Multiple Equilibria

Learning Objectives

By the end of this module, you will be able to:

Describe examples of systems involving two (or more) simultaneous chemical equilibria
Calculate reactant and product concentrations for multiple equilibrium systems
Compare dissolution and weak electrolyte formation

There are times when one equilibrium reaction does not adequately describe the system being studied. Sometimes we have more than one type of equilibrium occurring at once (for example, an acid-base reaction and a precipitation reaction).

The ocean is a unique example of a system with multiple equilibria, or multiple states of solubility equilibria working simultaneously. 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.

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

Slightly soluble solids derived from weak acids generally dissolve in strong acids, unless their solubility products are extremely small. For example, we can dissolve CuCO₃, FeS, and Ca₃(PO₄)₂ in HCl because their basic anions react to form weak acids (H₂CO₃, H₂S, and [latex]{\text{H}}_{2}{\text{PO}}_{4}^{-}[/latex] ). The resulting decrease in the concentration of the anion results in a shift of the equilibrium concentrations to the right in accordance with Le Châtelier’s principle.

Of particular relevance to us is the dissolution of hydroxylapatite, Ca₅(PO₄)₃OH, in acid. Apatites are a class of calcium phosphate minerals (Figure 2); a biological form of hydroxylapatite is found as the principal mineral in the enamel of our teeth. A mixture of hydroxylapatite and water (or saliva) contains an equilibrium mixture of solid Ca₅(PO₄)₃OH and dissolved Ca²⁺, [latex]{\text{PO}}_{4}^{3-}[/latex], and OH^– ions:

[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.

When exposed to acid, phosphate ions react with hydronium ions to form hydrogen phosphate ions and ultimately, phosphoric acid:

[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]

Hydroxide ion reacts to form water:

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

These reactions decrease the phosphate and hydroxide ion concentrations, and additional hydroxylapatite dissolves in an acidic solution in accord with Le Châtelier’s principle. Our teeth develop cavities when acid waste produced by bacteria growing on them causes the hydroxylapatite of the enamel to dissolve. Fluoride toothpastes contain sodium fluoride, NaF, or stannous fluoride [more properly named tin(II) fluoride], SnF₂. They function by replacing the OH^– ion in hydroxylapatite with F^– ion, producing fluorapatite, Ca₅(PO₄)₃F:

[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 resulting Ca₅(PO₄)₃F is slightly less soluble than Ca₅(PO₄)₃OH, and F^– is a weaker base than OH^–. Both of these factors make the fluorapatite more resistant to attack by acids than hydroxylapatite. 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.

When acid rain attacks limestone or marble, which are calcium carbonates, a reaction occurs that is similar to the acid attack on hydroxylapatite. The hydronium ion from the acid rain combines with the carbonate ion from calcium carbonates and forms the hydrogen carbonate ion, a weak acid:

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

Calcium hydrogen carbonate, Ca(HCO₃)₂, is soluble, so limestone and marble objects slowly dissolve in acid rain.

If we add calcium carbonate to a concentrated acid, hydronium ion reacts with the carbonate ion according to the equation:

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

(Acid rain is usually not sufficiently acidic to cause this reaction; however, laboratory acids are.) The solution may become saturated with the weak electrolyte carbonic acid, which is unstable, and carbon dioxide gas can be evolved:

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

These reactions decrease the carbonate ion concentration, and additional calcium carbonate dissolves. If enough acid is present, the concentration of carbonate ion is reduced to such a low level that the reaction quotient for the dissolution of calcium carbonate remains less than the solubility product of calcium carbonate, even after all of the calcium carbonate has dissolved.

Example 1: Prevention of Precipitation of Mg(OH)₂

Calculate the concentration of ammonium ion that is required to prevent the precipitation of Mg(OH)₂ in a solution with [Mg²⁺] = 0.10 M and [NH₃] = 0.10 M.

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Check Your Learning

Consider the two equilibria:

[latex]\text{ZnS}\left(s\right)\rightleftharpoons {\text{Zn}}^{\text{2+}}\left(aq\right)+{\text{S}}^{2-}\left(aq\right);{K}_{\text{sp}}=1\times {10}^{-27}[/latex]
[latex]2{\text{H}}_{2}\text{O}\left(l\right)+{\text{H}}_{2}\text{S}\left(aq\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]

and calculate the concentration of hydronium ion required to prevent the precipitation of ZnS in a solution that is 0.050 M in Zn²⁺ and saturated with H₂S (0.10 M H₂S).

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Therefore, precise calculations of the solubility of solids from the solubility product are limited to cases in which the only significant reaction occurring when the solid dissolves is the formation of its ions.

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). The reaction with silver bromide is:

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): [latex]\text{AgBr}\left(s\right)\rightleftharpoons {\text{Ag}}^{\text{+}}\left(aq\right)+{\text{Br}}^{-}\left(aq\right);{K}_{\text{sp}}=3.3\times {10}^{\text{-13}}[/latex]
Reaction (2): [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);{K}_{\text{f}}=4.7\times {10}^{13}[/latex]

In order for 1.00 g of AgBr to dissolve, the [Ag⁺] in the solution that results must be low enough for Q for Reaction (1) to be smaller than K_sp for this reaction. We reduce [Ag⁺] by adding [latex]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}[/latex] and thus cause Reaction (2) to shift to the right. We need the following steps to determine what mass of Na₂S₂O₃ is needed to provide the necessary [latex]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}[/latex] .

We calculate the [Br^–] produced by the complete dissolution of 1.00 g of AgBr (5.33 [latex]\times[/latex] 10^–3 mol AgBr) in 1.00 L of solution:
[latex]\left[{\text{Br}}^{-}\right]=5.33\times {10}^{\text{-3}}M[/latex]
We use [Br^–] and K_sp to determine the maximum possible concentration of Ag⁺ that can be present without causing reprecipitation of AgBr:[latex]\left[{\text{Ag}}^{+}\right]=6.2\times {10}^{-11}M[/latex]
We determine the [latex]\left[{S}_{2}{O}_{3}{}^{2-}\right][/latex] required to make [Ag⁺] = 6.2 [latex]\times[/latex] 10^–11 M after the remaining Ag⁺ ion has reacted with [latex]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}[/latex] according to the equation:[latex]{\text{Ag}}^{+}+2{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\rightleftharpoons \text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{\text{3-}}{K}_{\text{f}}=4.7\times {10}^{13}[/latex]Because 5.33 [latex]\times[/latex] 10^–3 mol of AgBr dissolves:
[latex]\left(5.33\times {10}^{-3}\right)-\left(6.2\times {10}^{\text{-11}}\right)=5.33\times {10}^{-3}\text{mol}\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{3-}[/latex]

Thus, at equilibrium: [latex]\left[\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{3-}\right][/latex] = 5.33 [latex]\times[/latex] 10^–3M, [Ag⁺] = 6.2 [latex]\times[/latex] 10^–11M, and Q = K_f = 4.7 [latex]\times[/latex] 10¹³:

[latex]{K}_{\text{f}}=\frac{\left[\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{\text{3-}}\right]}{\left[{\text{Ag}}^{+}\right]{\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right]}^{2}}=4.7\times {10}^{13}[/latex]

[latex]\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right]=1.4\times {10}^{-3}M[/latex]

When [latex]\left[{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right][/latex] is 1.4 [latex]\times[/latex] 10^–3M, [Ag⁺] is 6.2 [latex]\times[/latex] 10^–11M and all AgBr remains dissolved.
We determine the total number of moles of [latex]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}[/latex] that must be added to the solution. This equals the amount that reacts with Ag⁺ to form [latex]\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{3-}[/latex] plus the amount of free [latex]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}[/latex] in solution at equilibrium. To form 5.33 [latex]\times[/latex] 10^–3 mol of [latex]\text{Ag}{\left({\text{S}}_{2}{\text{O}}_{3}\right)}_{2}{}^{3-}[/latex] requires 2 [latex]\times[/latex] (5.33 [latex]\times[/latex] 10^–3) mol of [latex]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}[/latex]. In addition, 1.4 [latex]\times[/latex] 10^–3 mol of unreacted [latex]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}[/latex] is present (Step 3). Thus, the total amount of [latex]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}[/latex] that must be added is:[latex]2\times \left(5.33\times {10}^{-3}\text{mol}{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\right)+1.4\times {10}^{-3}\text{mol}{\text{S}}_{2}{\text{O}}_{3}{}^{2-}=1.21\times {10}^{-2}\text{mol}{\text{S}}_{2}{\text{O}}_{3}{}^{2-}[/latex]
We determine the mass of Na₂S₂O₃ required to give 1.21 [latex]\times[/latex] 10^–2 mol [latex]{\text{S}}_{2}{\text{O}}_{3}{}^{2-}[/latex] using the molar mass of Na₂S₂O₃:[latex]1.21\times {10}^{\text{-2}}\text{mol}{\text{S}}_{2}{\text{O}}_{3}{}^{2-}\times \frac{158.1\text{g}{\text{Na}}_{2}{\text{S}}_{2}{\text{O}}_{3}}{1\text{mol}{\text{Na}}_{2}{\text{S}}_{2}{\text{O}}_{3}}=1.9\text{g}{\text{Na}}_{2}{\text{S}}_{2}{\text{O}}_{3}[/latex]Thus, 1.00 L of a solution prepared from 1.9 g Na₂S₂O₃ dissolves 1.0 g of AgBr.

Check Your Learning

AgCl(s), silver chloride, is well known to have a very low solubility: Ag(s) ⇌ Ag⁺(aq) + Cl⁻(aq), K_sp = 1.77 × 10^–10. Adding ammonia significantly increases the solubility of AgCl because a complex ion is formed: Ag⁺(aq) + 2NH₃(aq) ⇌ Ag(NH₃)₂⁺(aq), K_f = 1.7 × 10⁷. 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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Dissolution versus Weak Electrolyte Formation

We can determine how to shift the concentration of ions in the equilibrium between a slightly soluble solid and a solution of its ions by applying Le Châtelier’s principle. For example, one way to control the concentration of manganese(II) ion, Mn²⁺, in a solution is to adjust the pH of the solution and, consequently, to manipulate the equilibrium between the slightly soluble solid manganese(II) hydroxide, manganese(II) ion, and hydroxide ion:

[latex]\text{Mn}{\left(\text{OH}\right)}_{2}\left(s\right)\rightleftharpoons {\text{Mn}}^{\text{2+}}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right){K}_{\text{sp}}=\left[{\text{Mn}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^{2}[/latex]

This could be important to a laundry because clothing washed in water that has a manganese concentration exceeding 0.1 mg per liter may be stained by the manganese. We can reduce the concentration of manganese by increasing the concentration of hydroxide ion. We could add, for example, a small amount of NaOH or some other base such as the silicates found in many laundry detergents. As the concentration of OH^– ion increases, the equilibrium responds by shifting to the left and reducing the concentration of Mn²⁺ ion while increasing the amount of solid Mn(OH)₂ in the equilibrium mixture, as predicted by Le Châtelier’s principle.

Example 3: Solubility Equilibrium of a Slightly Soluble Solid

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 in water at equilibrium?

MgCl₂
KOH
an acid
NaNO₃
Mg(OH)₂

Show Answer

The equilibrium among solid Mg(OH)₂ and a solution of Mg²⁺ and OH^– is:

[latex]\text{Mg}{\left(\text{OH}\right)}_{2}\left(s\right)\rightleftharpoons {\text{Mg}}^{\text{2+}}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)[/latex].

The reaction shifts to the left to relieve the stress produced by the additional Mg²⁺ ion, in accordance with Le Châtelier’s principle. In quantitative terms, the added Mg²⁺ causes the reaction quotient to be larger than the solubility product (Q > K_sp), and Mg(OH)₂ forms until the reaction quotient again equals K_sp. At the new equilibrium, [OH^–] is less and [Mg²⁺] is greater than in the solution of Mg(OH)₂ in pure water. More solid Mg(OH)₂ is present.
The reaction shifts to the left to relieve the stress of the additional OH^– ion. Mg(OH)₂ forms until the reaction quotient again equals K_sp. At the new equilibrium, [OH^–] is greater and [Mg²⁺] is less than in the solution of Mg(OH)₂ in pure water. More solid Mg(OH)₂ is present.
The concentration of OH^– is reduced as the OH^– reacts with the acid. The reaction shifts to the right to relieve the stress of less OH^– ion. In quantitative terms, the decrease in the OH^– concentration causes the reaction quotient to be smaller than the solubility product (Q < K_sp), and additional Mg(OH)₂ dissolves until the reaction quotient again equals K_sp. At the new equilibrium, [OH^–] is less and [Mg²⁺] is greater than in the solution of Mg(OH)₂ in pure water. More Mg(OH)₂ is dissolved.
NaNO₃ contains none of the species involved in the equilibrium, so we should expect that it has no appreciable effect on the concentrations of Mg²⁺ and OH^–. (As we have seen previously, dissolved salts change the activities of the ions of an electrolyte. However, the salt effect is generally small, and we shall neglect the slight errors that may result from it.)
The addition of solid Mg(OH)₂ has no effect on the solubility of Mg(OH)₂ or on the concentration of Mg²⁺ and OH^–. The concentration of Mg(OH)₂ does not appear in the equation for the reaction quotient:
[latex]Q=\left[{\text{Mg}}^{\text{2+}}\right]{\left[{\text{OH}}^{-}\right]}^{2}[/latex]
Thus, changing the amount of solid magnesium hydroxide in the mixture has no effect on the value of Q, and no shift is required to restore Q to the value of the equilibrium constant.

Check Your Learning

What is the effect on the amount of solid NiCO₃ that dissolves and the concentrations of Ni²⁺ and [latex]{\text{CO}}_{3}{}^{2-}[/latex] when each of the following are added to a mixture of the slightly soluble solid NiCO₃ and water at equilibrium?

Ni(NO₃)₂
KClO₄
NiCO₃
K₂CO₃
HNO₃ (reacts with carbonate giving [latex]{\text{HCO}}_{3}{}^{-}[/latex] or H₂O and CO₂)

Show Answer

Key Concepts and Summary

Several systems we encounter consist of multiple equilibria, systems where two or more equilibria processes are occurring simultaneously. Some common examples include acid rain, fluoridation, and dissolution of carbon dioxide in sea water. When looking at these systems, we need to consider each equilibrium separately and then combine the individual equilibrium constants into one solubility product or reaction quotient expression using the tools from the first equilibrium chapter. Le Châtelier’s principle also must be considered, as each reaction in a multiple equilibria system will shift toward reactants or products based on what is added to the initial reaction and how it affects each subsequent equilibrium reaction.

Exercises

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?

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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

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

Equilibria of Other Reaction Classes

Learning Objectives

Role of Fluoride in Preventing Tooth Decay

Example 1: Prevention of Precipitation of Mg(OH)₂

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Example 2: Multiple Equilibria

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Dissolution versus Weak Electrolyte Formation

Example 3: Solubility Equilibrium of a Slightly Soluble Solid

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Key Concepts and Summary

Exercises

Glossary

Candela Citations

Equilibria of Other Reaction Classes