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7.11 - Introduction to Solubility Equilibria

Overview

When you stir an ionic solid into water, it doesn't simply dissolve or not dissolve. It dissolves until the solution is saturated, and from then on, dissolving and precipitating happen at the same rate. That's an equilibrium, and its equilibrium constant is the solubility-product constant, KspK_{sp}. In this topic you'll write KspK_{sp} expressions, calculate solubility from KspK_{sp} and KspK_{sp} from solubility, and use KspK_{sp} to compare salts.

Dissolving is a reversible process

Put excess lead(II) iodide in water. Pb2+\mathrm{Pb^{2+}} and I−\mathrm{I^{-}} ions leave the surface of the solid and spread through the water. As their concentrations build up, dissolved ions hit the solid and stick back onto it more and more often. Eventually the two rates are equal:
PbI2(s)⇌Pb2+(aq)+2I−(aq)\mathrm{PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^{-}(aq)}
7.11 - Introduction to Solubility Equilibria
This is a saturated solution: it holds as much dissolved salt as it can at that temperature, with some solid left over. Ions keep dissolving and precipitating, but their concentrations stay constant. The dissolved ions are in a 1 : 2 ratio, the same ratio as in the formula PbI2\mathrm{PbI_2}.
The equilibrium constant for dissolving a salt is written like any other K. The solid is left out (pure solids never appear in Q or K), so only the ions remain, each raised to its coefficient:
Ksp=[Pb2+][I−]2K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{I^{-}}]^2
KspK_{sp} describes the extent of dissolving: the larger KspK_{sp} is, the more of the salt dissolves before equilibrium is reached, as long as you compare salts with the same formula type (see below).
Salt
Dissolution equation
KspK_{sp} expression
AgCl\mathrm{AgCl}
AgCl(s)⇌Ag+(aq)+Cl−(aq)\mathrm{AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)}
[Ag+][Cl−][\mathrm{Ag^{+}}][\mathrm{Cl^{-}}]
CaF2\mathrm{CaF_2}
CaF2(s)⇌Ca2+(aq)+2F−(aq)\mathrm{CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^{-}(aq)}
[Ca2+][F−]2[\mathrm{Ca^{2+}}][\mathrm{F^{-}}]^2
Ag2CrO4\mathrm{Ag_2CrO_4}
Ag2CrO4(s)⇌2Ag+(aq)+CrO42−(aq)\mathrm{Ag_2CrO_4(s) \rightleftharpoons 2Ag^{+}(aq) + CrO_4^{2-}(aq)}
[Ag+]2[CrO42−][\mathrm{Ag^{+}}]^2[\mathrm{CrO_4^{2-}}]
Fe(OH)3\mathrm{Fe(OH)_3}
Fe(OH)3(s)⇌Fe3+(aq)+3OH−(aq)\mathrm{Fe(OH)_3(s) \rightleftharpoons Fe^{3+}(aq) + 3OH^{-}(aq)}
[Fe3+][OH−]3[\mathrm{Fe^{3+}}][\mathrm{OH^{-}}]^3
Ca3(PO4)2\mathrm{Ca_3(PO_4)_2}
Ca3(PO4)2(s)⇌3Ca2+(aq)+2PO43−(aq)\mathrm{Ca_3(PO_4)_2(s) \rightleftharpoons 3Ca^{2+}(aq) + 2PO_4^{3-}(aq)}
[Ca2+]3[PO43−]2[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2
Polyatomic ions stay together when a salt dissolves: chromate stays CrO42−\mathrm{CrO_4^{2-}} and phosphate stays PO43−\mathrm{PO_4^{3-}}.

Molar solubility and KspK_{sp}

Molar solubility, ss, is the number of moles of a salt that dissolve per liter to make a saturated solution, in mol/L (M). Each mole that dissolves releases ions in the formula ratio, so every ion concentration can be written in terms of ss. For PbI2\mathrm{PbI_2}, [Pb2+][\mathrm{Pb^{2+}}] = ss and [I−][\mathrm{I^{-}}] = 2s2s:
Ksp=[Pb2+][I−]2=(s)(2s)2=4s3K_{sp} = [\mathrm{Pb^{2+}}][\mathrm{I^{-}}]^2 = (s)(2s)^2 = 4s^3
Formula type
Example
Ion concentrations
KspK_{sp} in terms of ss
AB
AgCl\mathrm{AgCl}
ss and ss
s2s^2
AB2\mathrm{AB_2} or A2B\mathrm{A_2B}
PbI2\mathrm{PbI_2}, Ag2CrO4\mathrm{Ag_2CrO_4}
ss and 2s2s
4s34s^3
AB3\mathrm{AB_3}
Fe(OH)3\mathrm{Fe(OH)_3}
ss and 3s3s
27s427s^4
A3B2\mathrm{A_3B_2}
Ca3(PO4)2\mathrm{Ca_3(PO_4)_2}
3s3s and 2s2s
108s5108s^5
Don't memorize the last column. Write each ion concentration in terms of ss and substitute, every time. To get the solubility in grams per liter, multiply ss by the molar mass, M\mathcal{M}.
The KspK_{sp} values in these notes are 25 °C values from the CRC Handbook of Chemistry and Physics. Tables in different books differ slightly, so on the exam, always use the value you're given.

Example 1: Silver bromide, a 1 : 1 salt

KspK_{sp} for AgBr\mathrm{AgBr} is 5.4×10−135.4 \times 10^{-13} at 25 °C. Calculate its molar solubility and its solubility in grams per liter.
AgBr(s)⇌Ag+(aq)+Br−(aq)\mathrm{AgBr(s) \rightleftharpoons Ag^{+}(aq) + Br^{-}(aq)}
Step 1: Each formula unit gives one Ag+\mathrm{Ag^{+}} and one Br−\mathrm{Br^{-}}, so [Ag+][\mathrm{Ag^{+}}] = [Br−][\mathrm{Br^{-}}] = ss:
Ksp=[Ag+][Br−]=s2=5.4×10−13K_{sp} = [\mathrm{Ag^{+}}][\mathrm{Br^{-}}] = s^2 = 5.4 \times 10^{-13}
Step 2: Solve for ss:
s=5.4×10−13=7.3×10−7 Ms = \sqrt{5.4 \times 10^{-13}} = 7.3 \times 10^{-7}\ \mathrm{M}
Step 3: Convert to grams per liter with M=187.77\mathcal{M} = 187.77 g/mol:
(7.3×10−7 molL)(187.77 gmol)=1.4×10−4 g/L\left(7.3 \times 10^{-7}\ \tfrac{\mathrm{mol}}{\mathrm{L}}\right)\left(187.77\ \tfrac{\mathrm{g}}{\mathrm{mol}}\right) = 1.4 \times 10^{-4}\ \mathrm{g/L}
Answer: The molar solubility is 7.3×10−77.3 \times 10^{-7} M. Only about 0.14 mg of AgBr\mathrm{AgBr} dissolves in a liter of water.

Example 2: Lead(II) iodide, a 1 : 2 salt

KspK_{sp} for PbI2\mathrm{PbI_2} is 9.8×10−99.8 \times 10^{-9}. Find its molar solubility and the concentration of each ion in a saturated solution.
Step 1: [Pb2+][\mathrm{Pb^{2+}}] = ss and [I−][\mathrm{I^{-}}] = 2s2s:
Ksp=(s)(2s)2=4s3=9.8×10−9K_{sp} = (s)(2s)^2 = 4s^3 = 9.8 \times 10^{-9}
Step 2: Divide by 4, then take the cube root:
s=9.8×10−943=2.45×10−93=1.3×10−3 Ms = \sqrt[3]{\frac{9.8 \times 10^{-9}}{4}} = \sqrt[3]{2.45 \times 10^{-9}} = 1.3 \times 10^{-3}\ \mathrm{M}
Step 3: [Pb2+][\mathrm{Pb^{2+}}] = ss = 1.3×10−31.3 \times 10^{-3} M and [I−][\mathrm{I^{-}}] = 2s2s = 2.7×10−32.7 \times 10^{-3} M. (Double the unrounded value, 1.348×10−31.348 \times 10^{-3}, then round.)
The 2 goes inside the parentheses because [I−][\mathrm{I^{-}}] is 2s2s, and then the whole thing is squared: (2s)2=4s2(2s)^2 = 4s^2, not 2s22s^2.

Comparing the solubility of different salts

KspK_{sp} ranks salts by solubility directly only when they have the same formula type (the same number of ions per formula unit). AgCl\mathrm{AgCl}, AgBr\mathrm{AgBr}, and AgI\mathrm{AgI} are all AB salts, and their KspK_{sp} values are 1.8×10−101.8 \times 10^{-10}, 5.4×10−135.4 \times 10^{-13}, and 8.5×10−178.5 \times 10^{-17}. So the solubility order is AgCl\mathrm{AgCl} > AgBr\mathrm{AgBr} > AgI\mathrm{AgI}.
When the formula types differ, the KspK_{sp} expressions have different exponents, so calculate ss for each salt and compare those.

Example 3: AgCl\mathrm{AgCl} versus Ag2CrO4\mathrm{Ag_2CrO_4}

KspK_{sp} for AgCl\mathrm{AgCl} is 1.8×10−101.8 \times 10^{-10}, and KspK_{sp} for Ag2CrO4\mathrm{Ag_2CrO_4} is 1.1×10−121.1 \times 10^{-12}. Which salt has the greater molar solubility in water?
AgCl:s2=1.8×10−10⇒s=1.3×10−5 M\mathrm{AgCl}: \quad s^2 = 1.8 \times 10^{-10} \quad \Rightarrow \quad s = 1.3 \times 10^{-5}\ \mathrm{M}Ag2CrO4:(2s)2(s)=4s3=1.1×10−12⇒s=6.5×10−5 M\mathrm{Ag_2CrO_4}: \quad (2s)^2(s) = 4s^3 = 1.1 \times 10^{-12} \quad \Rightarrow \quad s = 6.5 \times 10^{-5}\ \mathrm{M}
Answer: Ag2CrO4\mathrm{Ag_2CrO_4} is about 5 times more soluble, even though its KspK_{sp} is more than 100 times smaller. Its ss is cubed in the KspK_{sp} expression, so a small KspK_{sp} still gives a fairly large ss.

KspK_{sp} and the solubility rules

Unit 4 had one solubility rule to know: all sodium, potassium, ammonium, and nitrate salts are soluble in water (Topic 4.7). KspK_{sp} puts numbers on the words "soluble" and "insoluble":
  • KspK_{sp} > 1: soluble. The ion concentrations have to get large before the solution is saturated, so a lot of the salt dissolves. For NaCl\mathrm{NaCl}, KspK_{sp} (calculated from thermodynamic data) is about 38.
  • KspK_{sp} much less than 1: insoluble (more precisely, sparingly soluble). Only a tiny amount dissolves. AgCl\mathrm{AgCl}'s molar solubility of 1.3×10−51.3 \times 10^{-5} M works out to about 1.9 mg per liter.
7.11 - Introduction to Solubility Equilibria
KspK_{sp} values span more than 40 powers of ten, so two "insoluble" salts can still differ enormously.

Finding KspK_{sp} from molar solubility

The reverse calculation is a common lab task: measure how much of a salt is dissolved in a saturated solution, then calculate KspK_{sp}. Convert the measurement to molar solubility, write each ion concentration in terms of ss, and substitute into the KspK_{sp} expression.

Example 4: KspK_{sp} of calcium fluoride from lab data (AP-style)

A student stirs excess CaF2\mathrm{CaF_2} in distilled water at 25 °C until no more dissolves, then filters the mixture. She evaporates 1.000 L of the clear filtrate to dryness and collects 0.0160 g of CaF2\mathrm{CaF_2}. (a) Calculate the molar solubility of CaF2\mathrm{CaF_2}. (b) Calculate KspK_{sp}. (c) Some fine CaF2\mathrm{CaF_2} particles passed through the filter paper. Is the calculated KspK_{sp} too high or too low? Explain.
(a) Use M=78.08\mathcal{M} = 78.08 g/mol, and keep one extra digit until the end:
s=0.0160 g/L78.08 g/mol=2.049×10−4 M≈2.05×10−4 Ms = \frac{0.0160\ \mathrm{g/L}}{78.08\ \mathrm{g/mol}} = 2.049 \times 10^{-4}\ \mathrm{M} \approx 2.05 \times 10^{-4}\ \mathrm{M}
(b) [Ca2+][\mathrm{Ca^{2+}}] = ss = 2.049×10−42.049 \times 10^{-4} M and [F−][\mathrm{F^{-}}] = 2s2s = 4.098×10−44.098 \times 10^{-4} M:
Ksp=[Ca2+][F−]2=(2.049×10−4)(4.098×10−4)2=3.44×10−11K_{sp} = [\mathrm{Ca^{2+}}][\mathrm{F^{-}}]^2 = (2.049 \times 10^{-4})(4.098 \times 10^{-4})^2 = 3.44 \times 10^{-11}
That matches the table value, 3.45×10−113.45 \times 10^{-11}, within rounding.
(c) Too high. The extra particles add mass to the residue, so the measured solubility is too high. Since Ksp=4s3K_{sp} = 4s^3, the error in ss gets cubed, so even a small error in the mass makes a bigger error in KspK_{sp}.

Will a precipitate form?

To predict whether mixing two solutions forms a precipitate, calculate Q for the dissolution, written just like KspK_{sp} but using the ion concentrations right after mixing, and compare it with KspK_{sp}:
  • Q>KspQ > K_{sp}: more ions than a saturated solution can hold, so solid precipitates until Q=KspQ = K_{sp}.
  • Q<KspQ < K_{sp}: unsaturated, so no precipitate forms.
  • Q=KspQ = K_{sp}: exactly saturated.

Example 5: Silver nitrate in tap water

A student adds 1.00 mL of 0.0100 M AgNO3\mathrm{AgNO_3} to 99.0 mL of tap water that contains 2.0×10−42.0 \times 10^{-4} M Cl−\mathrm{Cl^{-}}. Does AgCl\mathrm{AgCl} precipitate? (Ksp=1.8×10−10K_{sp} = 1.8 \times 10^{-10})
Step 1: Find each concentration in the mixed solution (total volume 100.0 mL):
[Ag+]=(0.0100 M)(1.00 mL)100.0 mL=1.00×10−4 M[Cl−]=(2.0×10−4 M)(99.0 mL)100.0 mL=2.0×10−4 M[\mathrm{Ag^{+}}] = \frac{(0.0100\ \mathrm{M})(1.00\ \mathrm{mL})}{100.0\ \mathrm{mL}} = 1.00 \times 10^{-4}\ \mathrm{M} \qquad [\mathrm{Cl^{-}}] = \frac{(2.0 \times 10^{-4}\ \mathrm{M})(99.0\ \mathrm{mL})}{100.0\ \mathrm{mL}} = 2.0 \times 10^{-4}\ \mathrm{M}
Step 2: Calculate Q and compare:
Q=[Ag+][Cl−]=(1.00×10−4)(2.0×10−4)=2.0×10−8Q = [\mathrm{Ag^{+}}][\mathrm{Cl^{-}}] = (1.00 \times 10^{-4})(2.0 \times 10^{-4}) = 2.0 \times 10^{-8}
Answer: Q=2.0×10−8>Ksp=1.8×10−10Q = 2.0 \times 10^{-8} > K_{sp} = 1.8 \times 10^{-10}, so AgCl\mathrm{AgCl} precipitates and the solution turns cloudy. That's why silver nitrate solutions are always made with distilled water.

Common mistakes

  • Including the solid in KspK_{sp}. The expression has only the ions.
  • Forgetting the coefficient inside the parentheses. For PbI2\mathrm{PbI_2}, [I−][\mathrm{I^{-}}] = 2s2s, so the term is (2s)2=4s2(2s)^2 = 4s^2.
  • Comparing KspK_{sp} values across different formula types. Compare KspK_{sp} directly only for salts with the same number of ions. Otherwise, calculate ss.
  • Mixing up ss and an ion concentration. In saturated PbI2\mathrm{PbI_2}, ss = [Pb2+][\mathrm{Pb^{2+}}], but [I−][\mathrm{I^{-}}] = 2s2s.
  • Forgetting dilution when two solutions are mixed. Find the new concentrations in the total volume before you calculate Q.

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