StuffHour Exam 3 Date: March 10, Venue: Berchman 6-7:30PM

Chapter 18 --Skip polyprotic problems in 786-787

Chapter 19: 1,2,3,6,8,9,14,15,20,24,34,42,66,70,78,85,86,87---Read 815- top p.818. Skip Sample Problem 19.1 Resume reading p.820 ---Skip mid page 827- middle p.832.--Read Section 19.3--839. Skip pages 840-end.

Make-Up Exam Date: March 17, 2009Room: Time:

2

Ionic Equilibria in Aqueous Systems

1. The Common-Ion Effect in Acid-Base Equilibria

2. Buffers

3. Acid-Base Indicators

4. Neutralization Reactions and Titration Curves

5. Ksp or “equilbrium for precipitation reactions”

% ionization of an acid is the ratio of [H+] at equilibrium to [HA]initial X 100.

[H+]equilbrium

[HA]initalx 100%% ionization =

[OH-]equilbrium

[OH-]initalx 100%% ionizationb =

As the concentration of a weak acid (base) decreases, the % ionization (dissociation) of the acid (base) increases!

Ka =[H3O+][A-]

[HA]

Ka =n

H3O+ A-n

HAn

1 V

Rationale: larger solution volume accommodates more ions(analogous to pressure effects on gas equilibria when ∆ngas is non-zero)

Strong Acid

Weak Acid

Acid Molarity

% Io

niza

tion

Weak Acid

The common ion effect is a LeChatlier equilibrium shift that occurs upon addition of a an ion common to an acid or its conjugate base.

Suppose a weak acid equilibrium in water:

Now add sodium acetate salt to it. The acetate anion is “common” to the acetate produced by the acid dissociation.

What will happen to the acid equilibrium?

add H+ or add CH3COO-Na(s)

Either one shifts the equilibrium to reactants

CH3COOH (aq) H+ (aq) + CH3COO- (aq)

The presence of a common ion suppresses the ionization of a weak acid or a weak base.

Common Ion Effect (Strong acid + weak acid)• Consider a solution with 0.100M CH3CO2H and 0.100 M

HCl. Ka = 1.8 X 10-5

Initial AcAc(M)

Change (M)Equilibrium (M)

0.100 0.00

-x +x0.100 - x

0.100+x

x 0.100 + x

CH3CO2H + H2O <===> CH3CO2- + H3O+

Initial HCl(M)0.00

0.00 -

Ka = 1.8 X 10-5 = x (0.100 + x)0.100 - x

CH3CO2H(aq) + H2O <==> CH3CO2-(aq) + H3O+(aq)

HCl(g) + H2O <==> Cl-(aq) + H3O+(aq) Strong acid

Weak acid

Simplfy? 100 x Ka << 0.1

x (0.100)0.100 = 1.8 X 10-5Ka =

Analysis of the Common Ion Effect

X = 1.8 X 10-5

0

x

Initial (M)Change (M)Equilibrium (M)

0.100 0.00-x +x

0.100 - x x

CH3CO2H + H2O <==> CH3CO2- + H3O+

x = [H+] = [CH3COO-] = 1.34 X 10-3

Can we simplify?

100 x Ka << 0.1, so we can drop x

x2 = 1.8 X 10-5 X 0.100

Ka =x2

0.100 - x= 1.8 X 10-5.

Ka =x2

0.100= 1.8 X 10-5.

+x

Now Same Problem--But No HCl Added

Comparison with and without HCl

x = 1.34 X 10-3 x = 1.8 X 10-5

Weak Acid + Strong Acid

CH3CO2H + H2O <==> CH3CO2- + H3O+

--the addition of the strong acid or conjugate base to a weak acid suppresses the ionization of the

weak acid (it does not change Ka!)

x xSingle Solute ResultWeak Acid Only

The common ion effect make buffers possible• Buffers are solutions containing a weak

conjugate acid/base pair. – this combination resists a change in pH upon

addition of another acid or base.

• The magic combination is the right concentrations of :

– A weak acid and it’s conjugate base or

– A weak base and it’s conjugate acid

A buffer is a an aqueous solution made from from a weak acid-conjugate base pair that will resist pH change upon the addition of H+ or OH-.

equal concentrations of weak acid and its conjugate base

Add OH-

Add H3O+

after addition of H3O+

after addition of OH-

CH3CO2- + H3O+ <==> CH3CO2H + H2O

CH3CO2H + OH- <==> CH3CO2- + H2O

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3 (d) Na2HPO4/H3PO4 e)HCl/CH3COONa

(a) KF is a weak acid and F- is its conjugate base. It’s a buffer solution. (b) HBr is a strong acid, KBr is the salt of a strong base and strong acid. It can not be a buffer solution.(c) CO3

2- is the conjugate base of HCO3-. HCO3- is it

conjugate acid. It is a buffer solution.(d) HPO4

2- is a conjugate base of H3PO4. It is a conjugate acid--base pair and a buffer.

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3 (d) Na2HPO4/H3PO4

To be a buffer we need a weak acid-conjugate base pair or the same combination generated in-situ.

(e) CH3COO- is conjugate base of CH3COOH. The HCl will generate the acid. It is a buffer.

The “Buffer Equation” (Henderson-Hasselbach) is derived from the Ka expression for a weak acid.

pH = pKa + log [base][acid]

Ka =[H3O+][A-]

[HA]

[H+] =Ka [HA]

[A-]

-log [H+] = -log Ka - log [HA][A-]

-log [H+] = -log Ka + log [A-][HA]

pH = pKa + log[A-][HA]

HA(aq) + H2O(l) H3O+(aq) + A- (aq)

Solve for [H+]

Take -log of both sides

Move the minus sign into the log

standard equilibriumfor HA

Ka for acid HA

OR

The “Buffer Equation” (Henderson-Hasslebach) is key to making and understanding buffers.

pH = pKa + log [base][acid]

Know this equation and how to use it!

pH of the buffer

pKa of the acid

Molarity of conjugate base and the acid

pKa = !log(Ka)

pKa = !log

!Kw

Kb

"

Buffer Capacity and Range• Buffer range is the pH range over which a buffer works.– Buffer range is +/- 1 units from the pKa of the acid.

• Buffer capacity is the amount of acid (or base) that a buffer can neutralize before the buffer solution pH changes.

– Higher concentrations of [HA] and [A-] given more capacity. Ideal is when [HA] and [A-] are large and equal to each other.

Buffer Selection: To make a buffer chose the pH of the desired buffer and then search for a weak acid/conjugate base pair having a pKa within pH +/- 1.

Example: If the desired buffer pH is 5, then look for a weak acid with a pKa between 4-6 with 5 being the ideal!

We can also prepare alkaline buffer solutions by mixing a weak base with its conjugate acid just like we do with an acid-conjugate base buffer system.

B + H2O <==> BH+ + OH-

Kb =[BH+][OH-]

[B]

[OH-] =Kb [B][BH+]

-log [OH-] = -log Kb - log [B][BH+]

-log [H+] = -log Kb + log [BH+][B]

pOH = pKb + log [conjugate acid][base]

pOH = pKb + log[BH+][B]

Solve for [OH-]

Take -log of both sides

Move the minus sign into the log

standard equilibriumfor base B

Kb for base, B

Kb =[NH4+][OH- ]

[NH3 ]= 1.8 X 10-5

This ammonia/ammonium salt buffer solution would work well +/- 1 pH from 9.26 or from ~8.3-10.3!

--Buffers work best when the concentrations of the acid and conjugate base are large and equal!

NH3(aq) + H2O <===> NH4+(aq) + OH-(aq)

Show using a chemical equation or use LeChatlier’s principle and explain why a solution of aqueous NH3 NH4Cl constitutes a buffer solution. Kb of NH3 is 1.8 X 10-5. Over what pH range should this buffer work?

pKb = -log(1.8 x 10-5) = 4.74, pKa = 14.00 - 4.74 = 9.26

Buffers can be made in many equivalent ways---just make sure you get the weak acid-conjugate base pair!

Weak acid + strong base

Weak acid + salt of acid

Salt of acid + strong acid

Weak base + strong acid

Weak base + salt of base

Salt of base + strong base

What is the pH of a solution containing 0.30 M HCOOH (pKa = 3.77) and 0.52 M HCOOK (potassium formate)?

pH = pKa + log [conjugate base][acid]

pH = 3.77 + log [0.52][0.30]

pH = 3.77 + log [0.52][0.30]

pH = 3.77 +0.2388 = 4.00

You wish to prepare a 1.0L buffer with a pH of 3.90. A list of possible acids and their conjugate bases is shown below. Which combination should be selected? Suppose the total concentration of the buffer is 0.5M. What is the concentration of each component?

Acid Conjugate Base Ka

HSO4- SO4- 1.2 x 10-2

HCO2H HCO2- 1.8 x 10-5

HCO3- CO3-2 4.8 x 10-11

HF F- 6.6 x 10-4

You wish to prepare a 1.0L buffer with a pH of 3.90. A list of possible acids and their conjugate bases is shown below. Which combination should be selected? Suppose the total concentration of the buffer is 0.5M. What is the concentration of each component?

2. Use the buffer equation and determine the ratio of base/acid

pH = pKa + log[HCOO-]initial

[HCOOH]initial

3.90 = 4.74 + log [HCOO-]initial

[HCOOH]initial-0.84 = log [HCOO-]initial

[HCOOH]initial

Acid Conjugate Base Ka pKaHSO4- SO4- 1.2 x 10-2 1.92HCO2H HCO2- 1.8 x 10-5 4.74HCO3- CO3-2 4.8 x 10-11 10.32HF F- 6.6 x 10-4 3.18

1. Choose the buffer using pKa of the acid (+/- 1 unit away). CH3COOH -CH3COO- or HF/F are the only choice that fits this criteria.

You wish to prepare a 1.0L buffer with a pH of 3.90. A list of possible acids and their conjugate bases is shown below. Which combination should be selected? Suppose the total concentration of the buffer is 0.5M. What is the concentration of each component?

total concentration = 0.5M = [HCOOH] + [HCOO-]

[HCOO-] = .144[HCOOH]

0.5M = [HCOOH] + 0.144[HCOOH]

0.5M = 1.144[HCOOH]

[HCOOH] = .5M/1.144 = 0.437M

[HCOO-] = .5 - 0.437 = 0.063M

10-0.84 = [HCOO-]initial

[HCOOH]initial= .144

HCO3-(aq) + H2O(l) CO3

2-(aq) + H3O+(aq)Identify the equilibrium and the acid-conjugate base pair and Ka

pH = pKa + log [conjugate base][acid]

pKaHCO3 = - log(KaHCO3) = - log(4.7 x 10-11) = 10.33

An environmental chemist requires a carbonate buffer of pH 10.00. How many grams of Na2CO3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO3 to prepare the buffer? Ka of HCO3

- is 4.7 x 10-11.

CO32-

0.2010-0.33 =

10 = 10.33 + log [HCO3-][CO32-]

[HCO3-][CO32-]-0.33 = log

[CO32-] = 0.20 X 10-0.33 = 0.0935 M

mol CO32- = (1.5 L)(0.0935 mol CO32- /L) = 0.1402 moles CO32-

0.14 mole Na2CO3 X 105.99 gmol Na2CO3

= 15 g Na2CO3

[CO32-] = 0.20 X 10-0.33 = 0.0935 M

We need 1.5L of 0.0935M CO32- . It’s now a molarity problem.

mol Na2CO3 = 0.140 mol CO32- X = 0.14 mol Na2CO32-1 mol of CO32-1 mol Na2CO3

An environmental chemist needs a carbonate buffer of pH 10.00. How many grams of Na2CO3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO3 to prepare the buffer? Ka of HCO3

- is 4.7 x 10-11.

A) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. The pKb of NH3 is 4.75.

NH4+ (aq) + H2O H+ (aq) + NH3 (aq)

pH = pKa + log [NH3][NH4

+]

pKb = 4.75

pH = 9.25 + log [0.30][0.36]

= 9.17

pH = pKa + log [conjugate base][acid]

NH3 + H2O NH4+ (aq) + OH- (aq)

pKa = ?

pKw = pKa + pKb = 14

pKa = 14.00 - pKb pKa = 14.00 - 4.75 = 9.25

A) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system (pKa = 9.25 of NH4+).

pH = pKa + log [NH3][NH4

+]

NH4+ (aq) H+ (aq) + NH3 (aq) pKa = 9.25

pH = 9.25 + log [0.30][0.36]

= 9.17

pH = pKa + log [conjugate base][acid]

A titration is the controlled addition of one substance (titrant) to a known quantity of another substance (titrate) until the stoichiometric requirements are met.

Titrant is added slowlyuntil color change

A buret holds the titrant while the solution to be titrated is prepared in the flask. A dye indicator flags the end point.

1. Strong Acid-Strong Base --- easy to handle--no hydrolysis of either ion. Just use simple stoichiometric principles from Chem 7.

2. Weak Acid-Strong Base --- neutralization followed hydrolysis of the anion (the conjugate base of the weak acid).

3. Strong Acid-Weak Base --- neutralization followd by hydrolysis of the cation (the conjugate acid).

4. Weak Acid-Weak Base ---not typical

5. Polyprotic Acid Neutralization---more complex

Acid-Bases titrations can be classified into roughly 5-types. We care about one.

1. Strong Acid - Strong Base Titration

The key is to recognize strong acids and strong bases and apply your 100% dissociation, solution chemistry and adding volumes.

2-moles OH per mole base

1-moles OH per mole base

2-moles H+ per mole acid

1-moles H+ per mole acid

1. Strong Acid-Strong Base Titrations (0.100M each)

NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)

OH- (aq) + H+ (aq) H2O (l) net-ionic reaction

Suppose we titrate 40.00 mL of 0.100 M HCl with 0.1000 M NaOH and monitor the pH of the solution as a function of volume of NaOH added.

VISUALIZE WHAT IS HAPPENING

1) H+ from HCl is reacting with OH- from NaOH2) the total solution volume is changing as youadd NaOH: key to the calculation! 3) need to consider both together and arrive at equilibrium [H+]

1. Strong Acid-Strong Base Titrations (40 mL of 0.100M HCl)

mL OH- mol OH- mol H+ left Total Volume pH = -log [H]0.00 0.00 0.0040 mol 40 mL 1.0010.00 0.0010 0.0030 mol 50 mL 1.2220.00 0.0020 0.0020 mol 60 mL 1.4830.00 0.0030 0.0010 mol 70 mL 1.8539.00 0.0039 0.0001 mol 79.0 mL 2.8939.50 0.00395 0.00005 mol 79.5 mL 3.2039.75 0.003975 0.000025 mol 79.75 mL 3.5039.95 0.003995 0.000005 mol 79.95 mL 3.9040.00 0.0040 0.00 mol 80.00 mL 7.0040.01 0.004001 0.000001mol 80.01 mL 9.4040.05 0.004005 0.000005 mol 80.05 mL 9.8040.10 0.00401 0.00001 mol 80.10 mL 10.4045.00 0.0045 0.0005 mol 85.00 mL 11.76

Initial Moles of H+ = Volume x Molarity HCl = 0.040 L X 0.100M = 0.00400 mol

1. Strong Acid-Strong Base Titrations (0.100M each)

NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)OH- (aq) + H+ (aq) H2O (l) net-ionic reaction

pH

mL of NaOH added

• Equivalence point:– The point in the reaction at which both acid and base have

been consumed. Neither acid nor base is present in excess.

• End point:– The point at which the pH indicator changes color. It is not

identical to the equivalence point---but close.

• Titrant:– The solution of known concentration (often standardized)

and added to the solution of unknown concentration.

• Titration Curve:– Terminology for a graph or plot of Solution pH vs. volume

of titrant added to the unknown solution.

There are a number of terms used in acid-base titrations that we need to know.

Acid-Base Titration Curves

Many indicators will work in Strong-Strong Titration

A 0.100M NaOH solution is used to titrate 50.0mL of a 0.100M HCl. Calculate the pH of the solution after 0mL, 40mL and 50mL of titrant have been added. What indicator would you use?

Total moles HCl =

A 0.100M NaOH solution is used to titrate 50.0mL of a 0.100M HCl. Calculate the pH of the solution after 0mL, 40mL and 50mL of titrant have been added. What indicator would you use?

Solubility Equilibria (precipitates and equilibrium)

Stalagmite (g--from the ground) Stalactite (c--from the ceiling) Both are precipitated CaCO3 (limestone caves and caverns)

Calcium Oxalate = Kidney Stones

Precipitation Reactions Can Hurt:As of April 16, 2003, Don Winfield of Caledonia, Ontario, Canada, has produced and passed an excruciating 4,504 kidney stones, ranging in size from a grain of sand to a dried pea. Eventually, the kidney had to be removed. "At its worst," says Don, "I produced 22 stones in 24 hours... and 35 over a period of five days."

Large value of Ksp = very soluble (equilibrium is product-favored)

Small value Ksp = very insoluble (equilibrium is reactant-favored)

Chemists use the equilibrium constant, Ksp to characterize the equilibrium between a slightly soluble solid and its saturated aqueous solution.

Al(OH)3 (s) 1Al3+ (aq) + 3OH- (aq) Ksp = [Al3+][OH-]3

When the value of Ksp exceeded a precipitate will form!

Ksp = 1.3 x 10-33

The solubility rules are a “summary” of Ksp values.

Ksp is written like other equilibrium constants. The equilibrium is established when solution is “saturated” whereby no more solute will dissolve (a precipitate).

MgF2 (s) Mg2+(aq) + 2F-(aq) Ksp = [Mg2+][F-]2

Solid = precipitate not included in equilibrium constant

Molarity raised to stoichiometric coefficients don’t include solids in expression!

Balance equation keeping track of stoichiometry!

Write the ion-product expression for each of the following:

Ag2CO3 (s) 2Ag+(aq) + CO32-(aq)

Ca3(PO4)2 (s) 3Ca2+(aq) + 2PO43-(aq)

MgCO3(s) Mg2+(aq) + CO32-(aq)

Fe(OH)2(s) Fe2+(aq) + 2OH- (aq)

Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq)

Write the ion-product expression for each of the following:

Ag2CO3 (s) 2Ag+(aq) + CO32-(aq) Ksp = [Ag+]2[CO3

2-]

Ca3(PO4)2 (s) 3Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3[PO3

3-]2

Ksp = [Mg2+][CO32-]

Ksp = [Fe2+][OH-] 2

Ksp = [Ca2+]3[PO43-]2

MgCO3(s) Mg2+(aq) + CO32-(aq)

(b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq)

Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq)

Q = Ksp

Q < Ksp

Q > Ksp

Saturated solutionUnsaturated solution No precipitate

Supersaturated solution Precipitate will form

Compare Qsp with Ksp.

Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq)

Ksp = [Ag+]2[CO32-]Q = [Ag+]2[CO3

2-] before equilbrium after equilbrium

By comparing Q vs Ksp we can predict whether a precipitate forms.

Molar solubility (mol/L) is the molarity of solute that produces a saturated solution--or a precipitate. It is the value at which Ksp is determined.

Solubility (g/L) is the number of grams of solute per liter of solution that will dissolve to just produce a saturated solution--it is the concentration at which Ksp is determined.

Molar solubilities can be obtained from Ksp and vis versa

Solubility Terms Know

Types of Problems in Solubility

Molar Mass

ICETable Stoichiometry

Molar Mass

Stoichiometry

Molar solubility (mol/L) is the molarity of solute in a solution just needed to produce a saturated solution or a precipitate---and the value at which Ksp is determined.

Solubility (g/L) is the number of grams of solute per liter of solution that will dissolve to just produce a saturated solution--it is the concentration at which Ksp is determined.

Molar solubilities can be obtained from Ksp and vis versa

Precipitation Terms To Have In Your Pocket

Molar Mass

Molar Mass

ICETable Stoich

Stoich

1. Determining Ksp from solubilityWhen lead (II) fluoride (PbF2) is shaken with pure water at 250C, the solubility is found to be 0.64g/L. Calculate the molar solubility and then find the Ksp of PbF2.

Ksp = (2.6x10-3)(5.2x10-3)2 = 7.0x10-8

PbF2(s) Pb2+(aq) + 2F-(aq) Ksp = [Pb2+][F-]2

= 2.6x10-3 M PbF20.64gL soln 245.2g PbF2

mol PbF2Molar Solubility =

[Pb+2] = 2.6x10-3 M PbF21 mol Pb2+

= 2.6x10-3 M Pb2+

1 mol PbF2

[F-] = 2.6x10-3 M PbF2 1 mol PbF2

2 mol F-= 5.2x10-3 M F-

Key part is to be careful with stoichiometry!

Determining Ksp from solubility(a) Lead (II) sulfate is a key component in lead-acid car batteries. Its solubility in water at 250C is 4.25x10-3g/100mL solution. What is the Ksp of PbSO4?

PLAN: Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression.

Determining Ksp from solubility(a) Lead (II) sulfate is a key component in lead-acid car batteries. Its solubility in water at 250C is 4.25x10-3g/100mL solution. What is the Ksp of PbSO4?

Ksp = [Pb2+][SO42-] PbSO4(s) Pb2+(aq) + SO4

2-(aq)(a)

Ksp = [Pb2+][SO42-] = (1.40x10-4) (1.40x10-4) = 1.96x10-8

The stoichiometry for Pb2+ and SO4- are 1:1 with PbSO4

= 1.40x10-4M PbSO44.25x10-3g

100mL soln 303.3g PbSO4

1 mol PbSO41000mLL

Molar Solubility =

Calcium hydroxide is a major component of mortar, plaster and cement, and solutions of Ca(OH)2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH)2 in water at 25 oC if its Ksp is 6.5 x 10-6.

PLAN: Write a dissociation equation and Ksp expression; find the molar solubility (S) using a table.

Determining solubility from Ksp

Calcium hydroxide is a major component of mortar, plaster and cement, and solutions of Ca(OH)2 are used in industry as a cheap, strong base. Calculate the molar solubility of Ca(OH)2 in water at 25 oC if its Ksp is 6.5 x 10-6.

Ksp = (S)(2S)2 = 4S3 = 6.5 x 10-6

Ca(OH)2(s) Ca2+ + 2OH- -initial

changeequilibrium

--

0 0+S + 2SS 2S

concentration (M)

_________________________________________

Ksp = [Ca2+][OH-]2

_________________________________________

S = 3

!6.5! 10!6

4= 1.2! 10!2MCa(OH)2

pH and Solubility of Insoluble Salts

• The presence of a common ion decreases the solubility.• Insoluble bases dissolve in acidic solutions• Insoluble acids dissolve in basic solutions

Mg(OH)2 (s) Mg2+ (aq) + 2OH- (aq)

Ksp = [Mg2+][OH-]2 = 1.2 x 10-11

Ksp = (s)(2s)2 = 4s3

4s3 = 1.2 x 10-11

s = 1.4 x 10-4 M[OH-] = 2s = 2.8 x 10-4 MpOH = 3.55 pH = 10.45

At pH less than 10.45Lower [OH-]

OH- (aq) + H+ (aq) H2O (l)Increase solubility of Mg(OH)2

At pH greater than 10.45Raise [OH-]Decrease solubility of Mg(OH)2

LeChatlier’s Principle: The presence of a stress on a system will cause the system to react to relieve the stress.

A salt becomes less soluble if a common ion is added to an existing sparingly soluble solution(the common ion effect)

Ksp = [Pb2+][I-]2

The solubility of a insoluble ionic salt can be increased by the addition of a strong acid IF the anion of the salt is from a weak acid, or is OH-.

PbF2(s) Pb2+(aq) 2F-

Add HCl or HNO3

Anion of weak acid will be basic in water (hydrolysis) and react with acid if added

2F- + H2O <=> HF + OH-

H2O + Cl- or NO3-

equilibrium shifts

Adding H3O+ reacts with the anion of the weak acid CO32- the equilibrium to the right.

CaCO3(s) Ca2+(aq) + CO32-(aq)

CO32-(aq) + H3O+(aq) HCO3

- (aq) + H2O(l)

HCO3-(aq) + H3O+(aq) H2CO3(aq) + H2O(l)

H2CO3(aq) 2H2O(l) + CO2(g)

If the insoluble compound contains the anion of a weak acid, addition of H3O+ (from a strong acid) increases its solubility (LeChâtelier’s principle)

Predicting the Effect on Solubility of Adding Strong Acid

Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of these ionic compounds:

(a) Lead (II) bromide(b) Copper (II) hydroxide(c) Iron (II) sulfide

Write dissolution equations and consider how strong acid would affect the anion component.

(d) calcium fluoride(e) zinc sulfide

Predicting the Effect on Solubility of Adding Strong AcidWrite balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of these ionic compounds:

Br- is the anion of a strong acid.

No effect.(a) PbBr2(s) Pb2+(aq) + 2Br-(aq)

(b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)H+ will react with OH- shift to the equilibrium right and increase solubility.

(c) FeS(s) Fe2+(aq) + S2-(aq)

S2- is the anion of a weak acid and will react with water to produce OH-.

Both weak acids serve to increase the solubility of FeS.

FeS(s) + H2O(l) Fe2+(aq) + HS-(aq) + OH-(aq)