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Chapter 7: Problem 100

The solubility product of \(\mathrm{A}_{2} \mathrm{X}_{3}\) is \(1.08 \times 10^{-23}\). Its solubility will be: (a) \(1.0 \times 10^{-3} \mathrm{M}\) (b) \(1.0 \times 10^{-4} \mathrm{M}\) (c) \(1.0 \times 10^{-5} \mathrm{M}\) (d) \(1.0 \times 10^{-6} \mathrm{M}\)

Short Answer

Expert verified
(b) \(1.0 \times 10^{-4} \mathrm{M}\).

Step by step solution

01

Understand the Dissolution Reaction

The compound \( \text{A}_2\text{X}_3 \) dissociates in water as follows: \( \text{A}_2\text{X}_3 (s) \rightleftharpoons 2\text{A}^{3+} (aq) + 3\text{X}^{2-} (aq) \). Thus, 2 moles of \( \text{A}^{3+} \) ions and 3 moles of \( \text{X}^{2-} \) ions are produced for every mole of \( \text{A}_2\text{X}_3 \) that dissolves.
02

Express Solubility in Terms of S

Let \( s \) be the solubility of \( \text{A}_2\text{X}_3 \) in mol/L. Therefore, the concentrations at equilibrium will be \( [\text{A}^{3+}] = 2s \) and \( [\text{X}^{2-}] = 3s \).
03

Write the Solubility Product Expression

The solubility product \( K_{sp} \) is given by the expression \( K_{sp} = [\text{A}^{3+}]^2 [\text{X}^{2-}]^3 \). Substituting the concentrations, we have \( K_{sp} = (2s)^2 (3s)^3 \).
04

Simplify and Solve for S

Substitute and simplify: \( K_{sp} = 4s^2 \times 27s^3 = 108s^5 \). Thus, \( s^5 = \frac{K_{sp}}{108} \).
05

Plug in the Known Values

Substitute the known \( K_{sp} \) value: \( s^5 = \frac{1.08 \times 10^{-23}}{108} \).
06

Calculate the Solubility S

Solve for \( s \): \( s = \left(\frac{1.08 \times 10^{-23}}{108}\right)^{1/5} \). Simplifying gives \( s = 1.0 \times 10^{-4} \mathrm{M} \).

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Key Concepts

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

Dissolution Reaction
When a compound like \( \text{A}_2\text{X}_3 \) dissolves in water, it undergoes a **dissolution reaction**. This is a process where the solid form of the compound breaks down into its constituent ions in a solution. For \( \text{A}_2\text{X}_3 \), it dissociates into 2 moles of \( \text{A}^{3+} \) ions and 3 moles of \( \text{X}^{2-} \) ions. This dissociation can be represented by the chemical equation:
  • \( \text{A}_2\text{X}_3 (s) \rightleftharpoons 2\text{A}^{3+} (aq) + 3\text{X}^{2-} (aq) \)
The double arrows indicate that this reaction is reversible, meaning the ions can recombine to form the solid or remain in solution. Understanding this concept is crucial because it forms the basis for calculating solubility and equilibrium concentrations.
Equilibrium Concentrations
Equilibrium concentrations are pivotal in understanding how much of each ion is present when a solution reaches a state of balance. During the dissolution of \( \text{A}_2\text{X}_3 \), we let \( s \) represent the solubility, or how much of the solid dissolves. This gives equilibrium concentrations of:
  • \( [\text{A}^{3+}] = 2s \)
  • \( [\text{X}^{2-}] = 3s \)
At equilibrium, the concentration of ions does not change unless disturbed. This is an important point because these concentrations will be used to find the solubility product \( K_{sp} \), a constant specific to the dissolution process at a given temperature.
Solubility Calculations
Solubility calculations involve finding the maximum concentration of a solute that can dissolve in a solvent at equilibrium. By knowing the solubility product \( K_{sp} \), you can calculate solubility \( s \) in mol/L. For \( \text{A}_2\text{X}_3 \), the \( K_{sp} \) expression is:
  • \( K_{sp} = [\text{A}^{3+}]^2 [\text{X}^{2-}]^3 \)
  • Substituting give us \( K_{sp} = (2s)^2 \times (3s)^3 \)
  • Simplifies to \( K_{sp} = 108s^5 \)
Finally, by substituting the known \( K_{sp} \) value into the equation \( s^5 = \frac{K_{sp}}{108} \), you solve for \( s \) to find the solubility.
Chemical Equilibrium
**Chemical equilibrium** refers to the state where the rates of the forward and reverse reactions are equal, resulting in no net change in concentrations of reactants and products. For \( \text{A}_2\text{X}_3 \), at equilibrium:
  • There is continuous movement of ions between the solid state and the dissolved state
  • The concentrations of \( \text{A}^{3+} \) and \( \text{X}^{2-} \) remain constant
This equilibrium ensures that the solution has reached its maximum solubility under the given conditions. Understanding chemical equilibrium allows you to predict how changes, such as temperature or pressure, might affect solubility and ion concentration.

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Most popular questions from this chapter

Which of the following change will shift the reaction in forward direction: \(\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 21(\mathrm{~g})\) Take \(\Delta \mathrm{H}^{\circ}=+150 \mathrm{~kJ}\) (a) Increase in concentration of I (b) Increase in total pressure (c) Decrease in concentration of \(\mathrm{I}_{2}\) (d) Increase in temperature

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at 300 is kept in a closed container under one atmosphere. It is heated to 600 when \(20 \%\) by mass of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) decomposes to \(\mathrm{NO}_{2}\) (g). The resultant pressure is: (a) \(1.2 \mathrm{~atm}\) (b) \(2.4 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.0 \mathrm{~atm}\)

In which of the following reactions, the concentration of product is higher than the concentration of reactant at equilibrium? = equilibrium constant): (a) \(\mathrm{A} \rightleftharpoons \mathrm{B} ; \mathrm{K}=0.001\) (b) \(\mathrm{M} \rightleftharpoons \mathrm{N} ; \mathrm{K}=10\) (c) \(\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}=0.005\) (d) \(\mathrm{R} \rightleftharpoons \mathrm{P} ; \mathrm{K}=0.01\)

In the reaction \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{S}_{2}(\mathrm{~g})\) the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(0.5 \mathrm{~mol} \mathrm{~L}^{-1^{2}}\) and concentration of \(\mathrm{H}_{2}\) is \(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\) while concentration of \(\mathrm{S}_{2}\) is \(0.4 \mathrm{~mol} \mathrm{~L}^{-1}\) in one litre vessel. The value of equilibrium constant of the reaction is: (a) \(0.016\) (b) \(0.013\) (c) \(0.020\) (d) \(0.030\)

The equilibrium constant for the reaction, \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\) At temperature \(\mathrm{T}\) is \(4 \times 10^{-4}\). The value of \(\mathrm{K}_{\mathrm{c}}\) for the reaction \(\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) at the same temperature is: (a) \(4 \times 10^{-6}\) (b) \(2.5 \times 10^{2}\) (c) \(0.02\) (d) 50
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