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Chapter 18: Problem 24

A handbook lists the \(K_{\mathrm{sp}}\) values \(1.1 \times 10^{-10}\) for \(\mathrm{BaSO}_{4}\) and \(5.1 \times 10^{-9}\) for \(\mathrm{BaCO}_{3} .\) When saturated \(\mathrm{BaSO}_{4}(\mathrm{aq})\) is also made with \(0.50 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq}),\) a precipitate of \(\mathrm{BaCO}_{3}(\mathrm{s})\) forms. How do you account for this fact, given that \(\mathrm{BaCO}_{3}\) has a larger \(K_{\mathrm{sp}}\) than does \(\mathrm{BaSO}_{4} ?\)

Short Answer

Expert verified
Although \(\mathrm{BaCO_3}\) has a larger \(K_{sp}\) than \(\mathrm{BaSO_4}\), it precipitates in a solution of \(\mathrm{BaSO_4}\) and \(\mathrm{Na_2CO_3}\) due to the common ion effect where it reacts with the high concentration of \(CO_{3}^{2-}\) ions in the solution.

Step by step solution

01

Understanding Solubility Product Constant

Solubility Product Constant, \(K_{sp}\), is the product of the concentrations of the ions of a compound with each concentration raised to a power equal to the coefficient of the ion in the dissociation equation. It is used to describe saturated solutions of ionic compounds of relatively low solubility and it allows us to compare solubilities of different salts. A larger \(K_sp\) value means the compound is more soluble, whereas a smaller \(K_{sp}\) implies less solubility.
02

Understand the reaction

The barium salt \(\mathrm{BaSO_4}\) is relatively insoluble and has a small \(K_{sp}\) value. When it's in a solution containing the carbonate ion (\(CO_{3}^{2-}\)), which is provided by the \(\mathrm{Na_2CO_3}\), this ion can react with any free barium ions to form \(\mathrm{BaCO_3}\).
03

Explaining the Paradox

Even though \(\mathrm{BaCO_3}\) has a larger \(K_{sp}\) than \(\mathrm{BaSO_4}\) and therefore is more soluble, it will still precipitate in the presence of a high concentration of \(\mathrm{CO_3^{2-}}\) ions due to a high driving force of the reaction. Any free \(\mathrm{Ba^{2+}}\) ions in solution would prefer to form a compound with the \(\mathrm{CO_3^{2-}}\) ion rather than with the \(\mathrm{SO_4^{2-}}\) ion due to the common ion effect.
04

Common ion effect explanation

The common ion effect is an effect that suppresses the ionization of an ion in the presence of another ion in a common ion compound. In this case, it means that the presence of the carbonate ion (\(CO_{3}^{2-}\)) inhibits the dissolving of the \(\mathrm{BaCO_3}\), making it precipitate out, even though it has a larger \(K_{sp}\). Hence, the actual solubility of a salt is not just a function of its own \(K_{sp}\), but also depends on the other species present in the solution.

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Key Concepts

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

Common Ion Effect
When a solution contains a common ion, the ionization or dissolution of a salt is suppressed due to the increase in the concentration of that particular ion. For example, when
  • \(\text{BaSO}_4\) is placed in a solution with \(\text{Na}_2\text{CO}_3\),\(\text{CO}_3^{2-}\) ions are introduced into the solution.
  • This presence of additional \(\text{CO}_3^{2-}\) ions is a common ion effect that suppresses the dissolution of \(\text{BaCO}_3\).
Instead of dissolving, more \(\text{BaCO}_3\) precipitates in order to maintain the solubility equilibrium.
The common ion effect is crucial because it reveals that solubility is not just influenced by a compound's inherent properties but also by the ions present in the solution.
Saturated Solutions
A saturated solution is one where the maximum amount of solute has dissolved in a solvent at a given temperature. In such a state,
  • no additional solute can dissolve.
  • The solution is at equilibrium, where the rate of dissolution of the solute is equal to the rate of precipitation.
In the context of the original problem, imagine a saturated solution of \(\text{BaSO}_4\). Adding \(\text{Na}_2\text{CO}_3\) disrupts this equilibrium by introducing more ions into the solution.
Due to the common ion effect from \(\text{CO}_3^{2-}\) ions, \(\text{BaCO}_3\) precipitates out until the solution reaches a new equilibrium. Understanding saturation helps in determining how and why salts precipitate out in solutions.
Precipitation Reaction
A precipitation reaction is a process where soluble ions in different solutions combine to form an insoluble compound that precipitates out of the solution. This happens when
  • the ionic product of the reaction exceeds the solubility product constant \(K_{sp}\).
  • In the original problem, when \(\text{Ba}^{2+}\) ions in the solution mix with \(\text{CO}_3^{2-}\) ions from added \(\text{Na}_2\text{CO}_3\),
they form \(\text{BaCO}_3\) precipitate. This phenomenon occurs because the concentration of ions exceeds the product of its solubility limits.
It's important to understand precipitation reactions as they are foundational in predicting and explaining the formation of solids from solutions.

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Most popular questions from this chapter

Write \(K_{\text {sp }}\) expressions for the following equilibria. For example, for the reaction \(\mathrm{AgCl}(\mathrm{s}) \rightleftharpoons \mathrm{Ag}^{+}(\mathrm{aq})+\) \(\mathrm{Cl}^{-}(\mathrm{aq}), K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]\). (a) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (b) \(\operatorname{Ra}\left(\mathrm{IO}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Ra}^{2+}(\mathrm{aq})+2 \mathrm{IO}_{3}^{-}(\mathrm{aq})\) (c) \(\mathrm{Ni}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s}) \rightleftharpoons 3 \mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{PO}_{4}^{3-}(\mathrm{aq})\) (d) \(\mathrm{PuO}_{2} \mathrm{CO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{PuO}_{2}^{2+}(\mathrm{aq})+\mathrm{CO}_{3}^{2-}(\mathrm{aq})\)

The following \(K_{\mathrm{sp}}\) values are found in a handbook. Write the solubility product expression to which each one applies. For example, \(K_{\mathrm{sp}}(\mathrm{AgCl})=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=\) \(1.8 \times 10^{-10}\). (a) \(K_{\mathrm{sp}}\left(\mathrm{Cr} \mathrm{F}_{3}\right)=6.6 \times 10^{-11}\) (b) \(K_{\mathrm{sp}}\left[\mathrm{Au}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]=1 \times 10^{-10}\) (c) \(K_{\mathrm{sp}}\left[\mathrm{Cd}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=2.1 \times 10^{-33}\) (d) \(K_{\mathrm{sp}}\left(\mathrm{Sr} \mathrm{F}_{2}\right)=2.5 \times 10^{-9}\)

A solution is saturated with magnesium palmitate \(\left[\mathrm{Mg}\left(\mathrm{C}_{16} \mathrm{H}_{31} \mathrm{O}_{2}\right)_{2}, \text { a component of bathtub ring }\right] \mathrm{at}\) \(50^{\circ} \mathrm{C} .\) How many milligrams of magnesium palmitate will precipitate from \(965 \mathrm{mL}\) of this solution when it is cooled to \(25^{\circ} \mathrm{C} ?\) For \(\mathrm{Mg}\left(\mathrm{C}_{16} \mathrm{H}_{31} \mathrm{O}_{2}\right)_{2},\) \(K_{\mathrm{sp}}=4.8 \times 10^{-12}\) at \(50^{\circ} \mathrm{C}\) and \(3.3 \times 10^{-12}\) at \(25^{\circ} \mathrm{C}\).

Reaction (18.10), described in the Integrative Example, is called a carbonate transposition. In such a reaction, anions of a slightly soluble compound (for example, hydroxides and sulfates) are obtained in a sufficient concentration in aqueous solution that they can be identified by qualitative analysis tests. Suppose that \(3 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) is used and that an anion concentration of \(0.050 \mathrm{M}\) is sufficient for its detection. Predict whether carbonate transposition will be effective for detecting (a) \(\mathrm{SO}_{4}^{2-}\) from \(\mathrm{BaSO}_{4}(\mathrm{s}) ;\) (b) \(\mathrm{Cl}^{-}\) from \(\mathrm{AgCl}(\mathrm{s}) ;(\mathrm{c}) \mathrm{F}^{-}\) from \(\mathrm{MgF}_{2}(\mathrm{s})\).

Can the following ion concentrations be maintained in the same solution without a precipitate forming: \(\left[\left[\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}\right]^{3-}\right]=0.048 \mathrm{M},\left[\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\right]=0.76 \mathrm{M},\) and \(\left[\mathrm{I}^{-}\right]=2.0 \mathrm{M} ?\)
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