Solubility Equilibria

1

Chem 30S Review…Solubility Rules✦ Salts are generally more soluble in HOT water(Gases

are more soluble in COLD water)✦ Alkali Metal salts are very soluble in water. ✦ NaCl, KOH, Li3PO4, Na2SO4 etc...✦ Ammonium salts are very soluble in water.✦ NH4Br, (NH4)2CO3 etc…✦ Salts containing the nitrate ion, NO3

-, are very soluble in water.

✦ Most salts of Cl-, Br- and I- are very soluble in water - exceptions are salts containing Ag+ and Pb2+.

✦ soluble salts: FeCl2, AlBr3, MgI2 etc...✦ “insoluble” salts: AgCl, PbBr2 etc...

2

Dissolving a salt...✦ A salt is an ionic compound -

usually a metal cation bonded to a non-metal anion.

✦ The dissolving of a salt is an example of equilibrium.

✦ The cations and anions are attracted to each other in the salt.

✦ They are also attracted to the water molecules.

✦ The water molecules will start to pull out some of the ions from the salt crystal.

3

✦ At first, the only process occurring is the dissolving of the salt - the dissociation of the salt into its ions.

✦ However, soon the ions floating in the water begin to collide with the salt crystal and are “pulled back in” to the salt. (crystalization)

✦ Eventually the rate of dissociation is equal to the rate of crystalization.

✦ The solution is now “saturated”. It has reached equilibrium.

4

Solubility Equilibrium: Dissociation = Crystalization

In a saturated solution, there is no change in amount of solid precipitate at the bottom of the beaker.

Concentration of the solution is constant.

The rate at which the salt is dissolving into solution equals the rate of crystalization.

Dissolving NaCl in water

Na+ and Cl - ions surrounded by

water molecules

NaCl Crystal

5

Dissolving silver sulfate, Ag2SO4, in H2O

✦When silver sulfate dissolves it dissociates into ions. When the solution is saturated, the following equilibrium exists:

Ag2SO4 (s) ⇔ 2 Ag+ (aq) + SO42- (aq)

✦Since this is an equilibrium, we can write an equilibrium expression for the reaction:

Ksp = [Ag+]2 [SO42-]

✦Ksp (solubility product constant)

Notice that the Ag2SO4 is left out of the expression! Why?

Ag2SO4 is a solid and not included in the expression.

6

Writing solubility product expressions...✦ For each salt below, write a balanced equation

showing its dissociation in water.✦ Then write the Ksp expression for the salt.

Iron (III) hydroxide, Fe(OH)3

Nickel sulfide, NiS

Silver chromate, Ag2CrO4

Zinc carbonate, ZnCO3

Calcium fluoride, CaF2

7

Some Ksp Values

Note: These are experimentally determined, and maybe slightly different on a different Ksp table.

8

Calculate Solubility from Ksp✦ Ksp for AgCl = 1.8 x 10-10 , what is the

molar solubility in pure water?✦ AgCl (s) ⇔ Ag+ (aq) + Cl- (aq) ✦ Ksp = [Ag+] [Cl-] ✦ At equilibrium the concentrations of Ag+

and Cl- are equal✦ let x = [Ag+] and also [Cl-]✦ 1.8 x 10-10 = (x) (x)✦ 1.8 x 10-10 = x2

9

Calculate Solubility from Ksp

✦ 1.8 x 10-10 = x2

✦ Square root both sides✦ 1.3 x 10-5 = x = [Ag+] and [Cl-]✦ Applying the concentration values to

the balanced solubility equation (1:1:1 ratio) indicates the molar solubility is 1.3 x 10-5 mol/L

10

Calculate Solubility from Ksp 2✦ Lead iodide (PbI2) has a Ksp = 7.9 x 10-9 What is the molar solubility of PbI2 in water?✦ PbI2 (s) ⇔ Pb2+ (aq) + 2 I- (aq) ✦ Ksp = [Pb2+] [I-]2 ✦ let x = [Pb2+] and let 2x = [I-]

✦ 7.9 x 10- 9 = (x) (2x)2

11

Calculate Solubility from Ksp 2

✦ 7.9 x 10- 9 = (x) (4x2)✦ 7.9 x 10-9 = 4x3 ✦ 1.3 x 10- 3 = x = [Pb2+]

✦ Applying the concentration values to the balanced solubility equation (1:1:2 ratio) indicates the molar solubility is 1.3 x 10-3 mol/L

12

Algebraic form of Ksp

✦ For “salts” the algebraic form of the Ksp are …

✦ AB: (x) (x) = x2 = Ksp e.g. AgCl✦ AB2: (x) (2x)2 = 4x3 = Ksp e.g. PbI2✦ AB3: (x) (3x)3 = 27x4 = Ksp e.g. AlBr3

✦ A2B3: e.g. Au2O3

13

Calculate Ksp given Solubility ✦ The solubility of Ag2CrO4 is 6.7 x 10-5 mol/L

What is the Ksp? Ag2CrO4 (s) ⇌ 2 Ag+

(aq) + CrO42−

(aq)

Ksp = [Ag+]2 [CrO42−]

✦ The equilibrium concentrations of the ions are determined from the solubility of Ag2CrO4

✦ The ratios of the coefficients from the balanced solubility equation are used.

14

Calculate Ksp given Solubility✦ e.g. 1 Ag2CrO4 : 2 Ag+ ✦ there is a 1:2 ratio, ✦ if 6.7 x 10-5 mol/L is the Ag2CrO4 solubility the [Ag+]eq = 2 ( 6.7 x 10-5) = 1.3 x 10-4 M ✦ e.g. 1 Ag2CrO4 : 1 CrO4

2−

✦ there is a 1:1 ratio, if 6.7 x 10-5 mol/L is the Ag2CrO4 solubility

the [CrO42-]eq= 1( 6.7 x 10-5) = 6.7 x10-5 M

15

Calculate Ksp given Solubility✦ Apply the equilibrium values to the Ksp

Ksp = [Ag+]2 [CrO42−]

= (1.3 x 10-4)2 (6.7 x 10-5) = 1.1 x 10-12

16

Calculate Ksp given Solubility

✦ Ag2CrO4 (s) ⇔ 2 Ag+ (aq) + CrO4

2− (aq)

s 2s s

Ksp = [Ag+]2 [CrO4

2−] = (2s)2(s) = (4s2)(s) = 4s3

s = 6.7 x 10-5 mol/L

Ksp = 4(6.7 x 10-5 )3 = 1.1 x 10-12

equilm conc.solubility

17

More Calculate Ksp given Solubility

✦ Cu(OH)2 (s) Solubility 3.42 x 10-7

✦ Answer: Ksp = 1.6 x 10-19

✦ Fe(OH)3 (s) Solubility 9.9 x 10-11

✦ Answer: Ksp = 2.6 x 10-39

18

Ksp and Solubility✦ Generally, salts with very small solubility product

constants (Ksp) are only slightly soluble in water.

✦ When comparing the solubilities of two salts, you can sometimes simply compare the relative sizes of their Ksp values.

✦ This works if the salts have the same ratio of ions!

✦ For example… CuI has Ksp = 5.0 x 10-12 and CaSO4 has Ksp = 6.1 x 10-5. Since the Ksp for calcium sulfate is larger than that for the copper (I) iodide, we can say that calcium sulfate is more soluble.

19

But be careful...

Do you see the “problem” here??can’t compare Ksp to predict solubility for compounds with different ratios

20

Mixing Solutions - Will a Precipitate Form?p. 581

If 15 mL of 0.024-M lead (II) nitrate is mixed with 30 mL of 0.030-M potassium chromate - will a precipitate form?

Pb(NO3)2 (aq) + K2CrO4 (aq) ⇔ PbCrO4 (s) + 2 KNO3 (aq)

21

Pb(NO3)2 (aq) + K2CrO4 (aq) ⇔ PbCrO4 (s) + 2 KNO3 (aq)

Step 1: Is an insoluble salt formed? We can see that a double replacement reaction can

occur and produce PbCrO4. Since this salt has a low solubility (very small Ksp), it may precipitate from the mixture. The solubility equilibrium is:

PbCrO4 (s) ⇔ Pb2+ (aq) + CrO42- (aq)

Ksp = 2 x 10-16 = [Pb2+][CrO4

2-]

If a precipitate forms, it means the amount of ions in solution is greater than the equilibrium concentrations

This will happen only if product of the ion concentrations

(Qsp) > Ksp in our mixture.

22

Step 2: Find the concentrations of the ions that form the insoluble salt.

Since we are mixing two solutions in this example, the concentrations of the Pb2+ and CrO4

2- will be diluted. We have to do a dilution calculation!

Dilution: M1V1 = M2V2

Pb(NO3)2 : Pb2+ and the K2CrO4 : CrO42- ratios are 1:1

therefore:

[Pb2+] = M1V1 = (0.024M)(15 mL) = 0.0080 Pb2+

V2 (45 mL)

[CrO42-] = M1V1 = (0.030M)(30 mL) = 0.020 CrO4

2-

V2 (45 mL)

23

Step 3: Calculate IP (or Qsp) for the mixture. ion product (IP or Qsp) – the product of the molar conc.in an ionic solution (raised to an appropriate power) PbCrO4 (s) ⇔ Pb2+ (aq) + CrO4

2- (aq)

Qsp = [Pb2+][CrO42-] = (0.0080 M)(0.020 M)

Qsp = 1.6 x 10-4

24

Step 4: Compare Qsp to Ksp.Qsp = 1.6 x 10-4 > Ksp = 2 x 10-16

Since Qsp > Ksp, a precipitate will form when the two solutions are mixed!

Note: If Qsp = Ksp, the solution is saturated If Qsp < Ksp, the solution is unsaturated

Either way, no ppt will form!

25

25

Summary✦ How do you know if a ppt forms?✦ ion product > Ksp ppt. forms

✦ ion product = Ksp ppt. will✦ ion product < Ksp not form

26

Another ppt. problem with DILUTION!✦ e.g. 50mL of 0.0010M CaCl2 and 50mL of 0.010 M

Na2SO4 were mixed. Will a precipitate of CaSO4 be formed? Ksp of CaSO4 = 2.4 x 10-5

✦ CaCl2 (aq) + Na2SO4 (aq) ⇔ CaSO4 (s) + 2 NaCl (aq)

✦ Step 1: Calc. The new ion concentrations using the new volume.

✦ In dilution problems we use: Mi Vi = Mf Vf ✦ The new volume is the combination of the two

solutions i.e. 50 mL + 50 mL = 100 mL✦ When calculating concentrations volume must be in

litres (L) therefore we have 0.100 L

27

✦ The new (final) concentration is calculated as follows:

✦ [CaCl2] = = = 5.0 x 10-4 M✦ ✦ CaCl2 ⇔ Ca2+ + 2 Cl-

therefore [Ca2+] = 5.0 x 10-4 M (1:1 ratio)

✦ [NaSO4] = = = 5.0 x 10-3M

✦ Na2SO4 ⇔ 2 Na+ + SO42-

therefore [SO42-] = 5.0 x 10-3 M (1:1 ratio)

f

ii

VVM

100.0)050.0)(0010.0(

f

ii

VVM

100.0)050.0)(010.0(

28

✦ Step 2: calculate ion product✦ We are looking for the formation of CaSO4

precipitate therefore we use the following:✦ CaSO4 (s) ⇔ Ca2+ (aq) + SO4

2- (aq)

✦ ion product = [Ca2+] [SO42-]

✦ ion product = (5.0 x 10-4) (5.0 x 10-3) = 2.5 x 10-6

✦ The Ksp for CaSO4 is 2.4 x 10-5

✦ ion product Ksp 2.5 x 10-6 2.4 x 10-5

therefore no precipitate!<

29

The Common Ion EffectThe presence of a common ion in a solution

will lower the solubility of a salt.shift left ↑

AgCl (s) ⇔ Ag+ (aq) + Cl- (aq)

(adding NaCl)✦ LeChatelier’s Principle:

The addition of the common ion will shift the solubility equilibrium to the left. This means that salt precipitates in the solution and the solubility is lowered.e.g. the addition of NaCl to a AgCl

equilibrium lowers the solubility of the AgCl ( the common ion is Cl-)

30

The solubility of MgF2 in pure water is 2.6 x 10-4 mol/L. What happens to the solubility if we dissolve the MgF2 in a solution of NaF, instead of pure water?

The Common Ion Effect on Solubility

31

Common Ion Calc✦ Estimate the solubility of barium sulfate in a

0.020 M sodium sulfate solution. The Ksp for barium sulfate is 1.1 x 10-10.

✦ Write the equation and the equilibrium expression for the dissolving of barium sulfate.

✦ BaSO4(s) <=> Ba2+(aq) + SO42-(aq)✦ Ksp = [Ba2+][SO42-]✦ Make an "ICE" chart.✦ Let "x" represent the barium sulfate that

dissolves in the sodium sulfate solution expressed in moles per liter.

32

Common Ion Calc

BaSO4(s) ⇌ Ba2+(aq) SO42-(aq)

I All solid 0 0.020 M (from Na2SO4)

C - x dissolves + x + x

E Less solid x 0.020 M + x

33

Common Ion Calc✦ Substitute into the equilibrium expression and

solve for x. ✦ The assumption is that since x is going to be

very small (the solubility is reduced in the presence of a common ion), the term "0.020 + x" is the same as "0.020."

✦ (You can leave x in the term and use the quadratic equation or the method of successive approximations to solve for x, but it will not improve the significance of your answer.)

✦ 1.1 x 10-10 = [x][0.020 + x] = [x][0.020] ✦ x = 5.5 x 10-9 mol/L Reduced Solubility BaSO4

34

Common Ion CalcCalculate molar solubility if Ca(IO3)2 in 0.060 mol/L NaIO3. Ksp Ca(IO3)2 = 7.1 x 10-7 Ans: 2.0 x 10-4 mol/L

Calculate the molar solubility of silver chloride in a 0.20 M sodium chloride solution. Ksp AgCl = 1.6 x 10-10

Ans: 8.0 x 10 -10 mol/L

Calc the molar solublity of silver chloride in a 1.5 x 10-3 mol/L silver nitrate solution.Ksp AgCl = 1.6 x 10-10 Ans: 1.3 x 10-3 mol/L

35