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Ch.17 - Additional Aspects of Aqueous Equilibria
Brown - Chemistry: The Central Science 15th Edition
Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970Not the one you use?Change textbook
All textbooksBrown 15th EditionCh.17 - Additional Aspects of Aqueous EquilibriaProblem 98
Chapter 17, Problem 98

For each pair of compounds, use Ksp values to determine which has the greater molar solubility: (a) CdS or CuS (b) PbCO3 or BaCrO4 (c) Ni(OH)2 or NiCO3 (d) AgI or Ag2SO4.

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1
Identify the Ksp (solubility product constant) values for both PbCO3 and BaCrO4 from a reliable source such as a chemistry textbook or database.
Write the dissolution equations for each compound. For PbCO3, the equation is PbCO3(s) ⇌ Pb^2+(aq) + CO3^2-(aq). For BaCrO4, the equation is BaCrO4(s) ⇌ Ba^2+(aq) + CrO4^2-(aq).
Express the molar solubility (s) for each compound using the dissolution equations. For PbCO3, if s is the molar solubility, then [Pb^2+] = s and [CO3^2-] = s. For BaCrO4, [Ba^2+] = s and [CrO4^2-] = s.
Set up the Ksp expression for each compound using the concentrations derived from the molar solubility. For PbCO3, Ksp = [Pb^2+][CO3^2-] = s * s = s^2. For BaCrO4, Ksp = [Ba^2+][CrO4^2-] = s * s = s^2.
Compare the calculated s values (molar solubility) for each compound. The compound with the higher s value has the greater molar solubility.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is an equilibrium constant that quantifies the solubility of a sparingly soluble ionic compound. It is defined as the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced dissolution equation. A higher Ksp value indicates a greater solubility of the compound in water.
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Molar Solubility

Molar solubility refers to the maximum concentration of a solute that can dissolve in a given volume of solvent at equilibrium. It is typically expressed in moles per liter (M). To compare the molar solubility of two compounds, one can derive it from their Ksp values, allowing for a direct comparison of how much of each compound can dissolve in solution.
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Dissociation of Ionic Compounds

When ionic compounds dissolve in water, they dissociate into their constituent ions. The extent of this dissociation affects the Ksp and, consequently, the molar solubility. Understanding the stoichiometry of the dissociation reaction is crucial for calculating the Ksp and determining the molar solubility of each compound, as different compounds yield different numbers of ions upon dissolution.
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Related Practice
Textbook Question

The solubility-product constant for barium permanganate, Ba1MnO422, is 2.5 * 10-10. Assume that solid Ba1MnO422 is in equilibrium with a solution of KMnO4. What concentration of KMnO4 is required to establish a concentration of 2.0 * 10-8 M for the Ba2 + ion in solution?

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Textbook Question

The solubility of CaCO3 is pH dependent. (b) Use the Kb expression for the CO32- ion to determine the equilibrium constant for the reaction CaCO3(s) + H2O(l) ⇌ Ca2+(aq) + HCO3-(aq) + OH-(aq)

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Textbook Question

The solubility product constants of PbSO4 and SrSO4 are 6.3 * 10-7 and 3.2 * 10-7, respectively. What are the values of 3SO4 2 - 4, 3Pb2 + 4, and 3Sr2 + 4 in a solution at equilibrium with both substances?

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Textbook Question

Lead(II) carbonate, PbCO3, is one of the components of the passivating layer that forms inside lead pipes. (d) The EPA threshold for acceptable levels of lead ions in water is 15 ppb. Does a saturated solution of lead(II) carbonate produce a solution that exceeds the EPA limit?

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Textbook Question

Suppose you want to do a physiological experiment that calls for a pH 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid H2A 1Ka1 = 2 * 10-2; Ka2 = 5.0 * 10-72 or its sodium salts. You have available a 1.0 M solution of this acid and a 1.0 M solution of NaOH. How much of the NaOH solution should be added to 1.0 L of the acid to give a buffer at pH 6.50? (Ignore any volume change.)

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