Chapter 9: Problem 10
Calculate the solubility-product constant for each of the following substances, given that the molar concentrations of their saturated solutions are as indicated: (a) \(\mathrm{AgSeCN}\left(2.0 \times 10^{-8} \mathrm{M}\right.\); products are \(\mathrm{Ag}^{+}\)and \(\left.\mathrm{SeCN}^{-}\right)\). *(b) \(\mathrm{RaSO}_{4}\left(6.6 \times 10^{-6} \mathrm{M}\right)\). (c) \(\mathrm{Pb}\left(\mathrm{BrO}_{3}\right)_{2}\left(1.7 \times 10^{-1} \mathrm{M}\right)\). *(d) \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}\left(1.9 \times 10^{-3} \mathrm{M}\right)\).
Short Answer
Step by step solution
Understanding Solubility-Product Constant
Calculating Ksp for AgSeCN
Calculating Ksp for RaSO4
Calculating Ksp for Pb(BrO3)2
Calculating Ksp for Ce(IO3)3
Summary of Calculations
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chemical Equilibrium
Saturated Solutions
- Any slight increase in the solute concentration beyond the saturation point will lead the excess solute to precipitate out of the solution.
- A saturated solution of a substance like RaSO4 will have the exact concentration of dissolved ions required to start forming a solid.
Molar Solubility
- The higher the molar solubility, the more a substance can dissolve in a given amount of solvent.
- Using the molar solubility of Ce(IO3)3, you can determine the concentration of ions at equilibrium—an essential step in calculating the Ksp.
Dissolution Reactions
- When a compound like lead bromate (\[\text{Pb(BrO}_3)_2\]) dissolves, it separates into lead ions (\[\text{Pb}^{2+}\]) and bromate ions.
- The stoichiometry of dissolution reactions is crucial because it dictates the proportions of ions formed: for each mole of Pb(BrO3)2, one mole of Pb2+ and two moles of BrO3- are produced.