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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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Most popular questions from this chapter

Determine the molar solubility of lead(II) azide, \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2},\) in a buffer solution with \(\mathrm{pH}=3.00,\) given that \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}(\mathrm{s}) \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{N}_{3}^{-}(\mathrm{aq})\) \(K_{\mathrm{sp}}=2.5 \times 10^{-9}\) \(\mathrm{HN}_{3}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{N}_{3}^{-}(\mathrm{aq})\) \(K_{\mathrm{a}}=1.9 \times 10^{-5}\)

Adding \(1.85 \mathrm{g} \mathrm{Na}_{2} \mathrm{SO}_{4}\) to \(500.0 \mathrm{mL}\) of saturated aqueous \(\mathrm{BaSO}_{4}:\) (a) reduces \(\left[\mathrm{Ba}^{2+}\right] ;\) (b) reduces \(\left[\mathrm{SO}_{4}^{2-}\right]\); (c) increases the solubility of \(\mathrm{BaSO}_{4} ;\) (d) has no effect.

Which of the following saturated aqueous solutions would have the highest \(\left[\mathrm{Mg}^{2+}\right]\): (a) \(\mathrm{MgCO}_{3} ;\) (b) \(\mathrm{MgF}_{2};\) (c) \(\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2} ?\) Explain.

How would you expect the presence of each of the following solutes to affect the molar solubility of \(\mathrm{CaCO}_{3}\) in water: (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3} ;\) (b) \(\mathrm{HCl} ;\) (c) \(\mathrm{NaHSO}_{4}\) ? Explain.

Will AgI(s) precipitate from a solution with \(\left[\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}\right]=0.012 \mathrm{M}, \left[\mathrm{CN}^{-}\right]=1.05 \mathrm{M}, \) and \(\left[\mathrm{I}^{-}\right]=2.0 \mathrm{M} ?\) For \( \mathrm{AgI}, K_{\mathrm{sp}}=8.5 \times 10^{-17} ; =\) for \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}, K_{\mathrm{f}}=5.6 \times 10^{18}\).
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