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Calculate the aqueous solubility, in moles per liter, of each of the following. (a) \(\mathrm{BaCrO}_{4}, K_{\mathrm{sp}}=1.2 \times 10^{-10}\) (b) \(\mathrm{PbBr}_{2}, K_{\mathrm{sp}}=4.0 \times 10^{-5}\) (c) \(\mathrm{CeF}_{3}, K_{\mathrm{sp}}=8 \times 10^{-16}\) (d) \(\operatorname{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}, K_{\mathrm{sp}}=2.1 \times 10^{-20}\)

Short Answer

Expert verified
Repeat these steps for each compound. The calculations yield the solubility as follows: (a) \(1.1 \times 10^{-5}\) M, (b) \(2.0 \times 10^{-2}\) M, (c) \(2.0 \times 10^{-5}\) M, (d) \(2.5 \times 10^{-7}\) M.

Step by step solution

01

Write the dissolution equation

Each compound will dissociate into ions in solution. For the first compound, \(\mathrm{BaCrO}_{4}\):\[\mathrm{BaCrO}_{4}(s) \leftrightarrow \mathrm{Ba}^{2+}(aq) + \mathrm{CrO}_4^{2-}(aq)\]
02

Establish the solubility product expression

The solubility product is determined by the multiplication of the concentrations of the ions derived from the compound, each raised to the power of its stoichiometric coefficient. For \(\mathrm{BaCrO}_{4}\), the expression is:\[[K_{sp}] = [\mathrm{Ba}^{2+}][\mathrm{CrO}_4^{2-}]\]
03

Determine the numerical value of the expression

Let \(x\) be the solubility of \(\mathrm{BaCrO}_{4}\) in moles per liter. Because of the 1:1 stoichiometric ratio, the concentrations of both ions in solution are equal to \(x\). Thus:\[[K_{sp}] = (x)(x) = \(x^2\) = 1.2 \times 10^{-10}\]
04

Solve for solubility

From the above equation, we can solve for \(x\), which is the solubility of \(\mathrm{BaCrO}_{4}\) in moles per liter, by taking the square root of both sides:\[x = \(\sqrt{1.2 \times 10^{-10}}\)\]

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Key Concepts

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

Aqueous Solubility
Aqueous solubility refers to the maximum amount of a compound that can dissolve in water, forming a homogenous solution. In chemistry, it is often expressed in terms of moles per liter (mol/L). Determining the aqueous solubility of compounds is essential, particularly for compounds that form precipitates or in situations where solubility plays a critical role in determining the course of reactions.

When dealing with sparingly soluble salts, their solubility in water can be predicted using the solubility product constant, or Ksp. This constant provides insight into how much of the compound can dissolve in solution until an equilibrium between the solid and its ions is reached.

By setting up and solving the solubility product expression for each compound, one can predict their solubility values accurately. This involves using basic algebra and knowledge of dissolution reactions to solve for the unknown concentration of ions in the saturated solution.
Dissolution Equation
The dissolution equation depicts how a solid ionic compound disintegrates into its respective ions when dissolved in water. This is an essential step in understanding solubility because it visualizes how the compounds disassociate in solution.

For a compound like \(BaCrO_4\), the dissolution equation is written as: \[\mathrm{BaCrO}_{4}(s) \leftrightarrow \mathrm{Ba}^{2+}(aq) + \mathrm{CrO}_4^{2-}(aq)\] This equation demonstrates the balanced chemical reaction where one formula unit of barium chromate separates into one barium ion and one chromate ion in water.

Understanding the dissolution equation allows us to set up the solubility product expression, a critical step in quantifying the ionic concentrations and solving for solubility. Each ion’s coefficient in the dissolution equation directly affects the setup of the Ksp expression.
Ionic Concentrations
Ionic concentrations refer to the quantities of ions present in a solution. When an ionic compound like \(\mathrm{BaCrO}_{4}\) dissolves, it releases ions into the solution. The concentration of each ion is dependent primarily on the original solubility of the compound and its dissociation characteristics.

From the dissolution equation, we can predict that each mole of \(\mathrm{BaCrO}_{4}\) will produce one mole each of \(\mathrm{Ba}^{2+}\) and \(\mathrm{CrO}_4^{2-}\). Therefore, if the solubility of \(\mathrm{BaCrO}_{4}\) is \(x\) mol/L, then both ions will also have concentrations \(x\) mol/L in a saturated solution.

The equilibrium between the dissolved ions and the solid compound can be described mathematically using the solubility product expression, \([\mathrm{Ba}^{2+}] [\mathrm{CrO}_4^{2-}] = K_{sp}\). By solving this expression, we can find the direct quantification of these ionic concentrations.
Stoichiometric Coefficient
The stoichiometric coefficient in a dissolution equation is the number before molecules or ions, indicating how many moles of each are involved in the process. It is crucial for correctly establishing the solubility product expression.

For example, in the dissolution of \(\mathrm{BaCrO}_{4}\), the dissolution equation reveals a 1:1 molar ratio between \(\mathrm{Ba}^{2+}\) and \(\mathrm{CrO}_4^{2-}\). This 1:1 relationship is reflected directly in the Ksp expression where the concentrations of each ion raised to the power of their stoichiometric coefficients (which in this case is 1 for both ions).

If a stoichiometric coefficient differs, as seen in more complex compounds, it would multiply in polynomial fashion in the expression: for instance, if an ion had a coefficient of 2, of that ion's concentration would be squared in the Ksp calculation. Understanding stoichiometric coefficients is essential for accurate chemical calculations and comprehending the dynamics between reacting species in a solution.

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Most popular questions from this chapter

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.

Without performing detailed calculations, indicate whether either of the following compounds is appreciably soluble in \(\mathrm{NH}_{3}(\mathrm{aq}):(\mathrm{a}) \mathrm{CuS}, K_{\mathrm{sp}}=6.3 \times 10^{-36},\)(b) \(\mathrm{CuCO}_{3}, K_{\mathrm{sp}}=1.4 \times 10^{-10} .\) Also use the fact that \(K_{\mathrm{f}}\) for \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\) is \(1.1 \times 10^{13}\).

Will precipitation occur in the following cases? (a) \(0.10 \mathrm{mg}\) NaCl is added to \(1.0 \mathrm{L}\) of \(0.10 \mathrm{M}\) \(\mathrm{AgNO}_{3}(\mathrm{aq})\). (b) One drop \((0.05 \mathrm{mL})\) of \(0.10 \mathrm{M} \mathrm{KBr}\) is added to 250 mL of a saturated solution of AgCl. (c) One drop \((0.05 \mathrm{mL})\) of \(0.0150 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})\) is added to \(3.0 \mathrm{L}\) of a solution with \(2.0 \mathrm{mg} \mathrm{Mg}^{2+}\) per liter.

In the Mohr titration, \(\mathrm{Cl}^{-}(\mathrm{aq})\) is titrated with \(\mathrm{AgNO}_{3}(\text { aq })\) in solutions that are at about \(\mathrm{pH}=7\). Thus, it is suitable for determining the chloride ion content of drinking water. The indicator used in the titration is \(\mathrm{K}_{2} \mathrm{CrO}_{4}(\text { aq }) .\) A red-brown precipitate of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s})\) forms after all the \(\mathrm{Cl}^{-}\) has precipitated. The titration reaction is \(\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow \mathrm{AgCl}(\mathrm{s}) .\) At the equivalence point of the titration, the titration mixture consists of \(\mathrm{AgCl}(\mathrm{s})\) and a solution having neither \(\mathrm{Ag}^{+}\) nor \(\mathrm{Cl}^{-}\) in excess. Also, no \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s})\) is present, but it forms immediately after the equivalence point. (a) How many milliliters of \(0.01000 \mathrm{M} \mathrm{AgNO}_{3}(\mathrm{aq})\) are required to titrate \(100.0 \mathrm{mL}\) of a municipal water sample having \(29.5 \mathrm{mg} \mathrm{Cl}^{-} / \mathrm{L} ?\) (b) What is \(\left[\mathrm{Ag}^{+}\right]\) at the equivalence point of the Mohr titration? (c) What is \(\left[\mathrm{CrO}_{4}^{2-}\right]\) in the titration mixture to meet the requirement of no precipitation of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{s})\) until immediately after the equivalence point? (d) Describe the effect on the results of the titration if \(\left[\mathrm{CrO}_{4}^{2-}\right]\) were (1) greater than that calculated in part (c) or (2) less than that calculated? (e) Do you think the Mohr titration would work if the reactants were exchanged - that is, with \(\mathrm{Cl}^{-}(\text {aq })\) as the titrant and \(\mathrm{Ag}^{+}(\) aq) in the sample being analyzed? Explain.

Which of the following has the highest molar solubility? (a) \(\mathrm{MgF}_{2}, K_{\mathrm{sp}}=3.7 \times 10^{-8}\) \(\mathrm{MgCO}_{3}\), \(K_{\mathrm{sp}}=3.5 \times 10^{-8} ;(\mathrm{c}) \mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{\mathrm{sp}}=1 \times 10^{-25}\); (d) \(\mathrm{Li}_{3} \mathrm{PO}_{4}, K_{\mathrm{sp}}=3.2 \times 10^{-9}\).

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