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Chapter 17: Problem 52

(a) True or false: "solubility" and "solubility-product constant" are the same number for a given compound. (b) Write the expression for the solubility- product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Short Answer

Expert verified
(a) False; solubility and solubility-product constant are different. (b) MnCO_3: K_{sp} = [Mn^{2+}][CO_{3}^{2-}] , Hg(OH)_2: K_{sp} = [Hg^{2+}][OH^{-}]^{2} , Cu_3(PO_4)_2: K_{sp} = [Cu^{2+}]^{3}[PO_{4}^{3-}]^{2} .

Step by step solution

01

Understanding Solubility vs. Solubility-Product Constant

Solubility refers to the maximum amount of solute that can dissolve in a solvent at a given temperature to form a saturated solution, usually expressed in moles per liter (mol/L). The solubility-product constant ( K_{sp} ) is an equilibrium constant for the dissolution of an ionic compound into its constituent ions, representing the saturated solution's equilibrium condition. Thus, true or false? The statement 'solubility' and 'solubility-product constant' are the same number is false because they represent different concepts and are expressed with different units.
02

Expression for the Solubility-Product Constant of MnCO鈧�

To express the solubility-product constant, K_{sp} , for MnCO_{3} dissolving, consider its dissociation in water: MnCO_{3 (s)} ightarrow Mn^{2+}_{(aq)} + CO_{3}^{2-}_{(aq)} . The K_{sp} expression is the product of the concentrations of the ions at equilibrium: K_{sp} = [Mn^{2+}][CO_{3}^{2-}] .
03

Expression for the Solubility-Product Constant of Hg(OH)鈧�

Starting with the dissociation reaction for Hg(OH)_{2} : Hg(OH)_{2 (s)} ightarrow Hg^{2+}_{(aq)} + 2OH^{-}_{(aq)} . The K_{sp} expression is thus the product of the concentrations of the ions, each raised to the power of their coefficients: K_{sp} = [Hg^{2+}][OH^{-}]^{2} .
04

Expression for the Solubility-Product Constant of Cu鈧�(PO鈧�)鈧�

Lastly, for Cu_{3}(PO_{4})_{2} dissolving in water, the dissociation is: Cu_{3}(PO_{4})_{2 (s)} ightarrow 3Cu^{2+}_{(aq)} + 2PO_{4}^{3-}_{(aq)} . The K_{sp} expression will be: K_{sp} = [Cu^{2+}]^{3}[PO_{4}^{3-}]^{2} , accounting for the stoichiometry of the reaction.

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Key Concepts

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

Ionic Compounds
Ionic compounds are substances made up of charged particles called ions. These ions form when atoms gain or lose electrons, creating positive ions (cations) and negative ions (anions). Ionic compounds typically form a structured lattice arrangement that helps in stabilizing the charges.
  • Cations are positively charged ions. Common examples are sodium ions (\( \text{Na}^+ \)) and magnesium ions (\( \text{Mg}^{2+} \)).
  • Anions are negatively charged ions. Chloride (\( \text{Cl}^- \)) and sulfate (\( \text{SO}_{4}^{2-} \)) are examples.
The strength and stability of ionic compounds are mostly due to the strong electrostatic forces between oppositely charged ions.
Ionic compounds readily dissolve in water, creating a solution of ions that is capable of conducting electricity. These dissolved ions are crucial in reaction processes, such as those involving equilibrium constants.
Equilibrium Constant
The equilibrium constant is a number that expresses the balance between products and reactants in a chemical reaction at a set condition of temperature and pressure. In the context of ionic compounds and solubility, this constant refers to the solubility-product constant, \( K_{sp} \).
It quantifies the point at which the dissolution and precipitation of a compound are in balance. At this point, the rate at which the compound dissolves equals the rate at which it forms a solid again.
  • A larger \( K_{sp} \) indicates a compound is more soluble.
  • A smaller \( K_{sp} \) means less solubility.
This value is significant when predicting or calculating how much of a compound will dissolve in a solution and at what concentration.
Dissolution Reaction
A dissolution reaction occurs when an ionic compound breaks apart into its individual ions in a solvent, usually water. During this process, the solid compound becomes part of the solution:
For example, when calcium chloride (\( \text{CaCl}_2 \)) dissolves in water, it dissociates into calcium ions (\( \text{Ca}^{2+} \)) and chloride ions (\( \text{Cl}^- \)).
These ions are then free to interact with other substances in the solution or remain as they are:
The dissolution reaction can be influenced by several factors:
  • Temperature: Usually, solubility increases with temperature.
  • The nature of solvent: Polar solvents, like water, are particularly effective at dissolving ionic compounds.
  • Pressure: Usually effects gas solubility, but solid solubility is mostly unaffected by pressure changes.
Understanding the nature of dissolution is key in fields ranging from chemistry to environmental science, as it explains how substances interact in a solution.

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Most popular questions from this chapter

Derive an equation similar to the Henderson-Hasselbalch equation relating the pOH of a buffer to the \(\mathrm{p} K_{b}\) of its base component.

Consider a beaker containing a saturated solution of \(\mathrm{Pbl}_{2}\) in equilibrium with undissolved \(\mathrm{Pbl}_{2}(s)\). Now solid \(\mathrm{KI}\) is added to this solution. (a) Will the amount of solid \(\mathrm{PbI}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Pb}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of I' ions in solution increase or decrease?

Mathematically prove that the \(\mathrm{pH}\) at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to \(\mathrm{p} K_{a}\) for the acid.

A weak monoprotic acid is titrated with \(0.100 \mathrm{M} \mathrm{NaOH}\). It requires \(25.0 \mathrm{~mL}\) of the \(\mathrm{NaOH}\) solution to reach the equivalence point. After \(12.5 \mathrm{~mL}\). of base is added, the pH of the solution is 4.16 . Estimate the \(\mathrm{p} K_{a}\) of the weak acid.

In nonaqueous solvents, it is possible to react HF to create \(\mathrm{H}_{2} \mathrm{~F}^{+} .\) Which of these statements follows from this observation? (a) HF can act like a strong acid in nonaqueous solvents, (b) HF can act like a base in nonaqueous solvents, (c) HF is thermodynamically unstable, \((\mathbf{d})\) There is an acid in the nonaqueous medium that is a stronger acid than HE.
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