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Chapter 7: Problem 74

A solution contains a mixture of \(\mathrm{Ag}(0.10 \mathrm{M})\) and \(\mathrm{Hg}_{2}^{++}(0.10 \mathrm{M})\) 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 percentage of that metal ion is precipitated? \(\left[K_{s p}: \mathrm{AgI}=8.5 \times 10^{-17} ; \mathrm{Hg}_{2} \mathrm{I}_{2}=2.5 \times 10^{-26}\right]\)

Short Answer

Expert verified
AgI precipitates first at \([I^-] = 8.5 \times 10^{-16}\) M, with 99.83% of Ag precipitated.

Step by step solution

01

Understand the Equation for Precipitation

The ion product that causes precipitation can be determined by the solubility product constant ( \( K_{sp} \)). The equation for precipitation represents: \[ K_{sp,\ AgI} = [Ag^+][I^-] \] \[ K_{sp,\ Hg_2I_2} = [Hg_2^{2+}][I^-]^2 \] We have the initial concentrations and *K_sp* given, therefore we can use these to find the conditions under which one ion will precipitate before the other.
02

Calculate Iodide Concentration for AgI Precipitation

Calculate the concentration of iodide ions \([I^-]\) required to start precipitating silver iodide \(AgI\) almost completely using the solubility product: \[ K_{sp,\ AgI} = 8.5 \times 10^{-17} = (0.10) \cdot [I^-] \] Solving for \([I^-]\), we find: \[ [I^-] = \frac{8.5 \times 10^{-17}}{0.10} = 8.5 \times 10^{-16} \]
03

Calculate Iodide Concentration for Hg2I2 Precipitation

Similarly, calculate the concentration of iodide ions \([I^-]\) required to begin precipitating \(Hg_2I_2\) using the solubility product: \[ K_{sp,\ Hg_2I_2} = 2.5 \times 10^{-26} = (0.10) \cdot [I^-]^2 \] Solving for \([I^-]\), we find: \[ [I^-] = \sqrt{\frac{2.5 \times 10^{-26}}{0.10}} = \sqrt{2.5 \times 10^{-25}} \approx 5.0 \times 10^{-13} \]
04

Determine Which Precipitates First

Comparing the iodide concentrations calculated: - For \( AgI \), \([I^-] = 8.5 \times 10^{-16} \) - For \( Hg_2I_2 \), \([I^-] = 5.0 \times 10^{-13} \) Since \(8.5 \times 10^{-16} < 5.0 \times 10^{-13}\), \(AgI\) will precipitate first, indicating it will precipitate at a lower \([I^-]\) concentration.
05

Calculate the Percentage of AgI Precipitated

When \( [I^-] \) reaches the precipitation threshold for \( Hg_2I_2 \): Calculate the ion product for \(AgI\): \[ [Ag^+] = \frac{K_{sp,\ AgI}}{[I^-]_{above}} = \frac{8.5 \times 10^{-17}}{5.0 \times 10^{-13}} \approx 1.7 \times 10^{-4} \ M \] Percentage precipitated: \[ \text{Precipitated} \% = \left(\frac{0.10 - 1.7 \times 10^{-4}}{0.10}\right) \times 100\% \approx 99.83\% \]
06

Conclusion

The maximum concentration of iodide ion that allows almost complete precipitation of one of the metals is given by the limit for \(AgI\), and approximately \(99.83\%\) of the \(Ag^+\) ions are precipitated when \([I^-]\) is at the threshold for \(Hg_2^{2+}\) precipitation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solubility Product Constant
Understanding the Solubility Product Constant (K_sp) is crucial for predicting when a substance will start to precipitate out of a solution. It represents the equilibrium constant for the dissolution of a sparingly soluble compound. Each ionic compound in a solution has its specific K_sp value, which is determined experimentally and reflects solubility.

To calculate the point of precipitation, apply the concept of K_sp to the ionic concentrations in the solution. For example, for silver iodide (AgI), the solubility product is expressed as:
  • \( K_{sp,\ AgI} = [Ag^+][I^-] \)
Similarly, for mercury(I) iodide (Hg鈧侷鈧�), it is given by:
  • \( K_{sp,\ Hg_2I_2} = [Hg_2^{2+}][I^-]^2 \)
These equations help us determine the exact point when dissolution and precipitation achieve equilibrium. This balance influences how and when ions will separate from the solution, significantly impacting selective precipitation methods.
Ion Product
The Ion Product is a concept closely tied to K_sp. It refers to the product of the concentrations of the ions of a solute in solution, raised to the power of their coefficients in the balanced equation. The ion product allows us to predict whether a solution is unsaturated, saturated, or supersaturated.

Comparing the ion product to K_sp establishes if precipitation is expected:
  • If the Ion Product < K_sp, the solution is unsaturated, and no precipitation occurs.
  • If the Ion Product = K_sp, the solution is saturated, meaning it is at equilibrium and any more ions will start a precipitation reaction.
  • If the Ion Product > K_sp, the solution is supersaturated, leading to precipitation.
In our example, the calculation of ion product values for both AgI and Hg鈧侷鈧� enabled an understanding of which compound will precipitate first when iodide ions are added. By determining these points, we can efficiently perform selective precipitation in mixtures.
Precipitation Reaction
A Precipitation Reaction occurs when ions in aqueous solution form an insoluble compound upon mixing. It involves the transformation of a soluble compound into an insoluble solid (precipitate), driven by the interaction of various ionic species. The goal is often to separate specific ions from a mixture based on their precipitation behavior.

In the context of selective precipitation, this technique helps identify conditions where one metal ion precipitates while another remains soluble. For example, in our calculation, we determined the concentrations where silver iodide (AgI) precipitates before mercury(I) iodide (Hg鈧侷鈧�). Understanding this helps achieve effective separation based on the differential solubility of these metal ions.

In practice:
  • Add a precipitating agent to the solution.
  • Adjust conditions (e.g., concentration) to favor the precipitation of one specific ion before others.
Thus, precipitation reactions serve as a practical method to cleanly separate ions within complex mixtures by controlling their ionic interactions.

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Most popular questions from this chapter

An acidic buffer solution can be prepared by mixing the solutions of (a) ammonium acetate and acetic acid (b) ammonium chloride and ammonioum hydroxide (c) sulphuric acid and sodium sulphate (d) sodium chloride and sodium hydroxide.

The variation of equilibrium constant with temperature is given below: $$ \begin{array}{ll} {\text { Temperature }} & {\text { Equilibrium Constant }} \\ \mathrm{T}_{1}=25^{\circ} \mathrm{C} & \mathrm{K}_{1}=10 \\ \mathrm{~T}_{2}=100{ }^{\circ} \mathrm{C} & \mathrm{K}_{2}=100 \end{array} $$ The values of \(\Delta \mathrm{H}^{\circ}, \Delta \mathrm{G}^{\circ}\) at \(\mathrm{T}_{1}\) and \(\Delta \mathrm{G}^{\circ}\) at \(\mathrm{T}_{2}\) (in \(\mathrm{kj} \mathrm{mol}^{-1}\) ) respectively, are close to [use \(\left.\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]\) (a) \(28.4,-7.14\) and \(-5.71\) (b) \(0.64,-7.14\) and \(-5.71\) (c) \(28.4,-5.71\) and \(-14.29\) (d) \(0.64,-5.71\) and \(-14.29\)

A liquid is in equilibrium with its vapour at its boiling point. On the average, the molecules in the two phases have equal : (a) inter-molecular forces (b) potential energy (c) total energy (d) kinetic energy

The Haber's process for the formation of \(\mathrm{NH}_{3}\) at \(298 \mathrm{~K}\) is \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; \Delta H=-46.0 \mathrm{~kJ} ;\) Which of the following is the correct statement (a) The condition for equilibrium is $$ G_{\mathrm{N}_{2}}+3 G_{\mathrm{H}_{2}}=2 G_{\mathrm{NH}_{3}} $$ where \(G\) is Gibb's free energy per mole of the gaseous species measured at that partial pressure. (b) On adding \(\mathrm{N}_{2}\), the equilibrium will shift to forward direction because according to \(\mathrm{II}^{\text {nd }}\) law of thermodynamics, the entropy must increase in the direction of spontaneous reaction (c) The catalyst will increase the rate of forward reaction by 2 times and that of backward reaction by \(1.5\) times (d) None of these

The degree of dissociation of water at \(25^{\circ} \mathrm{C}\) is \(1.9 \times 10^{-7} \%\) and density is \(1.0 \mathrm{~g} \mathrm{~cm}^{-3} .\) The ionic constant for water is : (a) \(1.0 \times 10^{-10}\) (b) \(1.0 \times 10^{-14}\) (c) \(1.0 \times 10^{-16}\) (d) \(1.0 \times 10^{-8}\)
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