Problem 1
What is the difference between an ion product and an ion product constant?
Problem 2
Use the following equilibrium to demonstrate why the \(K_{\mathrm{sp}}\) expression does not include the concentration of \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) in the denominator: $$ \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3 \mathrm{Ba}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) $$
Problem 3
What is the common ion effect? How does Le Châtelier's principle explain it? Use the solubility equilibrium for \(\mathrm{AgCl}\) and the addition of \(\mathrm{NaCl}\) to the solution to illustrate the common ion effect.
Problem 4
With respect to \(K_{\mathrm{sp}}\), what conditions must be met if a precipitate is going to form in a solution?
Problem 5
What limits the accuracy and reliability of solubility calculations based on \(K_{\mathrm{sp}}\) values?
Problem 6
Why do we not use \(K_{\mathrm{sp}}\) values for soluble salts such as \(\mathrm{NaCl}\) ?
Problem 9
Is \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) a basic salt? Justify your answer. $$ \mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3 \mathrm{Ba}^{2+}(a q)+2 \mathrm{PO}_{4}^{3-}(a q) $$
Problem 13
On the basis of Le Châtelier's principle, explain how the addition of solid \(\mathrm{NH}_{4} \mathrm{Cl}\) to a beaker containing solid \(\mathrm{Mg}(\mathrm{OH})_{2}\) in contact with water is able to cause the \(\mathrm{Mg}(\mathrm{OH})_{2}\) to dissolve. Write equations for all of the chemical equilibria that exist in the solution after the addition of the \(\mathrm{NH}_{4} \mathrm{Cl}\).
Problem 14
Which of the following will be more soluble if acid is added to the mixture? Will adding base increase the solubility of any of these? Are any of their solubilities independent of the \(\mathrm{pH}\) of the solution? (a) \(\mathrm{ZnS},\) (b) \(\mathrm{Ca}(\mathrm{OH})_{2}\) (c) \(\mathrm{MgCO}_{3}\) (d) \(\mathrm{AgCl}\), (e) \(\mathrm{PbF}_{2}\)
Problem 15
Potassium oxide is readily soluble in water, but the resulting solution contains essentially no oxide ion. Explain, using an equation, what happens to the oxide ion.
