Ionic solids dissolve in water to form saturated solutions.

The extent to which they dissolve depends on the forces of attraction between the Ions.

As the charges increase, the forces get stronger, the solubility decreases.

As the ions get larger in size, the forces get weaker, the solubility increases.

Here is what happens when an ionic solid is dropped into water …...

Ca2+ Cl1-

Cl1- Ca2+ Cl1-

Cl1- Ca2+ Cl1-

Cl1-

Cl1-

Ca2+

Cl1-

Notice for every Ca2+ which enters the solution, 2 Cl1- ions also dissolve

This phenomena can be expressed using a chemical equation:CaCl2 (s) -----> Ca2+(aq) + 2 Cl1-(aq)Write the equation for the dissolution of a) boron sulfate.B2(SO4)3(s) ----> 2 B3+

(aq) + 3 SO42-

(aq)

b) sodium chromateNa2CrO4(s) ----> 2 Na1+

(aq) + CrO42-

(aq)

c) aluminum fluorideAlF3(s) ----> Al3+

(aq) + 3 F1-(aq)

Quantitative Aspects of Dissolution

If 4.0 g of aluminium nitrate is dissolved in 450 mL of water what are the concentrations of aluminium and nitrate ions?First find number of moles of Al(NO3)3(s) n=m/MM = 4.0 g /212.996 g/mol = 0.019 molSecond write the equation for dissolutionmol Al(NO3)3(s) ----> Al3+

(aq) + 3 NO31-

(aq)

initial 0.019 final 0 0.019 0.019 x 3 = 0.056 mol

Lastly convert mol to mol/L using C = n/V[Al(NO3)3] = 0.019/0.450 L = 4.2 x 10-2 molL-1

[Al3+] = 4.2 x 10-2 molL-1

[NO31-)] = 3 x 4.2 x 10-2 molL-1

= 1.3 x 10-1 molL-1

If 2.0 g of magnesium phosphate is dissolved in 4.50 L of water what are the concentrations of magnesium and phosphate ions?

Mg3(PO4)2(s) ----> 3 Mg2+(aq) + 2 PO4

3-(aq)

First Find n of Mg3(PO4)2(s) = m/MM= 2.0 g / 262.86 g/mol = 7.6x 10-3 molSecond find [Mg3(PO4)2] using C = n/V

=7.6 x 10-3 mol/ 4.5 L = 1.7 x 10-3 molL-1

[Mg2+] = 3 x [Mg3(PO4)2] = 5.1 x 10-3 molL-1

[PO43-] = 2 x [Mg3(PO4)2] = 3.4 x 10-3 molL-1

1.7 x 10-3 molL-1 5.1 x 10-3 molL-1 3.4 x 10-3 molL-1

Soluble, Slightly Soluble and InsolubleIn general, if more than 1 g of solute dissolves in 1.0 L of solution the substance is said to be soluble. There is no definite dividing line between slightly soluble and insoluble however these rules can be used to determine when solid precipitates result when 2 different ionic solutions are combined.

Solubility Rules for Aqueous SolutionsSoluble Compounds1. All compounds of the alkali metals

(Group 1A) are soluble.2. All salts containing NH4

1+, NO31-, ClO4

1-, ClO3

1-, and C2H3O21- are soluble.

3. All chlorides, bromides, and iodides are soluble except when combined with Ag1+, Pb2+ and Hg2

2+

4. All sulfates are soluble except those of Hg2

2+, Pb2+, Ca2+, Sr2+ and Ba2+.

Solubility Rules for Aqueous SolutionsInsoluble Compounds5. All hydroxides and all metal oxides are

insoluble except those of Group 1A and those of Ca2+, Sr2+, and Ba2+.

6. When metal oxides do dissolve, they react with water to form hydroxides. The oxide ion, O2-, does not exist in water for exampleNa2O(s) + H2O ------> 2NaOH(aq)

7. All compounds containing CO32-, PO4

3-, SO3

2-, S2- are insoluble except those of group 1A and NH4

1+.These rules will be provided for evaluations. They are found in the textbook. Page 471. See index under solubility rules.

Sample Question 1- Will a precipitate form if saturated solutions of lead(II) nitrate and sodium iodide are combined. Write a balanced chemical equation. This is a double displacement.First write down all positive and negative ionsPb2+ NO3

1- + Na1+ I1- next balance charges Pb(NO3)2 + NaIswap partners to produce products Pb(NO3)2 + NaI ===> Pb2+ I1-

+ Na1+ NO31-

balance charges and number of particlesPb(NO3)2(aq)+ 2NaI(aq) -> PbI2(s) + 2NaNO3 (aq)

rules 2 and 3 tell us PbI2 is a precipitate

Sample Question 2 - Will a precipitate form if saturated solutions of calcium bromide and silver nitrate are combined.First write down all positive and negative ionsCa2+ Br1- + Ag1+ NO3

1- next balance charges CaBr2 + AgNO3

1-

swap partners to produce products CaBr2 + AgNO3

===> Ca2+NO31-+ Ag1+Br1-

balance charges and number of particlesCaBr2(aq)+ 2AgNO3(aq) ->2AgBr(s)+Ca(NO3)2(aq)

rules 2 and 3 tell us AgBr is the precipitate

Precipitates are slightly soluble and establish an Equilibrium between the solid and its ions in solution. In a previous example PbI2 was a precipitate. The equilibrium is:PbI2(s) <=====> Pb2+(aq) + 2I1-(aq)

The equilibrium expression is:Ke = [Pb2+][I1-]2 remember in a heterogeneous system solids are not includedThis type of Ke has a special name it is called a solubility product or Ksp for short. The values of Ksp are temperature dependent and can be found in the textbook (802)

Here is a demonstration of an equilibrium between solid lead(II) iodide and its ions

Pb2+

I1-PbI2

Ksp values can be used to determine the amount of solute present in a saturated solution of slightly soluble material.Example - What is the concentration of each ion in a saturated solution of manganese(II) hydroxide at 25oCKsp = 1.6 x 10-13?

Mn(OH)2(s) Mn2+ (aq)

+ 2 OH1- (aq)

Let x = the amount of each ion which appears when the system is at equilibrium

X 2 X

Ksp = [Mn2+] [OH1-] 2 = 1.6 x 10-13

[x] [2x] 2 = 1.6 x 10-13

= 1.6 x 10-134x3

X = 3.4 x 10-5 molL-1 so [Mn2+],[OH1-] are

3.4 x 10-5 M 6.8 x 10-5 M

Find the mass of Mn(OH)2 in 68.5 L of saturated solution at 250C.It was previously determined the [Mn(OH)2] = 3.4 x 10-5 molL-1 so the

number of moles (n) = CV = 3.42 x 10-5 molL-1 x 68.5 L = 2.334 x 10-3 molThe mass of Mn(OH)2 can be determined using m = n x MM som = 2.334 x 10-3 mol x 88.95 g/mol m = 0.21 g of Mn(OH)2

What is the mass of Pb2+ in 69 L of a saturated solution of lead(II) fluoride at 25oC? Ksp = 3.6 x 10-8

PbF2(s) Pb2+ (aq)

+ 2 F1- (aq)

Let x = the amount of each ion which appears when the system is at equilibrium

X 2 X

Ksp = [Pb2+] [F1-] 2 = 3.6 x 10-8

[x] [2x] 2 = 3.6 x 10-8

= 3.6 x 10-84x3

X = 2.1 x 10-3 molL-1 so [Pb2+],[F1-] are

2.1 x 10-3 M 4.2 x 10-3 M

[Pb2+] = 2.1 x 10-3 molL-1 so the number of moles (n) = CV = 2.1 x 10-3 molL-1 x 69 L = 0.1435 molThe mass of Pb2+ can be determined using m = n x MM som = 0.1435 mol x 207.2 g/mol m = 3.0 x 10 g of Pb2+

Find the Ksp of Mg3(PO4)2 if a saturated solution is analyzed and found to contain 1.2 x 10-5 molL-1 of PO4

3-.

Mg3(PO4)2(s) 3Mg2+ (aq)

+ 2 PO43-

(aq)

Let x = the amount of each ion which appears when the system is at equilibrium

3X 2 X

2x = 1.2 x 10-5 X = 6.0 x 10-6

[Mg2+] = 3x = 1.8 x 10-5

Ksp = [Mg2+] 3[PO43-]2

Ksp = [1.8 x 10-5]3[1.2 x 10-5]2 = 8.4 x 10-25

1.8 x 10-5 M 1.2 x 10-5 M

455 mL of a saturated solution of Ag2CrO4 is analyzed and found to contain 1.48 mg of silver. Using this data find the Ksp of Ag2CrO4.

Ag2CrO4(s) 2Ag1+ (aq)

+ CrO42-

(aq)

First find C of Ag1+ using C = n/V where n = m/MM, n = 1.48 x 10-3 g/107.9 g/molC = 1.37 x 10-5 / 0.455 L = 3.01 x 10-5 = 2x

2X X

x = 1.5 x 10-5

Ksp = [Ag1+]2[CrO42-]

Ksp = [3.01 x 10-5]2[1.5 x 10-5 ] = 1.37 x 10-14

3.01 x 10-5 M 1.5 x 10-5 M

15 mL of a saturated solution of Ca(OH)2 is used to neutralize 0.45 g of a monoprotic standard acid with a molar mass of 204 g. Using this data find the Ksp of Ca(OH)2.

0.10 of potassium hydrogen phthalatedistilled water few drops phenolphthalein

Titrant - Ca(OH)2 solution of known concentration

When the solution turns pink the number# of mol of acid = # of mol of baseequivalence pt.

Equilibrium Problems

1. What volume of a saturated solution of Cd(OH)2, Ksp is 7.2 x 10-15, is required to neutralize 1.3 g of potassium hydrogen phthalate (M is 204.23 g/mol). (2.6 x 102 L).

2. What is the Ksp of yttrium carbonate if 14.5 L of a saturated solution is analyzed and found to contain 1.3 x 10-3 g of this solute? (1.0 x 10-31)

3. If the Ksp of silver phosphate is 8.9 x 10-17 what mass of silver would be found in 29.9 L of saturated solution? (0.41 g)

4. What is the solubility of strontium arsenate. Ksp is 4.3 x 10-19?(8.3 x 10-5 mol/L)

5. What mass of lithium would you find in 34.1 L of a saturated solution of lithium phosphate if the Ksp is 2.4 x 10-11? (0.69 g)

Ag2CO3(s) 2Ag1+ (aq)

+ CO32-

(aq)

Consider this equilibrium

If either the Ag1+ ion or the CO32- ion is

added to this equilibrium it will upset the equilibrium and a shift will occur in this directionThis shift will cause a decrease in the amount of Ag2CO3(s) capable of dissolving

It is impossible to add just one of these ions so electrolytes containing one of these ions have a second ion which doesn’t affect the equilibrium. This other ion is known as a spectator ion.Examples - AgNO3, K2CO3, AgC2H3O2, Na2CO3 are some of the salts which could be used to shift this equilibrium. Notice they are all soluble materials.

Ag2CO3(s) 2Ag1+ (aq)

+ CO32-

(aq)

Determine mass of Ag2CO3(s) which could dissolve in 3.5 L of a 0.15 molL-1 solution of sodium carbonate. The common ion is

initial 0 0.15 molL-1

Amount capable of dissolving

2x x

@ equilibrium 2x 0.15 + x

X is too small to affect the 0.15 so

CO32-

Ag2CO3(s)

Na1+

Na1+

Na1+

Na1+

CO32-

CO32-

The presence of sodium carbonate pushes the equilibrium in the backwards direction which reduces the amount of silver carbonate which can dissolve

Ag2CO3(s) 2Ag1+ (aq)

+ CO32-

(aq)

initial 0.15 molL-1

Amount capable of dissolving

2x x

@ equilibrium 2x 0.15 + x

Ksp = [Ag1+]2[CO32-]

8.1 x 10-12 = [2x]2[0.15]4x2 = 5.4 x 10-11 x = 3.67 x 10-6 molL-1

n = CV = 3.67 x 10-6 molL-1 x 3.5 L m=n x MM = 1.29 x 10-5 mol x 275.7 g/molm=3.5 x 10-3 g of Ag2CO3(s)

How many times more soluble is Ca(OH)2 in distilled water than in a 0.28 molL-1 solution of potassium hydroxide.(Ksp of Ca(OH)2 is 6.5 x 10-6)

Ca(OH)2(s) Ca2+ (aq)

+ 2 OH1- (aq)

initial 0 0.28 molL-1

Amount capable of dissolving

x 2x

@ equilibrium x 0.28 + 2x

First determine the solubility in the KOH

Since the amount of OH1- which enters the solution from dissolving the Ca(OH)2 is very small it is insignificant

Ca(OH)2 (s)

K1+

K1+

K1+

K1+

OH1-

OH1-

The presence of KOH pushes the equilibrium in the backwards direction which reduces the amount of Ca(OH)2 which can dissolve

OH1

OH1

Ca(OH)2(s) Ca2+ (aq)

+ 2 OH1- (aq)

initial 0.28 molL-1

Amount capable of dissolving

x 2x

@ equilibrium x 0.28 + 2x

The amount of Ca(OH)2 which dissolves is represented by x so solving it answers half of the question.

Ksp = x(0.28)2 so using tables for Kspx = 6.5 x 10-6 / (0.28)2 = 8.29 x 10-5 molL-1

Now determine the solubility in pure water

Ca(OH)2(s) Ca2+ (aq)

+ 2 OH1- (aq)

initial 0 molL-1

Amount capable of dissolving

x 2x

@ equilibrium x 2x

0 molL-1

Ksp = x(2x)2 so using tables for Ksp6.5 x 10-6 = 4x3 x = 1.18 x 10-2 molL-1

To determine how much more soluble it is in pure water take 1.18 x 10-2 / 8.29 x 10-5 = 140x’s

How many times more soluble is HgI2 in distilled water than in a 0.21 molL-1 solution of mercuric nitrate (Hg(NO3)2).1.1 x 10-28Common ion is

HgI2(s) Hg2+ (aq)

+ 2 I1- (aq)

initial 0.21 molL-1 0Amount capable of dissolving

x 2x

@ equilibrium 0.21 + x 2x

First determine the solubility in the Hg(NO3)2

Since the amount of Hg2+ which enters the solution from dissolving the HgI2 is very small it is insignificant

Hg2+

HgI2 (s)

Hg2+

Hg2+

The presence of Hg2+ pushes the equilibrium in the backwards direction which reduces the amount of mercuric iodide which can dissolve

NO31-

NO31-

NO31- NO3

1-

initial 0.21 molL-1 0Amount capable of dissolving

x 2x

@ equilibrium 0.21 + x 2x

HgI2(s) Hg2+ (aq)

+ 2 I1- (aq)

Ksp = 0.21(2x)2 so using tables for Ksp1.1 x 10-28 = (0.21) 4x2 x = 1.14 x 10-14 molL-1

Now determine the solubility in pure waterHgI2(s) Hg2+

(aq) + 2 I1-

(aq)

initial 0 molL-1

Amount capable of dissolving

x 2x

@ equilibrium x 2x

0 molL-1

Ksp = x(2x)2 so using tables for Ksp1.1 x 10-28 = 4x3 x = 3.02 x 10-10 molL-1

To determine how much more soluble it is in pure water take 3.02 x 10-10 /1.14 x 10-14 = 2.6 x 104x’s

When Do Precipitates Form?. When different solutions are combined sometimes an insoluble precipitate will appear. This happens if the quantities of specific ions become large enough to form solid ionic crystals. Mathematically this can be determined using the ion product from the Ksp expression for the suspected precipitate. To demonstrate this let’s try a problem

Will a precipitate form if 24 mL of a 0.024 molL-

1 solution of Pb(NO3)2 is combined with 84 mL of a 0.0024 molL-1 solution of CaI2?First write the balanced chemical equation assuming a double displacement reaction.Make sure to use the appropriate charges.Ca2+ + 2 I1- + Pb2++ 2 NO3

1-

How do these ions interact?What are some of the possible precipitates?Ca(NO3)2

and Pb I2

Check out the solubility rules to determine which of these is only slightly soluble

Solubility Rules for Aqueous SolutionsSoluble Compounds1. All compounds of the alkali metals

(Group 1A) are soluble.2. All salts containing NH4

1+, NO31-, ClO4

1-, ClO3

1-, and C2H3O21- are soluble.

3. All chlorides, bromides, and iodides are soluble except when combined with Ag1+, Pb2+ and Hg2

2+

4. All sulfates are soluble except those of Hg2

2+, Pb2+, Ca2+, Sr2+ and Ba2+.

Pb I2 is a possible precipitate but only if the [Pb2+][I1-]2 > Ksp of Pb I2

Where did this come from?Remember an insoluble substance establishes an equilibrium between a solid and the dissolved ions. For this precipitate the equilibrium is

PbI2(s) Pb2+ (aq)

+ 2 I1- (aq)

Ksp = [Pb2+][I1-]2, This expression shows the maximum amount of Pb2+and I1- in solution which can coexist.

Now determine the [Pb2+] and [I1-] Initially 24 mL of a 0.024 molL-1 solution of Pb(NO3)2 is combined with 84 mL of a 0.0024

molL-1 solution of CaI2, so [Pb2+] = 0.024

molL-1, [I1-] = 2 mol I1- x 0.0024 molL-1 since CaI2 ------> Ca2+ + 2I1-

When the solutions are combined there is a dilution occurring so to find the new concentration use C1V1 = C2V2. For [Pb2+]C1 = 0.024 molL-1,V1 = 24 mL, V2=(24 + 84)mLC2 = (0.024 molL-1 )(24 mL) / 108 mL

so [Pb2+] = 0.00533 molL-1.

For [I1-] C1 = 0.0048 molL-1,V1 = 84 mL, V2=(24 + 84)mLC2 = (0.0048 molL-1 ) (84 mL) / 108 mLso [I1-] = 3.73 x 10-3 molL-1.

A precipitate forms if [Pb2+][I1-]2 > Ksp ( 0.00533 molL-1 ) ( 3.73 x 10-3 molL-1 )2 = 7.4 x 10-8

since 7.4 x 10-8 > 8.5 x 10-9 Yes, a precipitate formsNow try this problem

Will a precipitate form if 62 mL of a 0.036 molL-

1 solution of Ca(NO3)2 is combined with 83 mL of a 0.0074 molL-1 solution of Na2CO3?First write the balanced chemical equation assuming a double displacement reaction.Make sure to use the appropriate charges.Ca2+ + CO3

2- + Na1++ 2 NO31-

How do these ions interact?What are some of the possible precipitates?CaCO3

and NaNO3

Check out the solubility rules to determine which of these is only slightly soluble

Some of the Solubility Rules

1. All compounds of the alkali metals (Group 1A) are soluble.

2. All salts containing NH41+, NO3

1-, ClO41-,

ClO31-, and C2H3O2

1- are soluble.3. All chlorides, bromides, and iodides are

soluble except when combined with Ag1+, Pb2+ and Hg2

2+

6. All compounds containing CO32-, PO4

3-, SO3

2-, S2- are insoluble except those of group 1A and NH4

1+.

Ca CO3 is a possible precipitate but only if the [Ca2+][CO3

2-] > KspWhere did this come from?Remember an insoluble substance establishes an equilibrium between a solid and the dissolved ions. For this precipitate the equilibrium is

CaCO3(s) Ca2+ (aq)

+ CO32-

(aq)

Ksp = [Ca2+][CO32-], This expression

shows the maximum amount of Ca2+and CO3

2- in solution which can coexist.

Now determine the [Ca2+] and [CO32-]

Initially 62 mL of a 0.036 molL-1 solution of Ca(NO3)2 is combined with 83 mL of a 0.0074

molL-1 solution of Na2CO3, so [Ca2+] = 0.036

molL-1, [CO32-] = 0.0074 molL-1

When the solutions are combined there is a dilution occurring so to find the new concentration use C1V1 = C2V2. For [Ca2+]C1 = 0.036 molL-1,V1 = 62 mL, V2=(62 + 83)mLC2 = (0.036 molL-1 )(62 mL) / 145 mL

so [Ca2+] = 0.0154 molL-1.

For [CO32-]

C1 = 0.0074 molL-1,V1 = 83 mL, V2=(62 + 83)mLC2 = (0.0074 molL-1 ) (83 mL) / 145 mLso [CO3

2-] = 4.24 x 10-3 molL-1.

A precipitate forms if [Ca2+][CO32-] > Ksp

(1.54 x 10-2molL-1 ) (4.24 x 10-3 molL-1 ) = 6.5 x 10-5

since 6.5 x 10-5 > 4.5 x 10-9 Yes, a precipitate forms