starts to come out of the solution. This brings out the importance of the sequence in which ions precipitate based on their Ksp values for selective precipitation processes.