🧪 Contents · Interactive Lab Chapter 15 · Equilibria of Other Reaction Classes
Chemistry · Chapter 15

Equilibria of Other Reaction Classes

Some salts pour into water and vanish. Others sit on the bottom and barely move. That gap runs water treatment, kidney stones, cave formation, film developing and the stomach scans done in a clinic. This chapter takes the equilibrium tools you already have and points them at solids. You will learn one constant, Ksp, that says how far a stubborn salt dissolves. Then you will push that balance around. Add a shared ion and solubility falls. Mix two solutions and a solid may drop out. Add ammonia or acid and a solid you could not budge goes clear in seconds.

By the end of this chapter you can

  • Write the Ksp expression for any slightly soluble salt, and say why the solid is left out.
  • Convert between Ksp and molar solubility for 1:1, 1:2 and 3:2 salts.
  • Predict how much a shared ion cuts the solubility of a salt, and work out the new value.
  • Compare the ion product Q with Ksp to say whether a solid will form on mixing.
  • Choose a reagent that drops one metal ion out of a mixture and leaves the other behind.
  • Use a formation constant Kf to find how much free metal ion is left after ligands are added.
  • Explain why aluminum and zinc hydroxides dissolve in both acid and strong base.
  • Add two equilibria, multiply their constants, and predict whether acid or a ligand will dissolve a solid.

Section 1The solubility product

What a saturated solution really is

Stir a spoon of table salt into water and it disappears. Stir in twenty spoons and some settles on the bottom. That leftover solid is not resting. Ions keep breaking away from its surface. Other ions keep landing back on it and locking in. When those two rates match, the solution is saturated. The amount dissolved stops changing, and the solid stops shrinking.

So a saturated solution is an equilibrium. For silver chloride it reads like this.

AgCl(s) ⇌ Ag+(aq) + Cl-(aq)

The equilibrium constant for a reaction of this kind has its own name. It is the solubility product constant, Ksp. You build it the usual way. Multiply the concentrations on the product side, each raised to its coefficient.

Ksp = [Ag+][Cl-] = 1.8 × 10-10

Look at what is missing. The solid is not in the expression. A pure solid has fixed crowding, so its activity is 1. A kilogram of AgCl and a single speck of AgCl give the same Ksp. Adding more solid cannot force more of it into solution. That one fact saves people a lot of wasted stirring.

A saturated solution, where ions leave the crystal surface at the same rate that ions return to it Saturation is a tie, not a stop undissolved AgCl ions leave ions leave ions return Ag+ Cl- Cl- Ag+ Cl- Cl- Ag+ The equilibrium AgCl(s) ⇌ Ag+(aq) + Cl-(aq) rate out = rate back in The constant Ksp = [Ag+][Cl-] = 1.8 × 10-10 No AgCl(s) term. A pure solid has activity 1. More solid in the beaker changes nothing.
Figure 15.1 In a saturated solution, ions leave the crystal face at the same rate they return. The dissolved amount holds steady. The solid itself never appears in the Ksp expression, because its activity is fixed at 1.

From Ksp to molar solubility

Molar solubility is the number of moles of a salt that dissolve in one liter of water to reach saturation. Call it s. The bridge between s and Ksp is the formula of the salt, so write the balanced equation first every time.

A 1:1 salt such as AgCl gives one cation and one anion per formula unit. Both ions sit at s, so Ksp = s2. A 1:2 salt such as PbI2 gives one Pb2+ and two I-. Now [Pb2+] = s and [I-] = 2s, so Ksp = (s)(2s)2 = 4s3. A 3:2 salt such as Ca3(PO4)2 gives ions at 3s and 2s, so Ksp = (3s)3(2s)2 = 108s5.

Three salt types and the solubility product formula each one gives: s squared, four s cubed, and one hundred eight s to the fifth The ion ratio writes the formula for you 1 : 1 AgCl, BaSO4 M+ X- one of each per unit [M+] = s   [X-] = s Ksp = s2 K = 1.8 × 10-10 s = 1.3 × 10-5 M 1 : 2 PbI2, CaF2 M2+ X- X- two anions per cation [M2+] = s   [X-] = 2s Ksp = (s)(2s)2 = 4s3 K = 1.4 × 10-8 s = 1.5 × 10-3 M 3 : 2 Ca3(PO4)2 M2+ M2+ M2+ X3- X3- [M2+] = 3s   [X3-] = 2s Ksp = (3s)3(2s)2 = 108s5 K = 1.3 × 10-32 s = 1.6 × 10-7 M
Figure 15.2 Three salt shapes, three formulas. The number in front of s comes from the subscripts in the salt. Lose that number and the answer is off by a lot.

Worked example · Ksp to solubility for a 1:2 salt

Lead(II) iodide has Ksp = 1.4 × 10-8 at 25 °C. Find its molar solubility, then its solubility in grams per liter. The molar mass is 461.0 g/mol.

Step 1 · balance the dissolvingPbI2(s) ⇌ Pb2+(aq) + 2 I-(aq)

Step 2 · label the ions with s[Pb2+] = s and [I-] = 2s.

Step 3 · build KspKsp = (s)(2s)2 = 4s3 = 1.4 × 10-8

Step 4 · solve for ss3 = 1.4 × 10-8 / 4 = 3.5 × 10-9, so s = (3.5 × 10-9)1/3 = 1.5 × 10-3 M

Step 5 · check it backward4(1.518 × 10-3)3 = 1.40 × 10-8. That matches.

Answer1.5 × 10-3 M. In mass terms, 1.518 × 10-3 mol/L × 461.0 g/mol = 0.70 g/L. Also note [I-] = 2s = 3.0 × 10-3 M, twice the lead.

The trip runs the other way just as well. Weigh how much salt a liter of saturated solution holds, convert to moles, then build Ksp from those numbers.

Worked example · solubility to Ksp

A liter of saturated barium sulfate holds 2.4 × 10-3 g of BaSO4 at 25 °C. The molar mass is 233.39 g/mol. Find Ksp.

Step 1 · go to moless = 2.4 × 10-3 g/L ÷ 233.39 g/mol = 1.03 × 10-5 M

Step 2 · read the ratioBaSO4(s) ⇌ Ba2+(aq) + SO42-(aq), so both ions are at 1.03 × 10-5 M.

Step 3 · multiplyKsp = (1.03 × 10-5)2 = 1.1 × 10-10

AnswerKsp = 1.1 × 10-10, which is the table value. A barium sulfate drink is safe to swallow for an X-ray because almost none of the toxic barium ever gets loose.

Why Ksp is not a ranking

Here is a trap. A smaller Ksp does not always mean a less soluble salt. Silver chloride has Ksp = 1.8 × 10-10 and dissolves to 1.3 × 10-5 M. Silver chromate has the smaller Ksp, 9.0 × 10-12, yet dissolves to 1.3 × 10-4 M. That is ten times more. The reason is the exponent. Ag2CrO4 releases three ions, so its constant is 4s3, and a cube shrinks a small number much harder than a square does. You may compare Ksp values head to head only when the salts share the same ion ratio.

Key points
  • Ksp is the equilibrium constant for a solid dissolving into its ions.
  • The solid never appears in the expression, so extra solid changes nothing.
  • Molar solubility s is moles dissolved per liter at saturation.
  • 1:1 gives s2, 1:2 gives 4s3, and 3:2 gives 108s5.
  • Compare Ksp values only for salts with the same ion ratio.
Where these numbers come from. Every Ksp and Kf in this chapter is a table value for 25 °C, matching the appendix tables in OpenStax Chemistry 2e. Tables differ in the last digit, and some list a different crystal form of the same salt. Always work from the table that is printed with the problem in front of you.
Interactive Dissolve a salt to saturation and connect Ksp with molar solubility. Open full ↑

Section 2The common ion effect

A shared ion shuts the door

So far the water was pure before the salt went in. Real water is rarely pure. Suppose you try to dissolve silver chloride in a solution that already holds 0.10 M NaCl. The chloride from the salt water counts. It is the same Cl- that AgCl would release.

An ion that two solutes share is a common ion. Le Chatelier's principle tells you what happens next. Raise the amount of a product and the equilibrium shifts back toward the reactant side. Here the reactant side is solid. So less AgCl dissolves. This drop is the common ion effect.

Silver chloride dissolving in pure water compared with 0.10 molar sodium chloride, where the shared chloride ion cuts solubility by about seven thousand times A shared ion pushes the balance back to solid Ag+ Cl- Ag+ Cl- pure water s = 1.3 × 10-5 M Cl- Cl- Cl- Cl- Cl- Cl- 0.10 M NaCl s = 1.8 × 10-9 M add Cl- more solid stays undissolved Solubility on a log scale pure water, 1.3 × 10-5 M 0.10 M NaCl, 1.8 × 10-9 M each gridline is one power of ten Ksp never moved. The split did. (s)(0.10) = 1.8 × 10-10
Figure 15.3 The same salt, two different waters. Chloride already present shoves the dissolving equilibrium back toward the solid. The bar scale is logarithmic, so each step is a factor of ten.

Nothing about Ksp changes. It is a constant at a fixed temperature. What changes is how the two concentrations split the product. If one ion is already large, the other must be tiny to keep the product equal to Ksp.

Worked example · silver chloride in salt water

Find the molar solubility of AgCl in 0.10 M NaCl. Ksp = 1.8 × 10-10.

Step 1 · set up the tableLet s be the moles of AgCl that dissolve per liter. Then [Ag+] = s and [Cl-] = 0.10 + s.

Step 2 · write KspKsp = (s)(0.10 + s) = 1.8 × 10-10

Step 3 · drop the tiny termIn pure water s was only 1.3 × 10-5 M, and here it will be smaller still. So 0.10 + s ≈ 0.10.

Step 4 · solves = 1.8 × 10-10 / 0.10 = 1.8 × 10-9 M

Step 5 · test the shortcutIs 1.8 × 10-9 small next to 0.10? It is about 2 parts in 100 million. The shortcut was fair.

Answer1.8 × 10-9 M. In pure water it was 1.3 × 10-5 M, so the salt water cut solubility by a factor of about 7,500.

The effect gets stronger with more ions

When the shared ion carries an exponent, the squeeze is harder. Lead(II) iodide in 0.10 M KI is a good case. Iodide enters Ksp squared, so the added iodide gets squared too.

Worked example · lead iodide in potassium iodide

Find the molar solubility of PbI2 in 0.10 M KI. Ksp = 1.4 × 10-8.

Step 1 · label the ions[Pb2+] = s and [I-] = 0.10 + 2s ≈ 0.10

Step 2 · write and solveKsp = (s)(0.10)2 = 0.010 s = 1.4 × 10-8, so s = 1.4 × 10-6 M

Step 3 · check the shortcut2s = 2.8 × 10-6 M, far below 0.10 M. Good.

Answer1.4 × 10-6 M, down from 1.5 × 10-3 M in pure water. That is a drop of about 1,100 times.

Note that the solubility is now set almost entirely by the added ion. The salt supplies so little of its own that you can treat the added amount as fixed. That shortcut fails only when the added ion is very dilute, close to the salt's own solubility. Then you must keep the full expression and solve it properly.

Two ways people use this on purpose

Chemists lean on the common ion effect to finish a job. After a precipitate forms, you wash it with a solution holding one of its own ions instead of plain water. Pure water would redissolve some of your product. A wash made of dilute sodium chloride loses almost none of a silver chloride sample.

The same trick controls teeth. Tooth enamel is mostly hydroxyapatite, which slowly dissolves in the mouth. Fluoride in toothpaste swaps in to build fluorapatite, a salt with a far smaller Ksp. Fluoride in the water also acts as a common ion for that solid. Both effects push the enamel equilibrium back toward solid.

One warning. Adding a salt with no shared ion does not help in the same way. It can even raise solubility a little, because a crowd of ions screens the charges and makes each ion act slightly less concentrated than it is. That is the salt effect, and it is small. Ignore it unless a problem asks for it.

Key points
  • A common ion is one that the salt and some other solute both supply.
  • Adding a common ion shifts the dissolving equilibrium back toward solid, so solubility drops.
  • Ksp does not change. Only the split between the two concentrations changes.
  • If the shared ion is squared in Ksp, the squeeze is far stronger.
  • When the added ion is much larger than s, treat it as fixed and skip the small term.
Interactive Add a shared ion and watch solubility drop. Open full ↑

Section 3Predicting precipitation

Q tells you which way the solid will go

Pour two clear solutions together and sometimes a cloud appears. A precipitate is a solid that forms inside a solution and settles out. You can predict it before you pour. Build the same product that Ksp uses, but with the concentrations you actually have. That number is the ion product, Q.

Q has the same form as Ksp. The only difference is that Q uses the moment's concentrations, which may be far from equilibrium. Then you compare.

ComparisonWhat the solution isWhat happens
Q < KspUnsaturatedNo solid forms. More salt could still dissolve.
Q = KspSaturatedAt equilibrium. Solid and ions hold steady together.
Q > KspSupersaturatedSolid forms until Q falls back to Ksp.
A number line comparing the ion product Q with the solubility product, showing unsaturated, saturated and supersaturated zones One comparison answers the question Q = Ksp Q < Ksp Q > Ksp unsaturated clear solution, no solid more salt could dissolve saturated at equilibrium supersaturated solid forms and settles out until Q falls back to Ksp Q slides left as solid drops out Example: mix 25.0 mL of 0.0020 M Pb(NO3)2 with 75.0 mL of 0.010 M KI → Q = 2.8 × 10-8, Ksp = 1.4 × 10-8
Figure 15.4 One number line decides the question. Ksp marks the saturation point. Any Q to the right of it means a solid comes out until Q slides back down to Ksp.

There is one step people skip. Mixing two solutions dilutes both of them. Before you build Q, find the new concentration of each ion in the combined volume. Use moles divided by total liters, or the shortcut of multiplying by the volume fraction.

Worked example · will lead iodide drop out?

You mix 25.0 mL of 0.0020 M Pb(NO3)2 with 75.0 mL of 0.010 M KI. Does PbI2 form? Ksp = 1.4 × 10-8.

Step 1 · total volume25.0 mL + 75.0 mL = 100.0 mL

Step 2 · dilute each ion[Pb2+] = 0.0020 M × (25.0 / 100.0) = 5.0 × 10-4 M
[I-] = 0.010 M × (75.0 / 100.0) = 7.5 × 10-3 M

Step 3 · build QQ = [Pb2+][I-]2 = (5.0 × 10-4)(7.5 × 10-3)2

Step 4 · do the arithmetic(7.5 × 10-3)2 = 5.625 × 10-5. Then (5.0 × 10-4)(5.625 × 10-5) = 2.8 × 10-8

Step 5 · compareQ = 2.8 × 10-8 and Ksp = 1.4 × 10-8. Q is about 2.0 times Ksp.

AnswerQ > Ksp, so yellow PbI2 precipitates. The margin is thin, so only a little solid appears before Q settles back to 1.4 × 10-8.

Pulling one ion out and leaving the other

Now suppose a solution holds two ions that both form solids with the same reagent. Their Ksp values are not equal, so they do not come out together. Add the reagent slowly and the less soluble salt forms first. Separating ions this way is selective precipitation. It is the backbone of the old qualitative analysis schemes, and it still drives metal recovery today.

The method is to find the reagent concentration each salt needs to just begin forming. Rearrange Ksp to solve for the reagent. Whichever salt needs less reagent starts first.

Worked example · iodide out, chloride left behind

A solution is 0.10 M in Cl- and 0.010 M in I-. You add silver nitrate drop by drop. Which solid forms first, and how much iodide is left when the second one starts? Ksp(AgCl) = 1.8 × 10-10, Ksp(AgI) = 8.3 × 10-17.

Step 1 · silver needed for AgI[Ag+] = Ksp / [I-] = 8.3 × 10-17 / 0.010 = 8.3 × 10-15 M

Step 2 · silver needed for AgCl[Ag+] = 1.8 × 10-10 / 0.10 = 1.8 × 10-9 M

Step 3 · rank themAgI needs far less silver, so pale yellow AgI forms first. White AgCl waits.

Step 4 · iodide left at the crossoverWhen [Ag+] reaches 1.8 × 10-9 M, [I-] = 8.3 × 10-17 / 1.8 × 10-9 = 4.6 × 10-8 M

AnswerAgI first. Only 4.6 × 10-8 M iodide remains, which is 4.6 × 10-6 of the original 0.010 M. Over 99.999% of the iodide is out before any chloride joins it. A clean split.

A separation works when the two Ksp values are far apart. Here they differ by seven powers of ten, so the gap is huge. Barium and calcium sulfates differ by about five powers of ten, which is still workable. Two salts with nearly equal Ksp values cannot be split this way.

Key points
  • Q has the same form as Ksp but uses the concentrations you have right now.
  • Q < Ksp means no solid. Q > Ksp means a solid forms.
  • Always dilute each ion for the combined volume before building Q.
  • Selective precipitation drops the salt with the smaller Ksp first.
  • Find the reagent amount each salt needs to start. The smaller amount goes first.
Worth knowing. Q > Ksp says a solid should form, not that it will form right away. A supersaturated solution can sit clear for hours with nothing to nucleate on. Scratch the glass or drop in one seed crystal and the solid appears at once.
Interactive Mix two solutions and compare Q with Ksp. Open full ↑

Section 4Complex ion formation

Ligands wrap a metal ion

Pour ammonia into a solution of silver nitrate and a white cloud of AgOH forms. Keep pouring and the cloud clears. Something took the silver and held it in solution. That something is a complex.

A complex ion is a central metal ion bonded to several small molecules or ions around it. Each attached partner is a ligand. A ligand always has a lone pair to donate, so it acts as a Lewis base. The metal ion accepts that pair, so it acts as a Lewis acid. Water itself is a ligand, which is why a dissolved metal ion is really an aquo complex like [Al(H2O)6]3+.

The number of donor atoms attached is the coordination number. Silver likes 2 and is linear. Zinc and copper often take 4. Many ions of the first transition row take 6 in an octahedral arrangement. Common ligands include NH3, CN-, OH-, Cl-, S2O32- and H2O.

Ammonia takes a silver ion into a linear complex, which pulls the silver chloride dissolving equilibrium further to the right A ligand steals the metal ion, so the solid keeps feeding it AgCl(s) Ksp = 1.8 × 10-10 white, stubborn in water tiny Ag+ Cl- + 2 NH3 Ag NH3 NH3 linear, coordination number 2 [Ag(NH3)2]+ Kf = 1.7 × 107 free Ag+ drops to 10-11 M, so more solid dissolves Add the two steps and multiply the two constants AgCl(s) + 2 NH3 ⇌ [Ag(NH3)2]+ + Cl-    K = KspKf = 3.1 × 10-3
Figure 15.5 Two ammonia molecules replace the water around a silver ion and form a linear complex. Because the free Ag+ is swept up, the AgCl equilibrium above it is pulled to the right and more solid dissolves.

Kf measures the grip

Complex formation is an equilibrium like any other. Its constant is the formation constant, Kf. It is sometimes called a stability constant.

Ag+(aq) + 2 NH3(aq) ⇌ [Ag(NH3)2]+(aq)   Kf = 1.7 × 107 Kf = [Ag(NH3)2+] / ([Ag+][NH3]2)

Formation constants are large, often huge. Kf for [Ag(CN)2]- is about 1 × 1021. A big Kf means the complex wins. Nearly every metal ion ends up inside one, and the free ion concentration drops to almost nothing. The reverse reaction has the flipped constant, Kd = 1/Kf, and is called the dissociation constant.

Worked example · how little free silver is left

You make a solution that is 1.0 × 10-3 M in total silver and 1.00 M in NH3. Find the free [Ag+]. Kf = 1.7 × 107.

Step 1 · assume the complex winsKf is enormous and ammonia is in large excess. So treat almost all the silver as complexed: [Ag(NH3)2+] = 1.0 × 10-3 M.

Step 2 · count the ammonia usedEach complex uses 2 NH3, so 2.0 × 10-3 M is tied up. Free [NH3] = 1.00 - 0.0020 = 0.998 M.

Step 3 · rearrange Kf[Ag+] = [Ag(NH3)2+] / (Kf[NH3]2)

Step 4 · plug in[Ag+] = 1.0 × 10-3 / (1.7 × 107 × 0.996) = 5.9 × 10-11 M

Answer5.9 × 10-11 M. Of every ten million silver ions added, fewer than one is still swimming free. That is why ammonia keeps silver in solution.

Complexes dissolve solids that water cannot

Now put the two ideas together. Solid AgCl sets up a tiny supply of Ag+. Ammonia pulls that Ag+ into a complex. The solid responds by releasing more, and the dissolving keeps going. Adding the two equations gives one overall reaction, and multiplying the two constants gives its constant.

AgCl(s) + 2 NH3(aq) ⇌ [Ag(NH3)2]+(aq) + Cl-(aq) K = Ksp × Kf = (1.8 × 10-10)(1.7 × 107) = 3.1 × 10-3

Worked example · silver chloride in 1.0 M ammonia

How many moles of AgCl dissolve in one liter of 1.0 M NH3? Use K = 3.06 × 10-3.

Step 1 · set upLet s be the moles dissolved. Then the complex and the chloride are each at s, and [NH3] = 1.0 - 2s.

Step 2 · write KK = s2 / (1.0 - 2s)2 = 3.06 × 10-3

Step 3 · take the square root of both sidess / (1.0 - 2s) = (3.06 × 10-3)1/2 = 0.0553

Step 4 · solve the lines = 0.0553 - 0.1106s, so 1.1106s = 0.0553 and s = 0.0498 M

Step 5 · check[NH3] = 1.0 - 0.0996 = 0.900. Then (0.0498)2 / (0.900)2 = 3.06 × 10-3. It holds.

Answer0.050 M, against 1.3 × 10-5 M in pure water. Ammonia raised the solubility about 3,700 times. Taking the square root worked here only because both sides were perfect squares.

Key points
  • A complex ion is a metal ion surrounded by ligands that donate lone pairs.
  • Kf is the formation constant, and it is usually very large.
  • A large Kf drives free metal ion concentration down to almost zero.
  • Coordination number is how many donor atoms attach. Silver takes 2, zinc 4, and iron often 6.
  • Add a dissolving step to a complexing step and multiply Ksp by Kf.
Interactive Add ligands and watch a complex ion form. Open full ↑

Section 5Amphoteric hydroxides

Some hydroxides go both ways

Add a little base to a solution of Al3+ and a white gel of Al(OH)3 drops out. Split that gel between two test tubes. Pour acid into one and the gel dissolves. Pour concentrated NaOH into the other and that gel dissolves too. A solid that reacts as a base with acid and as an acid with base is amphoteric.

The two reactions are different in kind, which is the whole point. With acid, the hydroxide ions in the solid get taken by H3O+, and the metal leaves as a simple cation.

Al(OH)3(s) + 3 H3O+(aq) → Al3+(aq) + 6 H2O(l)

With extra base, the metal takes on one more hydroxide as a ligand. The product is a soluble hydroxo complex.

Al(OH)3(s) + OH-(aq) ⇌ [Al(OH)4]-(aq)

Zinc behaves the same way, with a coordination number of 4 on the acid side and 4 hydroxides on the base side. Beryllium, chromium(III), tin(II) and lead(II) hydroxides are amphoteric as well. Magnesium and iron(III) hydroxides are not. They dissolve in acid and then just sit there in base, no matter how much you add.

The V-shaped solubility curve

Plot the total dissolved metal against pH and an amphoteric hydroxide traces a V. On the left, acid eats the hydroxide and the free cation rules. Each pH unit up cuts solubility hard, because [OH-] rises tenfold and enters Ksp raised to a power. On the right, the hydroxo complex takes over and solubility climbs again. The bottom of the V is the pH where the solid is hardest to dissolve.

Total dissolved zinc against pH for zinc hydroxide, a V shaped curve with free zinc ion on the acid side and the tetrahydroxozincate complex on the base side Zn(OH)₂ solubility against pH: a V with a floor near pH 10.1 10⁻¹ 10⁻³ 10⁻⁵ 10⁻⁷ 10⁻⁹ 2 4 6 8 10 12 14 pH total dissolved Zn (M) floor: 4.8 × 10⁻⁹ M at pH 10.1 acid side [Zn2+] = Ksp / [OH-]2 solid is a base here base side [Zn(OH)42-] = 0.126[OH-]2 solid is an acid here
Figure 15.6 Total dissolved zinc against pH for Zn(OH)2. The falling branch is free Zn2+. The rising branch is [Zn(OH)4]2-. The floor sits near pH 10.1, where only about 5 × 10-9 M zinc stays in solution.

Both branches come from constants you already know how to use. The left branch is plain Ksp. For Zn(OH)2, Ksp = 4.5 × 10-17, so [Zn2+] = Ksp / [OH-]2. The right branch is a coupled reaction, Ksp times Kf for the hydroxo complex.

Zn(OH)2(s) + 2 OH- ⇌ [Zn(OH)4]2-   K = KspKf

Worked example · zinc hydroxide in 0.10 M NaOH

Ksp(Zn(OH)2) = 4.5 × 10-17 and Kf([Zn(OH)4]2-) = 2.8 × 1015. Find the dissolved zinc in 0.10 M NaOH, and compare it with pure water.

Step 1 · combine the stepsZn(OH)2(s) + 2 OH- ⇌ [Zn(OH)4]2-, so K = (4.5 × 10-17)(2.8 × 1015) = 0.126

Step 2 · write the expressionK = [Zn(OH)42-] / [OH-]2 = 0.126

Step 3 · put in 0.10 M OH-[Zn(OH)42-] = 0.126 × (0.10)2 = 0.126 × 0.010 = 1.3 × 10-3 M

Step 4 · get the pure-water valueZn(OH)2 is a 1:2 salt, so Ksp = 4s3. Then s3 = 1.125 × 10-17 and s = 2.2 × 10-6 M.

Answer1.3 × 10-3 M in 0.10 M NaOH, against 2.2 × 10-6 M in water. Strong base raised zinc solubility about 560 times. Note that the hydroxide here is both a common ion and a ligand, and the ligand role wins.

Dilute base and concentrated base do opposite things to the same solid. A trace of OH- is a common ion, so it drives zinc out of solution. A flood of OH- is a ligand, so it pulls zinc back in. That is the V in one sentence.

Why this matters in a water plant

Water treatment removes metals by raising the pH until hydroxides fall out. The operator must stop at the floor of the V. Go past it and amphoteric metals such as zinc, aluminum and lead start redissolving as hydroxo complexes. For Zn(OH)2 the floor sits near pH 10.1, and for Al(OH)3 it sits near pH 6. A plant that overshoots sends metal back out in the finished water.

Key points
  • Amphoteric hydroxides dissolve in acid and in excess strong base.
  • Acid gives the free cation. Excess base gives a hydroxo complex such as [Al(OH)4]-.
  • Al, Zn, Cr(III), Be, Sn(II) and Pb(II) hydroxides are amphoteric. Mg and Fe(III) are not.
  • Solubility against pH makes a V, and the floor is the best pH for removing the metal.
  • A little OH- is a common ion. A lot of OH- is a ligand.
Interactive Add acid or base to a hydroxide and watch it dissolve either way. Open full ↑

Section 6Coupled equilibria

Two reactions, one constant

You have now seen the trick twice. A solid that will not dissolve gets a partner reaction that eats one of its ions. Then the solid has to keep feeding that reaction, so it dissolves. Two equilibria that share a species like this are coupled equilibria.

The bookkeeping is simple and it never changes. Add the two balanced equations and cancel the shared species. Multiply the two constants to get the constant of the sum.

If A ⇌ B has K1 and B ⇌ C has K2, then A ⇌ C has K = K1K2
Two coupled equilibria that share one ion, with the shared ion cancelling when the equations are added and the constants multiplied Add the equations, multiply the constants step 1 · the solid dissolves CaCO₃(s) ⇌ Ca2+ + CO₃2- Ksp = 8.7 × 10-9 step 2 · acid eats the anion CO₃2- + H₃O+ ⇌ HCO₃- + H₂O 1 / Ka2 = 2.1 × 1010 the shared ion CO₃2- step 1 makes it step 2 destroys it the sum, with CO₃2- cancelled CaCO₃(s) + H₃O+ ⇌ Ca2+ + HCO₃- + H₂O K = Ksp × (1 / Ka2) = (8.7 × 10-9)(2.1 × 1010) = 1.9 × 102
Figure 15.7 The shared ion is the link. One equilibrium makes it and the other destroys it. Adding the equations cancels the link, and the overall constant is the product of the two.

Watch the direction. If you have to flip an equation before adding, flip its constant to 1/K as well. If you double an equation, square its constant. These are the same rules you used for gas-phase equilibria.

Lever one: acid, for salts of weak-acid anions

Many stubborn salts carry an anion that is the conjugate base of a weak acid. Carbonate, phosphate, sulfide, fluoride, oxalate and hydroxide all qualify. Add acid and that anion gets protonated and leaves the equilibrium. Limestone is the classic case.

Worked example · acid on limestone

Find K for CaCO3(s) + H3O+ ⇌ Ca2+ + HCO3- + H2O. Use Ksp(CaCO3) = 8.7 × 10-9 and Ka2 for carbonic acid = 4.7 × 10-11.

Step 1 · the dissolving stepCaCO3(s) ⇌ Ca2+ + CO32-, Ksp = 8.7 × 10-9

Step 2 · the acid step, flippedHCO3- + H2O ⇌ CO32- + H3O+ has Ka2 = 4.7 × 10-11. Reverse it and the constant becomes 1/Ka2 = 2.13 × 1010.

Step 3 · add and cancelCO32- appears as a product in step 1 and a reactant in step 2, so it cancels. The sum is the reaction asked for.

Step 4 · multiplyK = Ksp / Ka2 = (8.7 × 10-9)(2.13 × 1010) = 1.9 × 102

AnswerK = 1.9 × 102. That is far above 1, so even dilute acid dissolves limestone easily. Rainwater carrying dissolved CO2 is weak acid, and over long spans it carves caves out of solid rock.

Push the pH lower and the bicarbonate turns into H2CO3, which falls apart into CO2 gas. The gas bubbles away and leaves the solution for good. Now a third process is pulling, and the limestone has no way to come back. That fizz is the test every geology field kit relies on.

Acid does nothing useful for a salt whose anion is a very weak base. Chloride, bromide, iodide and nitrate hold almost no protons at ordinary pH. So AgCl shrugs off nitric acid, while CaCO3 and Mg(OH)2 both give way at once.

Lever two: a ligand, for the metal side

If acid cannot touch the anion, go after the cation instead. A ligand with a large Kf removes the metal ion and the solid follows. This is how film and prints were fixed for a century. Sodium thiosulfate dissolves the unexposed silver bromide and leaves the image behind.

Worked example · thiosulfate on silver bromide

Ksp(AgBr) = 5.0 × 10-13 and Kf([Ag(S2O3)2]3-) = 2.9 × 1013. Find the molar solubility of AgBr in 1.0 M Na2S2O3.

Step 1 · add the two stepsAgBr(s) + 2 S2O32- ⇌ [Ag(S2O3)2]3- + Br-

Step 2 · multiply the constantsK = (5.0 × 10-13)(2.9 × 1013) = 14.5

Step 3 · set up with sK = s2 / (1.0 - 2s)2 = 14.5

Step 4 · square root and solves / (1.0 - 2s) = 3.808, so s = 3.808 - 7.616s. Then 8.616s = 3.808 and s = 0.442 M

Step 5 · check[S2O32-] = 1.0 - 0.884 = 0.116, and (0.442)2 / (0.116)2 = 14.5. It holds.

Answer0.44 M. In pure water AgBr reaches only 7.1 × 10-7 M, so the fixer raised solubility by a factor of about 600,000. Note that s is a large fraction of the thiosulfate, so you could not drop the 2s term here.

One caution. These overall constants are only as good as their parts. A K built from two table values carries the error of both. Treat an answer like 0.44 M as a sound estimate, not a measurement.

Key points
  • Coupled equilibria share a species. Add the equations and multiply the constants.
  • Flip an equation and its constant flips to 1/K. Double it and the constant is squared.
  • Acid dissolves salts of weak-acid anions such as carbonate, phosphate, sulfide and hydroxide.
  • Acid will not help with chloride, bromide, iodide or nitrate salts.
  • A ligand with a large Kf dissolves a solid by stripping out the metal ion.
  • Escaping gas such as CO2 adds a third pull that the system cannot undo.
Interactive Link two equilibria and see one pull the other. Open full ↑

The chapter in one breath

  • Ksp is the equilibrium constant for a solid dissolving, and the solid is never in it.
  • Molar solubility and Ksp convert through the salt's ion ratio: s2, 4s3 or 108s5.
  • Compare two Ksp values only when the salts have the same ion ratio.
  • A common ion pushes the equilibrium back to solid, so solubility falls, sometimes by thousands of times.
  • Build Q from the diluted concentrations and compare it with Ksp to predict a precipitate.
  • Selective precipitation drops the smaller-Ksp salt first and can split ions almost completely.
  • A ligand wraps a metal ion into a complex, and a large Kf leaves almost no free metal.
  • Amphoteric hydroxides dissolve in acid and in excess base, giving a V-shaped solubility curve.
  • Couple two equilibria by adding the equations and multiplying the constants, then read the overall K.

ReferenceWords worth knowing

These are the terms this chapter leans on. Say each one out loud and you will find the idea comes with it.

Amphoteric
Able to react as a base with acid and as an acid with base. Al(OH)3 and Zn(OH)2 both dissolve either way.
Common ion
An ion supplied by two different solutes in the same solution, such as Cl- from both AgCl and NaCl.
Common ion effect
The drop in solubility caused by adding an ion the salt already releases.
Complex ion
A central metal ion joined to several ligands, carrying an overall charge, such as [Ag(NH3)2]+.
Coordination number
How many donor atoms are bonded to the central metal ion. Silver usually takes 2 and zinc 4.
Coupled equilibria
Two equilibria that share a species, so shifting one shifts the other.
Dissociation constant (Kd)
The constant for a complex falling apart. It is 1/Kf.
Formation constant (Kf)
The equilibrium constant for a metal ion plus ligands forming a complex ion. Also called a stability constant.
Hydroxo complex
A complex in which the ligands are hydroxide ions, such as [Al(OH)4]-.
Ion product (Q)
The same product of ion concentrations as Ksp, but built from the amounts present right now.
Ligand
A molecule or ion that donates a lone pair to a metal ion. NH3, CN- and H2O are common ones.
Molar solubility
Moles of a solid that dissolve in one liter of solution to reach saturation. Its symbol here is s.
Precipitate
A solid that forms inside a solution and separates from it.
Qualitative analysis
A scheme that identifies the ions in an unknown by dropping them out in planned groups.
Salt effect
A small rise in solubility caused by unrelated ions screening charges in the solution.
Saturated solution
A solution holding all the solute it can at that temperature, with Q equal to Ksp.
Selective precipitation
Adding a reagent slowly so the least soluble salt forms first and one ion is separated from another.
Solubility product constant (Ksp)
The equilibrium constant for a slightly soluble solid breaking into its ions in water.
Supersaturated solution
A solution holding more dissolved solute than equilibrium allows, with Q above Ksp.
Unsaturated solution
A solution that could still dissolve more solute, with Q below Ksp.

Check yourself

1. Silver bromide has Ksp = 5.0 × 10-13. What is its molar solubility in pure water?
7.1 × 10-7 M. AgBr is 1:1, so Ksp = s2. Then s = (5.0 × 10-13)1/2 = 7.07 × 10-7 M. Check: (7.07 × 10-7)2 = 5.0 × 10-13.
2. Calcium fluoride has Ksp = 3.9 × 10-11. Find its molar solubility and the fluoride concentration at saturation.
s = 2.1 × 10-4 M and [F-] = 4.3 × 10-4 M. CaF2 is 1:2, so Ksp = 4s3. Then s3 = 9.75 × 10-12 and s = 2.14 × 10-4 M. Fluoride is 2s.
3. Why does the amount of undissolved solid never appear in a Ksp expression?
Because a pure solid has a fixed activity of 1. Its particles are packed at one unchanging density, so piling on more solid cannot raise anything in the expression. The surface area changes how fast equilibrium is reached, not where it lands.
4. Find the molar solubility of AgCl in 0.050 M NaCl. Ksp = 1.8 × 10-10.
3.6 × 10-9 M. [Cl-] ≈ 0.050 M, so s = Ksp / 0.050 = 1.8 × 10-10 / 0.050 = 3.6 × 10-9 M. That is roughly 3,700 times less than the 1.3 × 10-5 M it reaches in pure water.
5. Silver chromate has a smaller Ksp than silver chloride, yet it dissolves ten times more. How can both be true?
The two salts have different ion ratios. AgCl uses Ksp = s2 while Ag2CrO4 uses 4s3. A cube root of a tiny number is much larger than a square root of one, so you cannot rank them by Ksp alone.
6. You mix equal volumes of 2.0 × 10-3 M BaCl2 and 2.0 × 10-3 M Na2SO4. Does BaSO4 form? Ksp = 1.1 × 10-10.
Yes, easily. Equal volumes halve each one, so [Ba2+] = [SO42-] = 1.0 × 10-3 M. Then Q = (1.0 × 10-3)(1.0 × 10-3) = 1.0 × 10-6. That is about 9,100 times Ksp, so solid drops out.
7. A solution is 0.010 M in Ba2+ and 0.010 M in Ca2+. You add sulfate slowly. Which precipitates first, and what sulfate concentration starts it? Ksp(BaSO4) = 1.1 × 10-10, Ksp(CaSO4) = 2.4 × 10-5.
BaSO4 first, starting at 1.1 × 10-8 M sulfate. Divide each Ksp by 0.010. Barium needs 1.1 × 10-8 M and calcium needs 2.4 × 10-3 M, which is 220,000 times more. When calcium finally starts, [Ba2+] is down to 1.1 × 10-10 / 2.4 × 10-3 = 4.6 × 10-8 M.
8. Name the ligand and the coordination number in [Cu(NH3)4]2+, and give the charge on the central ion.
Ammonia is the ligand, the coordination number is 4, and copper is 2+. NH3 is neutral, so the whole 2+ charge sits on the copper. Each ammonia donates the lone pair on its nitrogen atom.
9. Work out K for AgCl(s) + 2 NH3 ⇌ [Ag(NH3)2]+ + Cl-. Use Ksp = 1.8 × 10-10 and Kf = 1.7 × 107.
K = 3.1 × 10-3. Add the dissolving step to the complexing step and multiply: (1.8 × 10-10)(1.7 × 107) = 3.06 × 10-3. K is still below 1, but with 1.0 M ammonia it is enough to dissolve about 0.050 mol per liter.
10. You have four fresh precipitates: Mg(OH)2, Al(OH)3, Fe(OH)3 and Zn(OH)2. Which dissolve when you add excess 6 M NaOH?
Al(OH)3 and Zn(OH)2. Both are amphoteric, so they pick up extra hydroxide and leave as [Al(OH)4]- and [Zn(OH)4]2-. Mg(OH)2 and Fe(OH)3 are not amphoteric and stay solid in any amount of base.
11. A buffer holds a solution at pH 10.00. How much Mg2+ can stay dissolved in contact with Mg(OH)2? Ksp = 2.1 × 10-13.
2.1 × 10-5 M. At pH 10.00, pOH = 4.00, so [OH-] = 1.0 × 10-4 M. Then [Mg2+] = Ksp / [OH-]2 = 2.1 × 10-13 / 1.0 × 10-8 = 2.1 × 10-5 M. Raise the pH two more units and it falls another factor of 10,000.
12. Nitric acid dissolves CaCO3 at once but leaves AgCl untouched. Explain the difference, and give K for the limestone reaction. Ksp(CaCO3) = 8.7 × 10-9, Ka2 = 4.7 × 10-11.
Carbonate is a strong enough base to grab protons. Chloride is not. K = 1.9 × 102. Dividing gives K = 8.7 × 10-9 / 4.7 × 10-11 = 185. Acid removes CO32- as HCO3-, then as CO2 gas, so the solid keeps feeding it. Cl- is the conjugate base of a strong acid, so H3O+ cannot take it away.