Coupled Equilibria

Learning Outcomes

Describe examples of systems involving two (or more) coupled chemical equilibria
Calculate reactant and product concentrations for coupled equilibrium systems

As discussed in preceding chapters on equilibrium, coupled equilibria involve two or more separate chemical reactions that share one or more reactants or products. This section of this chapter will address solubility equilibria coupled with acid-base and complex-formation reactions.

An environmentally relevant example illustrating the coupling of solubility and acid-base equilibria is the impact of ocean acidification on the health of the ocean’s coral reefs. These reefs are built upon skeletons of sparingly soluble calcium carbonate excreted by colonies of corals (small marine invertebrates).

Carbon dioxide in the air dissolves in sea water, forming carbonic acid (H₂CO₃). The carbonic acid then ionizes to form hydrogen ions and bicarbonate ions [latex]\left({\text{HCO}}_{3}{}^{-}\right)[/latex], which can further ionize into more hydrogen ions and carbonate ions [latex]\left({\text{CO}}_{3}{}^{2-}\right):[/latex]

[latex]\begin{array}{rll}{}{\text{CO}}_{2}\left(g\right)&\rightleftharpoons&{\text{CO}}_{2}\left(aq\right)\\{\text{CO}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}&\rightleftharpoons&{\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)\\{\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)&\rightleftharpoons&{\text{H}}^{\text{+}}\left(aq\right)+{\text{HCO}}_{3}{}^{-}\left(aq\right)\\{\text{HCO}}_{3}{}^{-}\left(aq\right)&\rightleftharpoons&{\text{H}}^{\text{+}}\left(aq\right)+{\text{CO}}_{3}{}^{2-}\left(aq\right)\end{array}[/latex]

The excess H⁺ ions make seawater more acidic. Increased ocean acidification can then have negative impacts on reef-building coral, as they cannot absorb the calcium carbonate they need to grow and maintain their skeletons (Figure 1). This in turn disrupts the local biosystem that depends upon the health of the reefs for its survival. If enough local reefs are similarly affected, the disruptions to sea life can be felt globally. The world’s oceans are presently in the midst of a period of intense acidification, believed to have begun in the mid-nineteenth century, and which is now accelerating at a rate faster than any change to oceanic pH in the last 20 million years.

This figure contains two photographs of coral reefs. In a, a colorful reef that includes hues of purple and pink corals is shown in blue green water with fish swimming in the background. In b, grey-green mossy looking coral is shown in a blue aquatic environment. This photo does not have the colorful appearance or fish that were shown in figure a.

Figure 1. Healthy coral reefs (a) support a dense and diverse array of sea life across the ocean food chain. But when coral are unable to adequately build and maintain their calcium carbonite skeletons because of excess ocean acidification, the unhealthy reef (b) is only capable of hosting a small fraction of the species as before, and the local food chain starts to collapse. (credit a: modification of work by NOAA Photo Library; credit b: modification of work by “prilfish”/Flickr)

Learn more about ocean acidification and how it affects other marine creatures.

You can view the transcript for “Ocean Acidification” here (opens in new window).

This site has detailed information about how ocean acidification specifically affects coral reefs.

The dramatic increase in solubility with increasing acidity described above for calcium carbonate is typical of salts containing basic anions (e.g., carbonate, fluoride, hydroxide, sulfide). Another familiar example is the formation of dental cavities in tooth enamel. The major mineral component of enamel is calcium hydroxyapatite (Figure 2), a sparingly soluble ionic compound whose dissolution equilibrium is

[latex]{\text{Ca}}_{5}{\left({\text{PO}}_{4}\right)}_{3}\text{OH}\left(s\right)\longrightarrow 5{\text{Ca}}^{\text{2+}}\left(aq\right)+3{\text{PO}}_{4}^{3-}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)[/latex]

This figure includes an image of two large light blue apatite crystals in a mineral conglomerate that includes white, grey, and tan crystals. The blue apatite crystals have a dull, dusty, or powdered appearance.

Figure 2. Crystal of the mineral hydroxylapatite, Ca₅(PO₄)₃OH, is shown here. Pure apatite is white, but like many other minerals, this sample is colored because of the presence of impurities.

This compound dissolved to yield two different basic ions: triprotic phosphate ions

[latex]{\text{PO}}_{4}{}^{3-}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{+}\longrightarrow {\text{H}}_{2}{\text{PO}}_{4}{}^{2-}+{\text{H}}_{2}\text{O}[/latex]
[latex]{\text{PO}}_{4}{}^{2-}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{+}\longrightarrow {\text{H}}_{2}{\text{PO}}_{4}{}^{-}+{\text{H}}_{2}\text{O}[/latex]
[latex]{\text{H}}_{2}{\text{PO}}_{4}{}^{-}+{\text{H}}_{3}{\text{O}}^{+}\longrightarrow {\text{H}}_{3}{\text{PO}}_{4}+{\text{H}}_{2}\text{O}[/latex]

and monoprotic hydroxide ions:

[latex]{\text{OH}}^{-}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{+}\longrightarrow 2{\text{H}}_{2}\text{O}[/latex]

Of the two basic productions, the hydroxide is, of course, by far the stronger base (it’s the strongest base that can exist in aqueous solution), and so it is the dominant factor providing the compound an acid-dependent solubility. Dental cavities form when the acid waste of bacteria growing on the surface of teeth hastens the dissolution of tooth enamel by reacting completely with the strong base hydroxide, shifting the hydroxyapatite solubility equilibrium to the right. Some toothpastes and mouth rinses contain added NaF or SnF₂ that make enamel more acid resistant by replacing the strong base hydroxide with the weak base fluoride:

[latex]\text{NaF}+{\text{Ca}}_{5}{\left({\text{PO}}_{4}\right)}_{3}\text{OH}\rightleftharpoons {\text{Ca}}_{5}{\left({\text{PO}}_{4}\right)}_{3}\text{F}+{\text{Na}}^{+}+{\text{OH}}^{-}[/latex]

The weak base fluoride ion reacts only partially with the bacterial acid waste, resulting in a less extensive shift in the solubility equilibrium and an increased resistance to acid dissolution. See the Chemistry in Everyday Life feature on the role of fluoride in preventing tooth decay for more information.

Role of Fluoride in Preventing Tooth Decay

As we saw previously, fluoride ions help protect our teeth by reacting with hydroxylapatite to form fluorapatite, Ca₅(PO₄)₃F. Since it lacks a hydroxide ion, fluorapatite is more resistant to attacks by acids in our mouths and is thus less soluble, protecting our teeth. Scientists discovered that naturally fluorinated water could be beneficial to your teeth, and so it became common practice to add fluoride to drinking water. Toothpastes and mouthwashes also contain amounts of fluoride (Figure 3).

Figure 3. Fluoride, found in many toothpastes, helps prevent tooth decay (credit: Kerry Ceszyk).

Unfortunately, excess fluoride can negate its advantages. Natural sources of drinking water in various parts of the world have varying concentrations of fluoride, and places where that concentration is high are prone to certain health risks when there is no other source of drinking water. The most serious side effect of excess fluoride is the bone disease, skeletal fluorosis. When excess fluoride is in the body, it can cause the joints to stiffen and the bones to thicken. It can severely impact mobility and can negatively affect the thyroid gland. Skeletal fluorosis is a condition that over 2.7 million people suffer from across the world. So while fluoride can protect our teeth from decay, the US Environmental Protection Agency sets a maximum level of 4 ppm (4 mg/L) of fluoride in drinking water in the US. Fluoride levels in water are not regulated in all countries, so fluorosis is a problem in areas with high levels of fluoride in the groundwater.

The solubility of ionic compounds may also be increased when dissolution is coupled to the formation of a complex ion. For example, aluminum hydroxide dissolves in a solution of sodium hydroxide or another strong base because of the formation of the complex ion Al(OH)4−.

An H atom is bonded to an O atom. The O atom has 2 dots above it and 2 dots below it. The O atom is bonded to an A l atom, which has three additional O atoms bonded to it as well. Each of these additional O atoms has 4 dots arranged around it, and is bonded to an H atom. This entire molecule is contained in brackets, to the right of which is a superscripted negative sign.

The equations for the dissolution of aluminum hydroxide, the formation of the complex ion, and the combined (net) equation are shown below. As indicated by the relatively large value of K for the net reaction, coupling complex formation with dissolution drastically increases the solubility of Al(OH)₃.

[latex]\text{Al(OH)}_3 (s) \rightleftharpoons \text{Al}^{3+} (aq) + 3\text{OH}^- (aq)[/latex] [latex]K_{sp} = 2 \times 10^{-32}[/latex]

[latex]\text{Al}^{3+} (aq) + 4\text{OH}^- (aq) \rightleftharpoons \text{Al(OH)}_4^{-} (aq)[/latex] [latex]K_{f} = 1.1 \times 10^{33}[/latex]

[latex]\text{Net: Al(OH)}_3 (s) + \text{OH}^- (aq) \rightleftharpoons \text{Al(OH)}_4^{-} (aq)[/latex] [latex]K = K_{sp} K_f = 22[/latex]

Example 1: Increased solubility in acidic solutions

Compute and compare the molar solublities for aluminum hydroxide, Al(OH)₃, dissolved in (a) pure water and (b) a buffer containing 0.100 M acetic acid and 0.100 M sodium acetate.

Solution

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Check Your Learning

What is the solubility of aluminum hydroxide in a buffer comprised of 0.100 M formic acid and 0.100 M sodium formate?

ANSWER:

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Example 2: Multiple Equilibria

Unexposed silver halides are removed from photographic film when they react with sodium thiosulfate (Na₂S₂O₃, called hypo) to form the complex ion [latex]\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{3-}[/latex] (K_f = 4.7 [latex]\times[/latex] 10¹³).

What mass of Na₂S₂O₃ is required to prepare 1.00 L of a solution that will dissolve 1.00 g of AgBr by the formation of [latex]\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{3-}?[/latex]

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Two equilibria are involved when AgBr dissolves in a solution containing the [latex]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}[/latex] ion:

Reaction 1 (dissolution): [latex]\text{AgBr}\left(s\right)\rightleftharpoons {\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Br}}^{-}\left(aq\right)\qquad{K}_{\text{sp}}=5.0\times {10}^{\text{-13}}[/latex]
Reaction 2 (complexation): [latex]{\text{Ag}}^{\text{+}}\left(aq\right)+{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\left(aq\right)\rightleftharpoons \text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{3-}\left(aq\right)\qquad{K}_{\text{f}}=4.7\times {10}^{13}[/latex]

First, calculate the concentration of bromide that will result when the 1.00 g of AgBr is completely dissolved via the cited complexation reaction:

[latex]1.00\text{ g AgBr }/(187.77\text{ g/mol})(1\text{ mol Br}^{-}/1\text{ mol AgBr})=0.00532\text{ mol Br}^{-}[/latex]

[latex]0.00532\text{ mol Br}^{-}/1.00\text{ L}=0.00532 M\text{ Br}^{-}[/latex]

Next, use this bromide molarity and the solubility product for silver bromide to calculate the silver ion molarity in the solution:

[latex]\left[{\text{Ag}}^{+}\right]=K_{\text{sp}}/ \left[\text{Br}^{-}\right]=5.0\times{10}^{-13}/0.00532=9.4\times {10}^{-11}M[/latex]

Based on the stoichiometry of the complex ion formation, the concentration of complex ion produced is

[latex]0.00532-9.4\times{10}^{-11}=0.00521M[/latex]

Use the silver ion and complex ion concentrations and the formation constant for the complex ion to compute the concentration of thiosulfate ion.

[latex]\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right]^{2}=\left[\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{\text{3-}}\right]/\left[\text{Ag}^{+}\right]K_{\text{f}}=0.00521/(9.6\times{10}^{-11})(4.7\times{10}^{13})=1.15\times{10}^{-6}[/latex]

[latex]\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right]=1.1\times{10}^{-3}M[/latex]

Finally, use this molar concentration to derive the required mass of sodium thiosulfate:

[latex](1.1\times {10}^{\text{-3}}\text{ mol}{\text{ S}}_{2}{\text{O}}_{3}{}^{2-}/\text{L})\times(1\text{ mol}{\text{ Na}}_{2}{\text{S}}_{2}{\text{O}}_{3}/1\text{ mol}{\text{ S}}_{2}{\text{O}}_{3}{}^{2-})\times(158.1\text{ g}{\text{ Na}}_{2}{\text{S}}_{2}{\text{O}}_{3}/\text{mol})=1.7\text{ g}[/latex]

Thus, 1.00 L of a solution prepared from 1.7 g Na₂S₂O₃ dissolves 1.0 g of AgBr.

Check Your Learning

AgCl(s), silver chloride, is well known to have a very low solubility:

[latex]\text{AgCl}(\text{s})\rightleftharpoons\text{Ag}^{+}(\text{aq}) + \text{Cl}^{-}(\text{aq})\qquad{K}_{\text{sp}}=1.77\times{10}^{-10}[/latex]

Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed:

[latex]\text{Ag}^{+}(\text{aq})+2\text{NH}_{3}(\text{aq})\rightleftharpoons\text{Ag(NH}_{3})_{2}{}^{+}(\text{aq})\qquad{K}_{\text{f}}=1.7\times{10}^{7}[/latex]

What mass of NH₃ is required to prepare 1.00 L of solution that will dissolve 2.00 g of AgCl by formation of Ag(NH₃)₂⁺?

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Key Concepts and Summary

Systems involving two or more chemical equilibria that share one or more reactant or product are called coupled equilibria. Common examples of coupled equilibria include the increased solubility of some compounds in acidic solutions (coupled dissolution and neutralization equilibria) and in solutions containing ligands (coupled dissolution and complex formation). The equilibrium tools from other chapters may be applied to describe and perform calculations on these systems.

Try It

A saturated solution of a slightly soluble electrolyte in contact with some of the solid electrolyte is said to be a system in equilibrium. Explain. Why is such a system called a heterogeneous equilibrium?
Calculate the equilibrium concentration of Ni²⁺ in a 1.0-M solution [Ni(NH₃)₆](NO₃)₂.
Calculate the equilibrium concentration of Zn²⁺ in a 0.30-M solution of [latex]\text{Zn}{\left(\text{CN}\right)}_{4}{}^{2-}[/latex].
Calculate the equilibrium concentration of Cu²⁺ in a solution initially with 0.050 M Cu²⁺ and 1.00 M NH₃.
Calculate the equilibrium concentration of Zn²⁺ in a solution initially with 0.150 M Zn²⁺ and 2.50 M CN^–.
Calculate the Fe³⁺ equilibrium concentration when 0.0888 mole of K₃[Fe(CN)₆] is added to a solution with 0.0.00010 M CN^–.
Calculate the Co²⁺ equilibrium concentration when 0.100 mole of [Co(NH₃)₆](NO₃)₂ is added to a solution with 0.025 M NH₃. Assume the volume is 1.00 L.
The equilibrium constant for the reaction [latex]{\text{Hg}}^{\text{2+}}\left(aq\right)+2{\text{Cl}}^{-}\left(aq\right)\rightleftharpoons {\text{HgCl}}_{2}\left(aq\right)[/latex] is 1.6 [latex]\times[/latex] 10¹³. Is HgCl₂ a strong electrolyte or a weak electrolyte? What are the concentrations of Hg²⁺ and Cl^– in a 0.015-M solution of HgCl₂?
Calculate the molar solubility of Sn(OH)₂ in a buffer solution containing equal concentrations of NH₃ and [latex]{\text{NH}}_{4}{}^{+}[/latex].
Calculate the molar solubility of Al(OH)₃ in a buffer solution with 0.100 M NH₃ and 0.400 M [latex]{\text{NH}}_{4}{}^{+}[/latex].
What is the molar solubility of CaF₂ in a 0.100-M solution of HF? K_a for HF = 7.2 [latex]\times[/latex] 10^–4.
What is the molar solubility of BaSO₄ in a 0.250-M solution of NaHSO₄? K_a for [latex]{\text{HSO}}_{4}{}^{-}[/latex] = 1.2 [latex]\times[/latex] 10^–2.
What is the molar solubility of Tl(OH)₃ in a 0.10-M solution of NH₃?
What is the molar solubility of Pb(OH)₂ in a 0.138-M solution of CH₃NH₂?
A solution of 0.075 M CoBr₂ is saturated with H₂S ([H₂S] = 0.10 M). What is the minimum pH at which CoS begins to precipitate?
[latex]\text{CoS}\left(s\right)\rightleftharpoons {\text{Co}}^{\text{2+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right){K}_{\text{sp}}=4.5\times {10}^{-27}[/latex][latex]{\text{H}}_{2}\text{S}\left(aq\right)+2{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons 2{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)K=1.0\times {10}^{\text{-26}}[/latex]
A 0.125-M solution of Mn(NO₃)₂ is saturated with H₂S ([H₂S] = 0.10 M). At what pH does MnS begin to precipitate?[latex]\text{MnS}\left(s\right)\rightleftharpoons {\text{Mn}}^{\text{2+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right){K}_{\text{sp}}=4.3\times {10}^{-22}[/latex][latex]{\text{H}}_{2}\text{S}\left(aq\right)+2{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons 2{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)K=1.0\times {10}^{-26}[/latex]
Calculate the molar solubility of BaF₂ in a buffer solution containing 0.20 M HF and 0.20 M NaF.
Calculate the molar solubility of CdCO₃ in a buffer solution containing 0.115 M Na₂CO₃ and 0.120 M NaHCO₃
To a 0.10-M solution of Pb(NO₃)₂ is added enough HF(g) to make [HF] = 0.10 M.
1. Does PbF₂ precipitate from this solution? Show the calculations that support your conclusion.
2. What is the minimum pH at which PbF₂ precipitates?
Calculate the concentration of Cd²⁺ resulting from the dissolution of CdCO₃ in a solution that is 0.250 M in CH₃CO₂H, 0.375 M in NaCH₃CO₂, and 0.010 M in H₂CO₃.
Both AgCl and AgI dissolve in NH₃.
1. What mass of AgI dissolves in 1.0 L of 1.0 M NH₃?
2. What mass of AgCl dissolves in 1.0 L of 1.0 M NH₃?
Calculate the volume of 1.50 M CH₃CO₂H required to dissolve a precipitate composed of 350 mg each of CaCO₃, SrCO₃, and BaCO₃.
Even though Ca(OH)₂ is an inexpensive base, its limited solubility restricts its use. What is the pH of a saturated solution of Ca(OH)₂?
What mass of NaCN must be added to 1 L of 0.010 M Mg(NO₃)₂ in order to produce the first trace of Mg(OH)₂?
Magnesium hydroxide and magnesium citrate function as mild laxatives when they reach the small intestine. Why do magnesium hydroxide and magnesium citrate, two very different substances, have the same effect in your small intestine. (Hint: The contents of the small intestine are basic.)
The following question is taken from a Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service. Solve the following problem: [latex]{\text{MgF}}_{2}\left(s\right)\rightleftharpoons {\text{Mg}}^{\text{2+}}\left(aq\right)+2{\text{F}}^{-}\left(aq\right)[/latex]In a saturated solution of MgF₂ at 18 °C, the concentration of Mg²⁺ is 1.21 [latex]\times[/latex] 10^–3M. The equilibrium is represented by the equation above.
1. Write the expression for the solubility-product constant, K_sp, and calculate its value at 18 °C.
2. Calculate the equilibrium concentration of Mg²⁺ in 1.000 L of saturated MgF₂ solution at 18 °C to which 0.100 mol of solid KF has been added. The KF dissolves completely. Assume the volume change is negligible.
3. Predict whether a precipitate of MgF₂ will form when 100.0 mL of a 3.00 [latex]\times[/latex] 10^–3–M solution of Mg(NO₃)₂ is mixed with 200.0 mL of a 2.00 [latex]\times[/latex] 10^–3–M solution of NaF at 18 °C. Show the calculations to support your prediction.
4. At 27 °C the concentration of Mg²⁺ in a saturated solution of MgF₂ is 1.17 [latex]\times[/latex] 10^–3M. Is the dissolving of MgF₂ in water an endothermic or an exothermic process? Give an explanation to support your conclusion.
Which of the following compounds, when dissolved in a 0.01-M solution of HClO₄, has a solubility greater than in pure water: AgBr, BaF₂, Ca₃(PO₄)₃, ZnS, PbI₂? Explain your answer.
What is the effect on the amount of solid Mg(OH)₂ that dissolves and the concentrations of Mg²⁺ and OH^– when each of the following are added to a mixture of solid Mg(OH)₂ and water at equilibrium?
1. MgCl₂
2. KOH
3. HClO₄
4. NaNO₃
5. Mg(OH)₂
What is the effect on the amount of CaHPO₄ that dissolves and the concentrations of Ca²⁺ and [latex]{\text{HPO}}_{4}{}^{-}[/latex] when each of the following are added to a mixture of solid CaHPO₄ and water at equilibrium?
1. CaCl₂
2. HCl
3. KClO₄
4. NaOH
5. CaHPO₄
Identify all chemical species present in an aqueous solution of Ca₃(PO₄)₂ and list these species in decreasing order of their concentrations. (Hint: Remember that the [latex]{\text{PO}}_{4}{}^{\text{3-}}[/latex] ion is a weak base.)
A volume of 50 mL of 1.8 M NH₃ is mixed with an equal volume of a solution containing 0.95 g of MgCl₂. What mass of NH₄Cl must be added to the resulting solution to prevent the precipitation of Mg(OH)₂?

Show Selected Solutions

2. [latex]{\text{Ni}}^{\text{2+}}\left(aq\right)+6{\text{NH}}_{3}\left(aq\right)\rightleftharpoons {\left[\text{Ni}{\left({\text{NH}}_{3}\right)}_{6}\right]}^{\text{2+}}\left(aq\right)\,\,\,\,\,\,\,{;}\,\,\,\,\,\,\,{K}_{\text{f}}=1.8\times {10}^{8}[/latex]

Let x be the change in concentration as Ni²⁺ dissociates. Because the initial Ni²⁺ concentration is 0, the concentration at any times is x:

[latex]1.8\times {10}^{8}=\frac{{\left[\text{Ni}{\left({\text{NH}}_{3}\right)}_{6}\right]}^{\text{2+}}}{\left[{\text{Ni}}^{\text{2+}}\right]{\left[{\text{NH}}_{3}\right]}^{6}}=\frac{\left(1.0-x\right)}{x{\left(6x\right)}^{6}}[/latex]

1.8 [latex]\times[/latex] 10⁸(46656x⁷) = 1.0 – x

8.40 [latex]\times[/latex] 10¹²(x²) = 1.0 – x

Since x is small in comparison with 1.0, drop x:

8.40 [latex]\times[/latex] 10¹²(x⁷) = 1.0

x⁷ = 1.19 [latex]\times[/latex] 10^–13

x = 0.014 M

4. Assume that all Cu²⁺ forms the complex whose concentration is 0.050 M and the remaining NH₃ has a concentration of 1.00 M – 4(0.050 M) = 0.80 M. The complex dissociates:

[latex]{\left[\text{Cu}{\left({\text{NH}}_{3}\right)}_{4}\right]}^{\text{2+}}\rightleftharpoons \left[{\text{Cu}}^{\text{2+}}\right]+4\left[{\text{NH}}_{3}\right][/latex]

Let x be the change in concentration of Cu²⁺ that dissociates:

	[Cu(NH₃)₄²⁺]	[Cu²⁺]	[NH₃]
Initial concentration (M)	0.050	0	0.80
Equilibrium (M)	0.050 − x	x	4x + 0.80

[latex]\frac{\left[\text{Cu}{\left({\text{NH}}_{3}\right)}_{4}{}^{\text{2+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]{\left[{\text{NH}}_{3}\right]}^{4}}=1.2\times {10}^{12}=\frac{0.050-x}{x{\left(4x+0.80\right)}^{4}}[/latex]

Assume that 4x is small when compared with 0.80 and that x is small when compared with 0.050:

(0.80)⁴ [latex]\times[/latex] 1.2 [latex]\times[/latex] 10¹²x = 0.050

x = 1.0 [latex]\times[/latex] 10^–13M

6. Set up a table listing initial and equilibrium concentrations for the reaction:

[latex]{\text{Fe}}^{\text{3+}}+6{\text{CN}}^{-}\rightleftharpoons {\left(\text{Fe}{\left(\text{CN}\right)}_{6}\right]}^{3-}{K}_{\text{f}}=1\times {10}^{44}[/latex]

Let x be the concentration of Fe³⁺ that dissociates when 0.0888 mol dissolves in 1.00 L of 0.00010 M CN^–. Assume no volume change upon dissolution:

	[Fe(CN)₆³⁻]	[Fe³⁺]	[CN⁻]
Initial concentration (M)	0.0888	0	0.00010
Equilibrium (M)	0.0888 − x	x	0.00010 − 6x

[latex]\frac{\left[\text{Fe}{\left(\text{CN}\right)}_{6}{}^{3-}\right]}{\left[{\text{Fe}}^{\text{3+}}\right]{\left[{\text{CN}}^{-}\right]}^{6}}=\frac{0.0888-x}{x{\left(0.000010 - 6x\right)}^{6}}=1\times {10}^{44}[/latex]

Assume that x is small when compared with the terms from which it is subtracted:

0.0888 = (0.00010)⁶(x)(1 [latex]\times[/latex] 10⁴⁴)

[latex]x=\frac{0.0888}{1\times {10}^{26}}=9\times {10}^{-22}M[/latex]

8. Let x be the change in the number of moles of Hg²⁺ that form per liter:

	[HgCl₂]	[Hg²⁺]	[Cl⁻]
Initial concentration (M)	0.015	0	0
Equilibrium (M)	0.015 − x	x	2x

[latex]\begin{array}{l}\frac{\left[{\text{HgCl}}_{2}\right]}{\left[{\text{Hg}}^{\text{2+}}\right]{\left[{\text{Cl}}^{-}\right]}^{2}}=K=1.6\times {10}^{13}\\ \frac{0.015-x}{\left(x\right){\left(2x\right)}^{2}}\approx \frac{0.015}{4{x}^{3}}=1.6\times {10}^{13}\end{array}[/latex]

x³ = 2.3 [latex]\times[/latex] 10^–16

x = 6.2 [latex]\times[/latex] 10^–6M = [Hg²⁺]

2x = 1.2 [latex]\times[/latex] 10^–5] M = [Cl^–]

The substance is a weak electrolyte because very little of the initial 0.015 M HgCl₂ dissolved.

10. [latex]{K}_{\text{b}}=\frac{\left[{\text{NH}}_{4}{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{NH}}_{3}\right]}=\frac{\left(0.400\right)\left[{\text{OH}}^{-}\right]}{\left(0.100\right)}=1.8\times {10}^{-5}[/latex]

[latex]\left[{\text{OH}}^{-}\right]=\frac{\left(0.100\right)\left(1.8\times {10}^{-5}\right)}{0.0400}\text{=}4.5\times {10}^{-5}[/latex]

K_sp = [Al³⁺][OH^–]³ = [Al³⁺](4.5 [latex]\times[/latex] 10^–5)³ = 1.9 [latex]\times[/latex] 10^–33

[Al³⁺] = 2.1 [latex]\times[/latex] 10^–20 (molar solubility)

12. Find the amount of [latex]{\text{SO}}_{4}{}^{2-}[/latex] present from K_a for the equilibrium:

[latex]{\text{HSO}}_{4}{}^{-}+{\text{H}}_{2}\text{O}\longrightarrow {\text{H}}_{3}{\text{O}}^{+}+{\text{SO}}_{4}{}^{2-}[/latex]

Let x be the change in [latex]\left[{\text{SO}}_{4}{}^{2-}\right][/latex] :

[latex]{K}_{\text{a}}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{SO}}_{4}{}^{2-}\right]}{\left[{\text{HSO}}_{4}{}^{-}\right]}=\frac{{x}^{2}}{0.250-x}=1.2\times {10}^{-2}[/latex]

Because K_a is too large to disregard x in the expression 0.250 – x, we must solve the quadratic equation:

x² + 1.2 [latex]\times[/latex] 10^–2x – 0.250(1.2 [latex]\times[/latex] 10^–2) = 0

[latex]x=\frac{-1.2\times {10}^{-2}\pm \sqrt{{\left(1.2\times {10}^{-2}\right)}^{2}+4\left(3.0\times {10}^{\text{-3}}\right)}}{2}=\frac{-1.2\times {10}^{-2}\pm 0.11}{2}=0.049M[/latex]

K_sp = [Ba²⁺] [latex]\left[{\text{SO}}_{4}{}^{2-}\right][/latex] = [Ba²⁺](0.049) = 1.08 [latex]\times[/latex] 10^–10

[Ba²⁺] = 2.2 [latex]\times[/latex] 10^–9 (molar solubility)

14. [latex]{\text{CH}}_{3}{\text{NH}}_{2}+{\text{H}}_{2}\text{O}\rightleftharpoons {\text{CH}}_{3}{\text{NH}}_{3}{}^{+}+{\text{OH}}^{-}[/latex]

[latex]{K}_{\text{b}}=\frac{\left[{\text{CH}}_{3}{\text{NH}}_{3}{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{3}{\text{NH}}_{2}\right]}=\frac{\left(x\right)\left(x\right)}{0.138-x}=4.4\times {10}^{-4}[/latex]

Solve the quadratic equation using the quadratic formula:

x² + 4.4 [latex]\times[/latex] 10^–4x – 0.138(4.4 [latex]\times[/latex] 10^–4) = 0

[latex]x=\frac{-4.4\times {10}^{-4}\pm \sqrt{{\left(4.4\times {10}^{-4}\right)}^{2}+4\left(6.07\times {10}^{\text{-5}}\right)}}{2}=\frac{-4.4\times {10}^{\text{-4}}\pm \sqrt{2.43\times {10}^{-4}}}{2}=\frac{0.0152}{2}=7.6\times {10}^{-3}M[/latex]

K_sp = [Pb²⁺][OH^–]² = [Pb²⁺](7.6 [latex]\times[/latex] 10^–3)² = 2.8 [latex]\times[/latex] 10^–16

[Pb²⁺] = 4.8 [latex]\times[/latex] 10^–12 (molar solubility)

16. Two equilibria are in competition for the ions and must be considered simultaneously. Precipitation of MnS will occur when the concentration of S^2– in conjunction with 0.125 M Mn²⁺ exceeds the K_sp of MnS. The [S^2–] must come from the ionization of H₂S as defined by the equilibrium:

[latex]{\text{H}}_{2}\text{S}\left(aq\right)+2{\text{H}}_{2}\text{O}\left(l\right)\longrightarrow 2{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)[/latex]

[latex]\frac{{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}^{2}\left[{\text{S}}^{2-}\right]}{\left[{\text{H}}_{2}\text{S}\right]}={K}_{1}{K}_{2}\left({\text{H}}_{2}\text{S}\right)=1.0\times {10}^{\text{-26}}[/latex]

As a saturated solution of H₂S is 0.10 M, this later expression becomes:

[latex]{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}^{2}\left[{\text{S}}^{2-}\right]=1.0\times {10}^{-27}[/latex]

From the equilibrium of MnS, the minimum concentration of S^2– required to cause precipitation is calculated as:

[latex]\text{MnS}\left(s\right)\longrightarrow {\text{Mn}}^{\text{2+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right)[/latex]

K_sp = [Mn²⁺][S^2–] = 4.3 [latex]\times[/latex] 10^–22

[latex]\left[{\text{S}}^{2-}\right]=\frac{4.3\times {10}^{-22}}{0.125}=3.44\times {10}^{-21}[/latex]

This amount of S^2– will exist in solution at a pH defined by the H₂S equilibrium:

[latex]{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}^{2}\left(3.44\times {10}^{-21}\right)=1.0\times {10}^{-27}[/latex]
[latex]{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}^{2}=2.91\times {10}^{-7}[/latex]
[latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=5.39\times {10}^{-4}M[/latex]
[latex]\text{pH}=-\text{log}\left[{\text{H}}_{3}{\text{O}}^{+}\right]=3.27[/latex]

18. Three equilibria are involved:

[latex]{\text{H}}_{2}{\text{CO}}_{3}+{\text{H}}_{2}\text{O}\longrightarrow {\text{H}}_{3}{\text{O}}^{+}+{\text{HCO}}_{3}{}^{-}{K}_{{\text{a}}_{1}}=4.3\times {10}^{-7}[/latex]
[latex]{\text{HCO}}_{3}{}^{-}+{\text{H}}_{2}\text{O}\longrightarrow {\text{H}}_{3}{\text{O}}^{+}+{\text{CO}}_{3}{}^{2-}{K}_{{\text{a}}_{2}}=7\times {10}^{-11}[/latex]
[latex]{\text{CdCO}}_{3}\longrightarrow {\text{Cd}}^{\text{2+}}+{\text{CO}}_{3}{}^{2-}{K}_{\text{sp}}=2.5\times {10}^{-14}[/latex]

First, find the pH of the buffer from the Henderson-Hasselbach equation. Then find [H₃O⁺]:

[latex]\text{pH}=\text{p}{K}_{\text{a}}+\mathrm{log}\frac{\left[{\text{CO}}_{3}{}^{2-}\right]}{\left[{\text{HCO}}_{3}{}^{-}\right]}[/latex]

10.155 + log [latex]\frac{0.115}{0.120}[/latex] = 10.155 – 0.018 = 10.137

In this case, several more significant figures are carried than justified so that the value of the log ratio is meaningful:

[latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=7.3\times {10}^{-11}[/latex]

Now, find the concentration of [latex]{\text{CO}}_{3}{}^{2-}[/latex] present in the buffer solution. Next, using [latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] and K_sp, calculate the concentration of Cd²⁺. This latter value represents the molar solubility. From [latex]{K}_{{a}_{1}}[/latex] determine [H₂CO₃]:

[latex]{K}_{{\text{a}}_{1}}=4.3\times {10}^{-7}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{HCO}}_{3}{}^{-}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}=\frac{\left(7.3\times {10}^{-11}\right)\left(0.120\right)}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}[/latex]

[H₂CO₃] = 2.05 [latex]\times[/latex] 10^–5

From [latex]{K}_{{\text{a}}_{1}}{K}_{{\text{a}}_{2}}[/latex] , find the concentration of [latex]{\text{CO}}_{3}{}^{2-}[/latex] :

[latex]\begin{array}{rll}{}{K}_{{\text{a}}_{1}}{K}_{{\text{a}}_{2}}&=&\left(4.3\times 10 - 7\right)\left(7\times 10 - 11\right)=\frac{\left[{\text{HCO}}_{3}{}^{-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}\times \frac{\left[{\text{CO}}_{3}{}^{2-}\right]\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{CO}}_{3}{}^{2-}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}=\frac{{\left(7.3\times {10}^{-11}\right)}^{2}\times \left[{\text{CO}}_{3}{}^{2-}\right]}{2.04\times {10}^{-5}}\\& =&3\times {10}^{-17}\end{array}[/latex]

[latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] = 0.115 M

K_sp = [Cd²⁺] [latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] = [Cd²⁺](0.115) = 3 [latex]\times[/latex] 10^–13

[latex]\left[{\text{Cd}}^{\text{2+}}\right]=\frac{3\times {10}^{-13}}{0.115}=3\times {10}^{-12}M[/latex]

20. For K_sp, (CdCO₃) = [Cd²⁺] [latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] = 2.5 [latex]\times[/latex] 10^–14, the amount of [latex]{\text{CO}}_{3}{}^{2-}[/latex] is governed by K_a of H₂CO₃, 4.3 [latex]\times[/latex] 10^–7, and K_a of [latex]{\text{HCO}}_{3}{}^{-}[/latex] , 7 [latex]\times[/latex] 10^–11, and [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] by K_a of acetic acid. First, calculate the [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] from K_a of acetic acid (HOAc):

[latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right]=\frac{{K}_{\text{a}}\left[\text{HOAc}\right]}{\left[{\text{OAc}}^{-}\right]}=\frac{1.8\times {10}^{-5}\left(0.250\right)}{\left(0.375\right)}=1.2\times {10}^{-5}M[/latex]

From this and K_a for H₂CO₃, calculate [latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] present. As [latex]\left[{\text{H}}_{3}{\text{O}}^{+}\right][/latex] is fixed by K_a of acetic acid:

[latex]{K}_{\text{a}}\left({\text{H}}_{2}{\text{CO}}_{3}\right)=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{HCP}}_{3}{}^{-}\right]}{\left[{\text{H}}_{2}{\text{CO}}_{3}{}^{-}\right]}[/latex]

[latex]\left[{\text{H}}_{2}{\text{CO}}_{3}{}^{-}\right]=\frac{{K}_{\text{a}}\left[{\text{H}}_{2}{\text{CO}}_{3}\right]}{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}=\frac{4.3\times {10}^{-7}\left[0.010\right]}{\left[1.2\times {10}^{-5}\right]}=3.58\times {10}^{-4}M[/latex]

From [latex]{K}_{\text{a}}\left({\text{HCO}}_{3}{}^{-}\right)[/latex] we obtain [latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] :

[latex]{K}_{a}=7\times {10}^{-11}=\frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]\left[{\text{CO}}_{3}{}^{2-}\right]}{\left[{\text{HCO}}_{3}{}^{-}\right]}=\frac{1.2\times {10}^{-17}\left[{\text{CO}}_{3}{}^{2-}\right]}{\left[3.58\times {10}^{-4}\right]}\text{ }[/latex]

[latex]\left[{\text{CO}}_{3}{}^{2-}\right]=\frac{7\times {10}^{-11}\times 3.58\times {10}^{-4}}{1.2\times {10}^{-5}}=2.09\times {10}^{-9}[/latex]

From the solubility product:

K_sp = [Cd²⁺] [latex]\left[{\text{CO}}_{3}{}^{2-}\right][/latex] = 2.5 [latex]\times[/latex] 10^–14 = Cd²⁺ (2.09 [latex]\times[/latex] 10^–9)

[latex]\left[{\text{Cd}}^{\text{2+}}\right]=\frac{2.5\times {10}^{-14}}{2.09\times {10}^{-9}}=1\times {10}^{-5}M[/latex]

22. [latex]{\text{CaCO}}_{3}:0.350\cancel{\text{g}}\times \frac{1\text{mol}}{100.09\cancel{\text{g}}}=3.50\times {10}^{-3}\text{mol}[/latex]

[latex]{\text{SrCO}}_{3}:0.350\cancel{\text{g}}\times \frac{1\text{mol}}{147.63\cancel{\text{g}}}=2.37\times {10}^{-3}\text{mol}[/latex]

[latex]{\text{BaCO}}_{3}:0.350\cancel{\text{g}}\times \frac{1\text{mol}}{197.34\cancel{\text{g}}}=1.77\times {10}^{-3}\text{mol}[/latex]

Total: 7.64 [latex]\times[/latex] 10^–3 mol
CaCO₃: K_sp = 4.8 [latex]\times[/latex] 10^–9
SrCO₃: K_sp = 9.42 [latex]\times[/latex] 10^–10
BaCO₃: K_sp = 8.1 [latex]\times[/latex] 10^–9

Solubilities are approximately equal.

When the three solids dissolve, 7.64 [latex]\times[/latex] 10^–3 mol of metal ions and 7.64 [latex]\times[/latex] 10^–3 mol of [latex]{\text{CO}}_{3}{}^{2-}[/latex] are initially present. The [latex]{\text{CO}}_{3}{}^{2-}[/latex] reacts with CH₃CO₂H to form [latex]{\text{HCO}}_{3}{}^{-}[/latex] :

[latex]{\text{ CO}}_{3}{}^{2-}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)\rightleftharpoons {\text{HCO}}_{3}{}^{-}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)[/latex]

[latex]K=\frac{\left[{\text{HCO}}_{3}{}^{-}\right]\left[{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\right]}{\left[{\text{CO}}_{3}{}^{2-}\right]\left[{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right]}\times \frac{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}{\left[{\text{H}}_{3}{\text{O}}^{+}\right]}=\frac{{K}_{\text{a}}\left(\text{for}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right)}{{K}_{\text{a}}\left(\text{for}{\text{HCO}}_{3}{}^{-}\right)}=\frac{1.8\times {10}^{-5}}{7\times {10}^{-11}}=2.6\times {10}^{5}[/latex]

This K value is large, so virtually all [latex]{\text{CO}}_{3}{}^{2-}[/latex] undergoes this reaction and approximately 7.64 [latex]\times[/latex] 10^–3 mol of [latex]{\text{HCO}}_{3}{}^{-}[/latex] forms. The [latex]{\text{HCO}}_{3}{}^{-}[/latex] reacts with CH₃CO₂H:

[latex]{\text{HCO}}_{3}{}^{-}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)\longrightarrow {\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{-}\left(aq\right)[/latex]

[latex]K=\frac{{K}_{\text{a}}\left(\text{for}{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\right)}{{K}_{\text{a}}\left(\text{for}{\text{HCO}}_{3}{}^{-}\right)}=\frac{1.8\times {10}^{-5}}{4.3\times {10}^{-7}}=42[/latex]

This reaction is virtually complete as well (K is large).

For each mol of [latex]{\text{CO}}_{3}{}^{2-}[/latex] produced, 2 mol of CH₃CO₂H is required for conversion to H₂CO₃. Thus 15.3 [latex]\times[/latex] 10^–3 mol CH₃CO₂H is needed:

[latex]\text{Volume}=\frac{15.3\times {10}^{-3}\text{mol}}{1.50\text{mol}{\text{L}}^{-1}}=0.0102\text{L}\left(10.2\text{mL}\right)[/latex]

24. There are two equilibria involved: [latex]\begin{array}{l}\text{Mg}{\left(\text{OH}\right)}_{2}\left(s\right)\rightleftharpoons {\text{Mg}}^{\text{2+}}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right){K}_{\text{sp}}=1.5\times {10}^{-11}\\ {\text{CN}}^{-}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightleftharpoons \text{HCN}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\left({K}_{\text{b}}=\frac{{K}_{\text{w}}}{{K}_{\text{a}}}=\frac{1.0\times {10}^{-14}}{4\times {10}^{-10}}=2.5\times {10}^{-5}\right)\end{array}[/latex] The Mg(NO₃)₂ dissolves, and [Mg²⁺] = 0.010 M.

[Mg²⁺][OH^–]² = K_sp = 1.5 [latex]\times[/latex] 10^–11

(0.010)[OH^–]² = 1.5 [latex]\times[/latex] 10^–11

[OH^–] = 3.9 [latex]\times[/latex] 10^–5M

We need to add enough CN^– to make [OH^–] = 3.9 [latex]\times[/latex] 10^–5M. Both OH^– and HCN come from CN^–, so [OH^–] = [HCN]:

[latex]\begin{array}{l}\frac{\left[\text{HCN}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CN}}^{-}\right]}={K}_{\text{b}}=2\times {10}^{-5}\\ \frac{{\left(3.9\times {10}^{-5}\right)}^{2}}{\left[{\text{CN}}^{-}\right]}=2.5\times {10}^{-5}\end{array}[/latex]

[CN^–] = 6.1 [latex]\times[/latex] 10^–5M

mol NaCN = mol HCN + mol CN^– = 3.9 [latex]\times[/latex] 10^–5 + 6.1 [latex]\times[/latex] 10^–5 = 1.0 [latex]\times[/latex] 10^–4

[latex]\text{mass}\left(\text{NaCN}\right)=1.0\times {10}^{-4}\text{mol}\times \frac{49.007\text{g}}{1\text{mol}}=5\times {10}^{-3}\text{g}[/latex]

26. The answers are as follows:

K_sp = [Mg²⁺][F^–]² = (1.21 [latex]\times[/latex] 10^–3)(2 [latex]\times[/latex] 1.21 [latex]\times[/latex] 10^–3)² = 7.09 [latex]\times[/latex] 10^–9
K_sp = [Mg²⁺][F^–]² = [x][0.100 + 2x]² = 7.09 [latex]\times[/latex] 10^–9Assume that 2x is small when compared with 0.100 M.
0.100x = 7.09 [latex]\times[/latex] 10^–9x = [MgF₂] = 7.09 [latex]\times[/latex] 10^–7M
The value 7.09 [latex]\times[/latex] 10^–7M is quite small when compared with 0.100 M, so the assumption is valid.
Determine the concentration of Mg²⁺ and F^– that will be present in the final volume. Compare the value of the ion product [Mg²⁺][F^–]² with K_sp. If this value is larger than K_sp, precipitation will occur.
0.1000 L [latex]\times[/latex] 3.00 [latex]\times[/latex] 10^–3M Mg(NO₃)₂ = 0.3000 L [latex]\times[/latex] M Mg(NO₃)₂M Mg(NO₃)₂ = 1.00 [latex]\times[/latex] 10^–3M
0.2000 L [latex]\times[/latex] 2.00 [latex]\times[/latex] 10^–3M NaF = 0.3000 L [latex]\times[/latex] M NaF
M NaF = 1.33 [latex]\times[/latex] 10^–3M
ion product = (1.00 [latex]\times[/latex] 10^–3)(1.33 [latex]\times[/latex] 10^–3)² = 1.77 [latex]\times[/latex] 10^–9This value is smaller than K_sp, so no precipitation will occur.
MgF₂ is less soluble at 27 °C than at 18 °C. Because added heat acts like an added reagent, when it appears on the product side, the Le Châtelier’s principle states that the equilibrium will shift to the reactants’ side to counter the stress. Consequently, less reagent will dissolve. This situation is found in our case. Therefore, the reaction is exothermic.

28. Effect on amount of solid Mg(OH)₂, [Mg²⁺], [OH^–]:

increase, increase, decrease;
increase, decrease, increase;
decrease, increase, decrease;
no effect predicted;
increase, no effect, no effect

30. [H₂O] > [Ca²⁺] > [latex]\left[{\text{PO}}_{4}{}^{3-}\right][/latex] > [latex]{\text{HPO}}_{4}{}^{2-}[/latex] = [OH^–] > [latex]\left[{\text{HPO}}_{4}{}^{-}\right][/latex]

Glossary

coupled equilibrium: system characterized by more than one state of balance between a slightly soluble ionic solid and an aqueous solution of ions working simultaneously

Module 15: Equilibria of Other Reaction Classes

Learning Outcomes

Role of Fluoride in Preventing Tooth Decay

Example 1: Increased solubility in acidic solutions

Solution

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ANSWER:

Example 2: Multiple Equilibria

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Key Concepts and Summary

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Glossary

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