Solubility

John A. Schreifels

Chemistry 212

Chapter 12-2

Types of Solution Solution – homogeneous mixture of two or more

substances of ions or molecules. E.g. NaCl (aq)1) Solvent = component which is the component in

greater amount.2) Solute = component which is present in the smaller

amount.– Gaseous = gases are completely miscible in each other.– Liquid = gas, liquid or solid solute dissolved in solute.– Solid = mixture of two solids that are miscible in each other to

form a single phase. Colloid – appears to be a homogeneous mixture, but

particles are much bigger, but not filterable. E.g. Fog, smoke, whipped cream, mayonnaise, etc.

Suspension: larger particle sizes, filterable. E.g. mud, freshly squeezed orange juice.

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.

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. (precipitation)

Eventually the rate of dissociation is equal to the rate of precipitation.

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

Solubility Equilibrium: Dissociation = Precipitation

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.

Dissolving NaCl in water

Na+ and Cl - ions surrounded by water molecules

NaCl Crystal

An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.

A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.

nonelectrolyte weak electrolyte strong electrolyte4.1

Precipitation Reactions

Mix two aqueous solutions made by dissolving ionic compounds in water.

If a reaction happens, a precipitate (solid) is formed.

Predicting Products of Precipitation Reactions

1) Ionic Compounds are Strong Electrolytes –Determine charge on all ions of reactants

2) Using Ion Charges; Predict formula of products. ( + ion of one reactant forms compound with – ion of other reactant)

3) Balance Equation4) Determine is product is solid or

aqueous solution

Solubility Rules for Common Ionic CompoundsIn water at 250C

Soluble Compounds Exceptions

Compounds containing alkali metal ions and NH4

+THESE ARE INSOLUABLE

NO3-, HCO3

-, ClO3-

Cl-, Br-, I- Halides of Ag+, Hg22+, Pb2+

SO42- Sulfates of Ag+, Ca2+, Sr2+, Ba2+,

Hg2+, Pb2+

Insoluble Compounds Exceptions

CO32-, PO4

3-, CrO42-, S2- Compounds containing alkali

metal ions and NH4+

OH- Compounds containing alkali metal ions and Ba2+

4.2

Predicting Products of Precipitation Reactions (Cont)

5) Determine spectator ions (Ions that are still dissolved in water in the product)

6) Write net ionic equation (Only shows ions involved in forming solid)

Precipitation Reactions

Precipitate – insoluble solid that separates from solution

molecular equation

ionic equation

net ionic equation

Pb2+ + 2NO3- + 2Na+ + 2I- PbI2 (s) + 2Na+ + 2NO3

-

Na+ and NO3- are spectator ions

PbI2

Pb(NO3)2 (aq) + 2NaI (aq) PbI2 (s) + 2NaNO3 (aq)

precipitate

Pb2+ + 2I- PbI2 (s)

4.2

AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)

Ag+ + NO3- + Na+ + Cl- AgCl (s) + Na+ + NO3

-

Ag+ + Cl- AgCl (s)4.2

Write the net ionic equation for the reaction of silver nitrate with sodium chloride.

Acids

Have a sour taste. Vinegar owes its taste to acetic acid. Citrusfruits contain citric acid.

React with certain metals to produce hydrogen gas.

React with carbonates and bicarbonates to produce carbon dioxide gas

Have a bitter taste.

Feel slippery. Many soaps contain bases.

Bases

4.3

Acids

Produce H+ (proton) or (H3O)+ when dissolved in water

Proton donor

HNO3 (aq) + H2O (l) H3O+ (aq) + (NO3)- (aq)

HNO3 (aq) H+ (aq) + (NO3)- (aq)H2O

Monoprotic acids; Produce one H+ when dissolved in water

HNO3 H+ + NO3- Strong electrolyte, strong acid

Diprotic acids; Produce two H+ when dissolved in water

H2SO4 2 H+ + SO4-2 Strong electrolyte, strong acid

Triprotic acids; Produce three H+ when dissolved in water

H3PO4 3 H+ + PO4-3 Weak electrolyte, weak acid

4.3

Bases

•Produce (OH)- when dissolved in water•Proton (H+) acceptor

F- (aq) + H2O (l) <-> HF (aq) + (OH)- (aq)

Na(OH) (s) -----> Na+ (aq) + (OH)- (aq)H2O

Neutralization Reaction

Acid + Base -> Salt + H2O

Acid + Carbonate -> Salt + CO2(g) + H2O (l)

Carbonate; Contains

(CO3)-2 or (HCO3)-

Chalk; Ca(CO3)

Displacement Reactions – Metal Displaces H from acid or water

Metal + Acid -> Salt + H2 (g)

Metal + Water -> Base + H2(g)

Use Activity Series to Know if a Reaction Will Happen

Solution Stoichiometry

The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.

M = molarity =moles of solute

liters of solution

What mass of KI is required to make 500. mL ofa 2.80 M KI solution?

volume KI moles KI grams KIM KI M KI

500. mL = 232 g KI166 g KI

1 mol KIx

2.80 mol KI

1 L solnx

1 L

1000 mLx

4.5

4.5

Acid/Base Titrations

Experimental technique that determines the concentration (in Molarity) of an acid (or base)

This is based upon an acid/base neutralization reaction.– ACID +BASE -> SALT + H2O

Base (or acid) is added until there is the same amount (same # moles) of base and acid.

TitrationsIn a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

Equivalence point – the point at which the reaction is complete

Indicator – substance that changes color at (or near) the equivalence point

Slowly add baseto unknown acid

UNTIL

the indicatorchanges color

4.7

Fig. 4.17a,b

Acid-Base Titrations

Introductory Chemistry 2/e by N Tro, Prentice Hall, 2006, pg 480

Base; (OH)-

Acid; H+

Acid + Base -> Salt + H2O

At the endpoint of an acid/base titration….

Moles acid = Moles base

(MV)acid = (MV)base

Note– If solid; moles = mass/ MM– If aqueous solution; moles = MV

What volume of a 1.420 M NaOH solution isRequired to titrate 25.00 mL of a 4.50 M H2SO4 solution?

4.7

WRITE THE CHEMICAL EQUATION!

volume acid moles acid moles base volume base

H2SO4 + 2NaOH 2H2O + Na2SO4

M

acid

rx

coef.

M

base

4.50 mol H2SO4

1000 mL solnx

2 mol NaOH

1 mol H2SO4

x1000 ml soln

1.420 mol NaOHx25.00 mL = 158 mL

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.

Dilution

Add Solvent

Moles of solutebefore dilution (i)

Moles of soluteafter dilution (f)=

MiVi MfVf=4.5

How would you prepare 60.0 mL of 0.2 MHNO3 from a stock solution of 4.00 M HNO3?

MiVi = MfVf

4.5

Vi =MfVf

Mi

=0.200 x 0.06

4.00= 0.003 L = 3 mL

3 mL of acid + 57 mL of water = 60 mL of solution

Dissolving silver sulfate, Ag2SO4, in water

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-]

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

Since K is always calculated by just multiplying concentrations, it is called a “solubility product” constant - Ksp.

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

Try Problems 1 - 8

Some Ksp Values

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

Calculating Ksp of Silver Chromate

A saturated solution of silver chromate, Ag2CrO4, has [Ag+] = 1.3 x 10-4 M. What is the Ksp for Ag2CrO4?

Ag2CrO4 (s) 2 Ag+ (aq) + CrO42- (aq)

---- ----

1.3 x 10-4 M

Ksp = [Ag+]2[CrO42-]

Ksp = (1.3 x 10-4 )2 (6.5 x 10-5) = 1.1 x 10-12

Calculating the Ksp of silver sulfate The solubility of silver sulfate is 0.014 mol/L. This

means that 0.0144 mol of Ag2SO4 will dissolve to make 1.0 L of saturated solution. Calculate the value of the equilibrium constant, Ksp for this salt.

Ag2SO4 (s) 2 Ag+ (aq) + SO42- (aq)

--- ---

+ 2s + s

2s s

Ksp = [Ag+]2[SO42-] = (2s)2(s) = (4s2)(s) = 4s3

We know: s = 0.0144 mol/L

Ksp = 4(0.0144)3 = 1.2 x 10-5

Calculating solubility, given Ksp The Ksp of NiCO3 is 1.4 x 10-7 at 25°C. Calculate its

molar solubility.

NiCO3 (s) Ni2+ (aq) + CO32- (aq)

--- ---

+ s + s

s s

Ksp = [Ni2+][CO32-]

1.4 x 10-7 = s2

s = = 3.7 x 10-4 M710x4.1

Other ways to express solubility... We just saw that the solubility of nickel (II) carbonate

is 3.7 x 10-4 mol/L. What mass of NiCO3 is needed to prepare 500 mL of saturated solution?

0.022 g of NiCO3 will dissolve to make 500 mL solution.

Try Problems 9 - 26

g 0.022 NiCO mol 1

g 118.72 x

L 0.500 x

L1NiCO mol10x3.7

3

34

Calculate the solubility of MgF2 in water. What mass will dissolve in 2.0 L of water?

MgF2 (s) Mg2+ (aq) + 2 F- (aq)

---- ----

+ s + 2s

s 2s

Ksp = [Mg2+][F-]2 = (s)(2s)2 = 4s3

Ksp = 7.4 x 10-11 = 4s3

s = 2.6 x 10-4 mol/L

22

24

MgF g 0.032 MgF mol 1

g 62.31 x

L 2.0 x

L1MgF mol10x2.6

Solubility and pH Calculate the pH of a saturated solution of silver

hydroxide, AgOH. Refer to the table in your booklet for the Ksp of AgOH.

AgOH (s) Ag+ (aq) + OH- (aq)

---- ----

+ s + s

s s

Ksp = 2.0 x 10-8 = [Ag+][OH-] = s2

s = 1.4 x 10-4 M = [OH-]

pOH = - log (1.4 x 10-4) = 3.85

pH = 14.00 - pOH = 10.15

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

Calculate the solubility of MgF2 in a solution of 0.080 M NaF.

MgF2 (s) Mg2+ (aq) + 2 F- (aq)

---- 0.080 M

+ s + 2s

s 0.080 + 2s

Ksp = 7.4 x 10-11 = [Mg2+][F-]2 = (s)(0.080 + 2s)2

Since Ksp is so small…assume that 2s << 0.080

7.4 x 10-11 = (s)(0.080)2

s = 1.2 x 10-8 mol/L

Explaining the Common Ion Effect

The presence of a common ion in a solution will lower the solubility of a salt.

LeChatelier’s Principle:

The addition of the common ion will shift the solubility equilibrium backwards. This means that there is more solid salt in the solution and therefore the solubility is lower!

Ksp and Solubility Generally, it is fair to say that salts with very small

solubility product constants (Ksp) are only sparingly soluble in water.

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

This works if the salts have the same number 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.

But be careful...

Salt KspSolubility(mol/L)

CuS 8.5 x 10-45 9.2 x 10-23

Ag2S 1.6 x 10-49 3.4 x 10-17

Bi2S3 1.1 x 10-73 1.0 x 10-15

Do you see the “problem” here??

Mixing Solutions - Will a Precipitate Form?

If 15 mL of 0.024-M lead 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)

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

Step 1: Is a sparingly soluble salt formed?

We can see that a double replacement reaction can occur and produce PbCrO4. Since this salt has a 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+][CrO42-]

If a precipitate forms, it means the solubility equilibrium has shifted BACKWARDS.

This will happen only if Qsp > Ksp in our mixture.

Step 2: Find the concentrations of the ions that form the sparingly soluble 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: C1V1 = C2V2

[Pb2+] =

[CrO42-] =

2

2

11 Pb M 0.0080 mL) (45

mL) M)(15 (0.024

VVC

-24

2

11 CrO M 0.020 mL) (45

mL) M)(20 (0.030

VVC

Step 3: Calculate Qsp for the mixture.

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

Qsp = 1.6 x 10-4

Step 4: Compare Qsp to Ksp.

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

Note: If Qsp = Ksp, the mixture is saturated

If Qsp < Ksp, the solution is unsaturated

Either way, no ppte will form!