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Chapter 17: Problem 49

For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility. (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion.

Short Answer

Expert verified
(a) True - The solubility of a slightly soluble salt can be expressed in units of moles per liter (M or mol/L). (b) False - The solubility product of a slightly soluble salt is not always simply the square of the solubility, as it depends on the stoichiometries of the constituent ions. (c) False - The solubility of a slightly soluble salt can be affected by the presence of a common ion, as explained by the common ion effect. (d) True - The solubility product of a slightly soluble salt is independent of the presence of a common ion, as it is a constant value that only depends on the temperature and the nature of the salt.

Step by step solution

01

Statement (a): The solubility of a slightly soluble salt can be expressed in units of moles per liter.

This statement is true. The solubility of a salt is a measure of the maximum amount of the salt that can dissolve in a given amount of solvent at a particular temperature. It is typically expressed in units of moles per liter (M or mol/L).
02

Statement (b): The solubility product of a slightly soluble salt is simply the square of the solubility.

This statement is false in general. The solubility product (K_sp) of a slightly soluble salt is the product of the concentrations of its constituent ions, each raised to the power of their stoichiometric coefficients in the balanced dissolution reaction. For a simple salt like \(AB\), which dissolves according to the reaction: \(AB \rightleftharpoons A^+ + B^-\), the solubility product is indeed the square of the solubility, because the stoichiometric coefficients are both 1 and the concentrations of the ions are equal: \(K_{sp} = [A^+][B^-]\). However, for salts with different stoichiometries, such as \(AB_2\), the solubility product is not simply the square of the solubility: \(K_{sp} = [A^+][B^-]^2\).
03

Statement (c): The solubility of a slightly soluble salt is independent of the presence of a common ion.

This statement is false. The solubility of a slightly soluble salt can be affected by the presence of a common ion, as explained by the common ion effect. The common ion effect is a consequence of Le Chatelier's principle: when a reaction is at equilibrium, adding a product will shift the equilibrium towards the reactants. Thus, if a common ion is introduced into the solution, the solubility of the slightly soluble salt can decrease, because the reaction will shift towards the undissolved salt to counteract the increase in product concentration.
04

Statement (d): The solubility product of a slightly soluble salt is independent of the presence of a common ion.

This statement is true. The solubility product (K_sp) of a slightly soluble salt is a constant value that only depends on the temperature and the nature of the salt. It is not affected by the presence of a common ion, although the solubility of the salt may be affected, as explained in Statement (c). The solubility product provides information about the relative solubilities of different salts and can be used to predict whether a precipitation reaction will occur or not, given the concentrations of the ions in the solution.

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Most popular questions from this chapter

A 10.0-mL sample of \(0.250 \mathrm{MHNO}_{3}\) solution is titrated with $0.100 M$ KOH solution. Calculate the pH of the solution after the following volumes of base have been added: (a) \(20.0 \mathrm{~mL}\), (b) \(24.9 \mathrm{~mL}\) (c) \(25.0 \mathrm{~mL}\) (d) \(25.1 \mathrm{~mL}\), (e) \(30.0 \mathrm{~mL}\).

The solubility of two slightly soluble salts of \(\mathrm{M}^{2+}, \mathrm{MA}\) and \(\mathrm{MZ}_{2}\), is the same, $4 \times 10^{-4} \mathrm{~mol} / \mathrm{L} .(\mathbf{a})$ Which has the larger numerical value for the solubility product constant? (b) In a saturated solution of each salt in water, which has the higher concentration of \(\mathrm{M}^{2+} ?(\mathbf{c})\) If you added an equal volume of a solution saturated in MA to one saturated in \(\mathrm{MZ}_{2}\), what would be the equilibrium concentration of the cation, \(\mathrm{M}^{2+}\) ?

The solubility of \(\mathrm{CaCO}_{3}\) is pH dependent. (a) Calculate the molar solubility of \(\mathrm{CaCO}_{3}\left(K_{s p}=4.5 \times 10^{-9}\right)\) neglecting the acid-base character of the carbonate ion. (b) Use the \(K_{b}\) expression for the \(\mathrm{CO}_{3}^{2-}\) ion to determine the equilibrium constant for the reaction $$ \mathrm{CaCO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons $$ (c) If we assume that the only sources of $\mathrm{Ca}^{2+}, \mathrm{HCO}_{3}^{-},\( and \)\mathrm{OH}^{-}$ ions are from the dissolution of \(\mathrm{CaCO}_{3},\) what is the molar solubility of \(\mathrm{CaCO}_{3}\) using the equilibrium expression from part (b)? (d) What is the molar solubility of \(\mathrm{CaCO}_{3}\) at the \(\mathrm{pH}\) of the ocean (8.3)\(?(\mathbf{e})\) If the \(\mathrm{pH}\) is buffered at \(7.5,\) what is the molar solubility of \(\mathrm{CaCO}_{3} ?\)

Which of the following solutions is a buffer? (a) A solution made by mixing \(50 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) formic acid \((\mathrm{HCOOH})\) and \(250 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{KOH},(\mathbf{b})\) A solution made by mixing \(50 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) formic acid \((\mathrm{HCOOH})\) and \(25 \mathrm{~mL}\) of \(0.200 M\) nitric acid \(\left(\mathrm{HNO}_{3}\right),(\mathbf{c})\) A solution made by mixing $50 \mathrm{~mL}\( of \)0.200 \mathrm{M}\( potassium formate \)(\mathrm{HCOOK})$ and \(25 \mathrm{~mL}\) of \(0.200 \mathrm{MKNO}_{3},(\mathbf{d})\) A solution made by mixing \(50 \mathrm{~mL}\) of \(0.200 M\) formic acid \((\mathrm{HCOOH})\), and $25 \mathrm{~mL}\( of \)0.200 \mathrm{MKOH}$.

Using the value of \(K_{s p}\) for \(\mathrm{Ag}_{2} \mathrm{~S}, K_{a 1}\) and \(\mathrm{K}_{a 2}\) for \(\mathrm{H}_{2} \mathrm{~S},\) and $K_{f}=1.1 \times 10^{5}\( for \)\mathrm{AgCl}_{2}^{-},$ calculate the equilibrium constant for the following reaction: $$ \mathrm{Ag}_{2} \mathrm{~S}(s)+4 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}^{+}(a q) \rightleftharpoons 2 \mathrm{AgCl}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{~S}(a q) $$
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