APPLICATIONS OF APPLICATIONS OF AQUEOUS EQUILIBRIAAQUEOUS EQUILIBRIA

REACTIONS AND EQUILIBRIA REACTIONS AND EQUILIBRIA INVOLVING ACIDS, BASES, INVOLVING ACIDS, BASES,

AND SALTSAND SALTS

Common IonsCommon Ions

Common ion effect Common ion effect - The - The addition addition

of an ion already present of an ion already present (common) (common)

in a system causes equilibrium in a system causes equilibrium to to

shift away from the common ion.shift away from the common ion.

General IdeaGeneral Idea

A + B <=> C + D A + B <=> C + D

If "C" is increased, the equilibrium of If "C" is increased, the equilibrium of

reaction will shift to the reactants reaction will shift to the reactants and and

thus, the amount dissociated thus, the amount dissociated

decreases.decreases.

For example, the addition of For example, the addition of

concentrated HCl to a saturated concentrated HCl to a saturated

solution of NaCl will cause some solution of NaCl will cause some

solid NaCl to precipitate out of solid NaCl to precipitate out of solution.solution.

The NaCl has become less soluble The NaCl has become less soluble

because of the addition of additional because of the addition of additional

chloride ion. This can be explained chloride ion. This can be explained

by the use of LeChatelier's Principle.by the use of LeChatelier's Principle.

NaClNaCl(s) (s) Na Na++(aq)(aq) + Cl + Cl--(aq)(aq)

The addition of a common ion to The addition of a common ion to a a

solution of a weak acid makes solution of a weak acid makes the the

solution less acidic. solution less acidic.

HCHC22HH33OO22 H H++ + C + C22HH33OO22--

If we add NaCIf we add NaC22HH33OO22, equilibrium , equilibrium

shifts to undissociated HCshifts to undissociated HC22HH33OO22, ,

raising pH. The new pH can be raising pH. The new pH can be

calculated by putting the calculated by putting the

concentration of the anion into concentration of the anion into

the Kthe Kaa equation and solving for equation and solving for

the new [Hthe new [H++].].

Understanding common ion Understanding common ion

problems aides understanding problems aides understanding of of

buffer solutions, acid-base buffer solutions, acid-base

indicators and acid-base indicators and acid-base titration titration

problems.problems.

Example problemExample problem

Determine the [HDetermine the [H33OO++] and ] and [C[C22HH33OO22

--] ]

in 0.100 M HCin 0.100 M HC22HH33OO22. .

The KThe Kaa for acetic acid is 1.8 x for acetic acid is 1.8 x 1010-5-5. .

Now determine the [HNow determine the [H33OO++] and ] and

[C[C22HH33OO22--] in a solution that is ] in a solution that is

0.100 M 0.100 M

in both acetic acid and in both acetic acid and hydrochloric hydrochloric

acid. acid.

Summary Summary

1. The common ion causes a 1. The common ion causes a huge huge

decrease in the concentration of decrease in the concentration of the the

acetate ion. acetate ion.

2. Look at the % dissociation of 2. Look at the % dissociation of

acetate ion in each situation. acetate ion in each situation.

(the common ion really (the common ion really decreased decreased

the amount of ionization!)the amount of ionization!)

Exercise 1Exercise 1 Acidic Acidic Solutions Containing Solutions Containing Common IonsCommon Ions

If we know that the equilibrium If we know that the equilibrium

concentration of Hconcentration of H++ in a 1.0 in a 1.0 MM HF HF

solution is 2.7 X 10solution is 2.7 X 10-2-2 MM, and the , and the

percent dissociation of HF is percent dissociation of HF is 2.7% ...2.7% ...

Exercise 1, cont.Exercise 1, cont.

Calculate [HCalculate [H++] and the percent ] and the percent

dissociation of HF in a solution dissociation of HF in a solution

containing 1.0 containing 1.0 MM HF (K HF (Kaa = 7.2 X = 7.2 X 1010-4-4) )

and 1.0 and 1.0 MM NaF. NaF.

SolutionSolution

[H[H++] = 7.2 X 10] = 7.2 X 10-4-4 MM

0.072% 0.072%

Buffered SolutionsBuffered Solutions

Solutions that resist changes in Solutions that resist changes in pH pH

when either OHwhen either OH-- or H or H++ ions are ions are

added. added.

Example:Example:

NHNH33/NH/NH44++ buffer system buffer system

Addition of strong acid: Addition of strong acid:

HH++ + NH + NH33 NH NH44++

Addition of strong base: Addition of strong base:

OHOH-- + NH + NH44++ NH NH33 + H + H22OO

Usually contain a weak acid and its Usually contain a weak acid and its

salt or a weak base and its salt. salt or a weak base and its salt.

Pure water has no buffering Pure water has no buffering

capacity---acids and bases added to capacity---acids and bases added to

water directly affect the pH of the water directly affect the pH of the

solution. solution.

Would acetic acid and sodium Would acetic acid and sodium

acetate be a buffer system? acetate be a buffer system?

Look at the components and Look at the components and how it how it

functions----functions----

ExampleExample

HCHC22HH33OO2 2 / C/ C22HH33OO22--

buffer systembuffer system

Addition of strong acid: Addition of strong acid:

HH++ + C + C22HH33OO22- - HC HC22HH33OO22

Addition of strong base: Addition of strong base:

OHOH-- + HC + HC22HH33OO22

HH22O + CO + C22HH33OO22--

Sure looks like an effective Sure looks like an effective buffer buffer

to me!!to me!!

Buffer capacityBuffer capacity

The amount of acid or base that The amount of acid or base that can can

be absorbed by a buffer system be absorbed by a buffer system

without a significant change in without a significant change in pH. pH.

In order to have a large In order to have a large buffer buffer

capacity, a solution should capacity, a solution should have have

large concentrations of large concentrations of both both

buffer components.buffer components.

Exercise 2Exercise 2 The The pH of a Buffered Solution pH of a Buffered Solution IIA buffered solution contains 0.50A buffered solution contains 0.50M M

acetic acid (HCacetic acid (HC22HH33OO22, K, Kaa = 1.8 X 10 = 1.8 X 10--

55) )

and and

0.50 0.50 MM sodium acetate (NaC sodium acetate (NaC22HH33OO22). ).

Calculate the Calculate the

pH of this pH of this

solution.solution.

SolutionSolution

pH = 4.74pH = 4.74

Exercise 3Exercise 3 pH pH Changes in Buffered Changes in Buffered SolutionsSolutionsCalculate the change in pH that Calculate the change in pH that

occurs when 0.010 mol solid occurs when 0.010 mol solid NaOH NaOH

is added to the buffered is added to the buffered solution solution

described in Sample Exercise 2. described in Sample Exercise 2.

Compare this pH Compare this pH

change with that change with that

which occurs which occurs

when 0.010 mol when 0.010 mol

solid NaOH is solid NaOH is

added to 1.0 L of added to 1.0 L of

water.water.

SolutionSolution

+ 5.00 pH units+ 5.00 pH units

The buffered solution resists The buffered solution resists

changes in pH more than changes in pH more than water.water.

One way to calculate the pH of a One way to calculate the pH of a

buffer system is with thebuffer system is with the Henderson- Henderson-

Hasselbach Hasselbach equation.equation.

pH = pKpH = pKaa + log + log [base][base]

[acid] [acid]

pH = pKpH = pKaa + log + log [A[A--]]

[HA][HA]

Remember conjugate acid/base pairs!Remember conjugate acid/base pairs!

For a particular buffering For a particular buffering system, system,

all solutions that have the all solutions that have the same same

ratio of ratio of [A[A--]/[HA] have the same ]/[HA] have the same pH.pH.

Optimum buffering occurs when Optimum buffering occurs when

[HA] = [A[HA] = [A--] and the pK] and the pKaa of the of the

weak acid used should be as close weak acid used should be as close

as possible to the desired pH of the as possible to the desired pH of the

buffer system.buffer system.

The Henderson-Hasselbach (HH) The Henderson-Hasselbach (HH)

equation needs to be used equation needs to be used

cautiously. cautiously.

It is sometimes used as a quick, It is sometimes used as a quick, easy easy

equation to plug numbers into. equation to plug numbers into.

A KA Kaa or K or Kbb problem requires a problem requires a

greater understanding of the greater understanding of the factors factors

involved and can always be involved and can always be used used

instead of the HH equation. instead of the HH equation.

This equation is only valid for This equation is only valid for solutions that contain weak solutions that contain weak monoprotic acids and their salts or monoprotic acids and their salts or weak bases and their salts. The weak bases and their salts. The buffered solution cannot be too buffered solution cannot be too

dilute and the Kdilute and the Kaa/K/Kbb cannot be too cannot be too large.large.

Hints for Solving Hints for Solving Buffer Problems:Buffer Problems:

(Still use RICE to begin!)(Still use RICE to begin!)

Determine major species involved Determine major species involved

initially.initially.

If chemical reaction occurs, write If chemical reaction occurs, write

equation and solve stoichiometry equation and solve stoichiometry in in

moles, then change to molarity. moles, then change to molarity.

** this is the only extra work!!!** ** this is the only extra work!!!**

Write equilibrium equation.Write equilibrium equation.

Set up equilibrium expression (KSet up equilibrium expression (Kaa

or Kor Kbb) or HH equation.) or HH equation.

Solve.Solve.

Check logic of answer.Check logic of answer.

ExampleExample

A solution is 0.120 M in acetic acid A solution is 0.120 M in acetic acid

and 0.0900 M in sodium acetate. and 0.0900 M in sodium acetate.

Calculate the [HCalculate the [H++] at equilibrium. ] at equilibrium.

The KThe Kaa of acetic acid is 1.8 x 10 of acetic acid is 1.8 x 10-5-5..

HCHC22HH33OO22 H H++ + C + C22HH33OO22--

Initial 0.120 0 0.0900Initial 0.120 0 0.0900

Change -Change -xx + +x x + +xx

Equil. 0.120 - Equil. 0.120 - x x xx 0.0900 0.0900 + + xx

KKaa = = x x (0.0900 +(0.0900 + x x) ) xx (0.0900)(0.0900) = =

0.120 - 0.120 - xx 0.120 0.120

1.8 x 101.8 x 10-5-5

xx = 2.4 x 10 = 2.4 x 10-5-5 M M

[H[H++] = 2.4 x 10] = 2.4 x 10-5-5

Using the Henderson-Using the Henderson-Hasselbach equation:Hasselbach equation:

pKpKaa = -log 1.8 x 10 = -log 1.8 x 10-5-5 = 4.74 = 4.74

pH = 4.74 + log (0.0900/0.120) = pH = 4.74 + log (0.0900/0.120) = 4.624.62

[H[H++] = antilog (-4.62) = 2.4 x 10] = antilog (-4.62) = 2.4 x 10-5-5

ExampleExample

Calculate the pH of the above Calculate the pH of the above buffer buffer

system when 100.0 mL of 0.100 M system when 100.0 mL of 0.100 M

HCl is added to 455 mL of solution.HCl is added to 455 mL of solution.

Here is where all of that Here is where all of that

stoichiometry comes in handy! stoichiometry comes in handy!

0.100 L HCl x 0.100 M = 0.0100 mol H0.100 L HCl x 0.100 M = 0.0100 mol H++

0.455 L C0.455 L C22HH33OO22- - x 0.0900 M = x 0.0900 M =

0.0410 mol C0.0410 mol C22HH33OO22--

0.455 L HC0.455 L HC22HH33OO2 2 x 0.120 M = x 0.120 M =

0.0546 mol HC0.0546 mol HC22HH33OO22

HH++ + C + C22HH33OO22-- HC HC22HH33OO22

Before: Before:

0.0100 mol 0.0410 mol 0.0546 mol 0.0100 mol 0.0410 mol 0.0546 mol

After: After:

0 0.0310 mol 0.0646 mol0 0.0310 mol 0.0646 mol

*remember limiting reagent!*remember limiting reagent!

0.0310 mol acetate / 0.555 L 0.0310 mol acetate / 0.555 L solution = 0.0559 M acetatesolution = 0.0559 M acetate

0.0646 mol acetic acid / 0.555 L 0.0646 mol acetic acid / 0.555 L solution = 0.116 M acetic acidsolution = 0.116 M acetic acid

*recalculate molarity! *recalculate molarity!

Now do the regular Now do the regular equilibrium equilibrium

like in the earlier chapter! like in the earlier chapter!

HCHC22HH33OO22 H H++ + C + C22HH33OO22--

Initial 0.116 M 0 Initial 0.116 M 0 0.0559 M0.0559 M

Change -Change -x x + +x x + +xx

Equil. 0.116 - Equil. 0.116 - xx x x 0.0559 0.0559 + + xx

KKaa = 1.8 x 10 = 1.8 x 10-5-5 = =

xx(0.0559+(0.0559+xx)) x(0.0559)x(0.0559)

- 0.116-- 0.116-xx 0.116 0.116

x = 3.74 x 10x = 3.74 x 10-5-5 M M

pH = 4.43pH = 4.43

Or you could have used HH Or you could have used HH

equation! equation!

Your choice!Your choice!

Same answer!Same answer!

Exercise 4Exercise 4 The pH The pH of a Buffered Solution IIof a Buffered Solution II

Calculate the pH of a solution Calculate the pH of a solution

containing 0.75 containing 0.75 MM lactic acid lactic acid

(K(Kaa = 1.4 X 10 = 1.4 X 10-4-4) and 0.25 ) and 0.25 MM

sodium lactate.sodium lactate.

Lactic acid (HCLactic acid (HC33HH55OO33) is a common ) is a common

constituent of biologic systems. constituent of biologic systems.

For example, it is found in milk For example, it is found in milk and and

is present in human muscle tissue is present in human muscle tissue

during exertion.during exertion.

SolutionSolution

pH = 3.38 pH = 3.38

Exercise 5Exercise 5 The The pH of a Buffered Solution pH of a Buffered Solution IIIIIIA buffered solution contains 0.25 A buffered solution contains 0.25 MM

NHNH33 (K (Kbb = 1.8 X 10 = 1.8 X 10-5-5) and 0.40 ) and 0.40 MM

NHNH44Cl. Cl.

Calculate the pH of this solution.Calculate the pH of this solution.

SolutionSolution

pH = 9.05pH = 9.05

Exercise 6Exercise 6 Adding Adding Strong Acid to a Buffered Strong Acid to a Buffered Solution ISolution ICalculate the pH of the solution Calculate the pH of the solution

that that

results when 0.10 mol gaseous results when 0.10 mol gaseous HCl HCl

is added to 1.0 L of the buffered is added to 1.0 L of the buffered

solution from Sample Exercise 5.solution from Sample Exercise 5.

SolutionSolution

pH = 8.73pH = 8.73

Exercise 7Exercise 7 Adding Adding Strong Acid to a Buffered Strong Acid to a Buffered Solution IISolution IICalculate the change in pH that Calculate the change in pH that

occurs when 0.010 mol occurs when 0.010 mol gaseous HCl gaseous HCl

is added to 1.0 L of each of the is added to 1.0 L of each of the

following solutions:following solutions:

Solution A: Solution A:

5.00 5.00 MM HC HC22HH33OO22 and and

5.00 5.00 MM NaC NaC22HH33OO22

Solution B: Solution B:

0.050 0.050 MM HC HC22HH33OO22 and and

0.050 0.050 MM NaC NaC22HH33OO22

SolutionSolution

A: noneA: none

B: -0.18B: -0.18

Preparing Buffer SolutionsPreparing Buffer Solutions

Usually use (redundant?) 0.10 M Usually use (redundant?) 0.10 M

to 1.0 M solutions of reagents & to 1.0 M solutions of reagents &

choose an acid whose Kchoose an acid whose Kaa is near is near

the [Hthe [H33OO++] concentration we ] concentration we

want. want.

The pKThe pKaa should be as close to should be as close to the the

pH desired as possible. Adjust pH desired as possible. Adjust

the ratio of weak A/B and its the ratio of weak A/B and its salt salt

to fine tune the pH.to fine tune the pH.

It is the relative # of moles of It is the relative # of moles of

acid/CB or base/CA that is acid/CB or base/CA that is

important since they are in the important since they are in the

same solution and share the same solution and share the same same

solution volume. (HH equation solution volume. (HH equation

makes this relatively painless.)makes this relatively painless.)

This allows companies to make This allows companies to make

concentrated versions of buffer concentrated versions of buffer

and let the customer dilute--this and let the customer dilute--this

will not affect the # of moles will not affect the # of moles

present--just the dilution of those present--just the dilution of those

moles.moles.

Buffers are needed in all types of Buffers are needed in all types of

places; places;

especially in biology when doing especially in biology when doing

any type of molecular biology any type of molecular biology work.work.

Exercise 8 Preparing a Exercise 8 Preparing a BufferBuffer

A chemist needs a solution A chemist needs a solution buffered buffered

at pH 4.30 and can choose from at pH 4.30 and can choose from the the

following acids (and their following acids (and their sodium sodium

salts):salts):

a. chloroacetic acid (Ka. chloroacetic acid (Kaa = 1.35 X 10 = 1.35 X 10-3-3))

b. propanoic acid (Kb. propanoic acid (Kaa = 1.3 X 10 = 1.3 X 10-5-5))

c. benzoic acid (Kc. benzoic acid (Kaa = 6.4 X 10 = 6.4 X 10-5-5))

d. hypochlorous acid (Kd. hypochlorous acid (Kaa = 3.5 X 10 = 3.5 X 10-8-8))

Calculate the ratio [HA]/[ACalculate the ratio [HA]/[A--] ]

required for each system to required for each system to yield a yield a

pH of 4.30. pH of 4.30.

Which system will work best?Which system will work best?

SolutionSolution

A: 3.7 X 10A: 3.7 X 10-2-2

B: 3.8B: 3.8

C: 0.78C: 0.78

D: 1.4 X 10D: 1.4 X 1033

Benzoic acid works bestBenzoic acid works best

ACID-BASEACID-BASETITRATIONSTITRATIONS

TitrantTitrant

solution of known solution of known concentration concentration

(in buret)(in buret)

The titrant is added to a solution of The titrant is added to a solution of

unknown concentration until the unknown concentration until the

substance being analyzed is just substance being analyzed is just

consumed.consumed.

(stoichiometric point or (stoichiometric point or equivalence equivalence

point)point)

The equivalence point is when The equivalence point is when

moles of acid = moles of base.moles of acid = moles of base.

The endpoint of the titration is The endpoint of the titration is

when the indicator changes when the indicator changes color. color.

This entire concept is known as This entire concept is known as

volumetric analysis! volumetric analysis!

If the indicator has been chosen If the indicator has been chosen

properly, the two will appear to properly, the two will appear to be be

the same. the same.

pH or Titration CurvepH or Titration Curve

Plot of pH as a function of the Plot of pH as a function of the

amount of titrant added.amount of titrant added.

Very beneficial in calculating Very beneficial in calculating the the

equivalence point. equivalence point.

Be able to sketch and label Be able to sketch and label these these

for all types of situations. for all types of situations.

TYPES OF ACID-BASE TYPES OF ACID-BASE

TITRATIONS TITRATIONS

Strong Acid + Strong Strong Acid + Strong BaseBase

Net ionic reaction: Net ionic reaction:

HH++ + OH + OH-- H H22OO

The pH is easy to calculate because The pH is easy to calculate because

all reactions go to completion. all reactions go to completion.

At the equivalence point, the At the equivalence point, the

solution is neutral. solution is neutral.

(pH = 7.00)(pH = 7.00)

No equilibrium here ; No equilibrium here ; stoichiometrystoichiometry

Before Before Equivalence Equivalence

Point:Point:

pH determined pH determined

by taking the log by taking the log

of the moles of of the moles of

HH++ left after left after

reaction divided by reaction divided by

total volume in total volume in

container.container.

Weak Acid with Strong Weak Acid with Strong BasBasee

These These problems problems

are easily are easily broken broken

down into two down into two steps:steps:

Stoichiometry ProblemStoichiometry Problem

After reaction, you must know After reaction, you must know

concentration of all substances concentration of all substances left.left.

Equilibrium ProblemEquilibrium Problem

The position of the weak acid The position of the weak acid

equilibrium must be equilibrium must be determined. determined.

Often these are referred to as a Often these are referred to as a

series of "buffer" problems.series of "buffer" problems.

Points to Ponder:Points to Ponder:

The reaction of a strong base The reaction of a strong base with with

a weak acid is assumed to go a weak acid is assumed to go to to

completion. completion.

Before the equivalence point, Before the equivalence point, the the

concentration of weak acid concentration of weak acid

remaining and the conjugate remaining and the conjugate base base

formed are determined.formed are determined.

Halfway to the equivalence Halfway to the equivalence

point, [HA] = [Apoint, [HA] = [A--] ]

At this halfway point, KAt this halfway point, Kaa = [H = [H++]]

So, the pH = pKSo, the pH = pKaa

**Remembering this can be a **Remembering this can be a

real time saver! real time saver!

At the equivalence At the equivalence

point, the pH > 7. point, the pH > 7.

The anion of the The anion of the

acid remains in acid remains in

solution and it is a solution and it is a

basic salt.basic salt.

The pH at the equivalence point The pH at the equivalence point is is

determined by the Kdetermined by the Kaa. .

The smaller the KThe smaller the Kaa value, the value, the

higher the pH at the equivalence higher the pH at the equivalence

point.point.

After the equivalence point, the pH After the equivalence point, the pH is is

determined directly by excess OHdetermined directly by excess OH-- in in

solution. solution.

A simple pH calculation can be A simple pH calculation can be made made

after the stoichiometry is done.after the stoichiometry is done.

Strong Acids with Weak Strong Acids with Weak BasesBases

Problems of Problems of this this

type work very type work very

similar to the similar to the

weak acid with weak acid with

strong base.strong base.

Before the equivalence point, a Before the equivalence point, a

weak base equilibria exists. weak base equilibria exists.

Calculate the stoichiometry and Calculate the stoichiometry and then the weak base equilibria. then the weak base equilibria.

The equivalence point will always The equivalence point will always

be less than 7 because of the be less than 7 because of the

presence of an acidic salt. presence of an acidic salt.

After equivalence point, the pH is After equivalence point, the pH is

determined by excess [Hdetermined by excess [H++] in ] in solution.solution.

Example Titration Example Titration ProblemsProblems

What is the pH at each of the What is the pH at each of the following points in the following points in the titration of 25.00 mL of 0.100 titration of 25.00 mL of 0.100 M HCl by 0.100 M NaOH? M HCl by 0.100 M NaOH?

1. Before the addition of any NaOH? 1. Before the addition of any NaOH?

2. After the addition of 24.00 mL of 2. After the addition of 24.00 mL of 0.100 M NaOH? 0.100 M NaOH?

3. At the equivalence point.3. At the equivalence point.

4. After the addition of 26.00 mL of 4. After the addition of 26.00 mL of 0.100 M NaOH?0.100 M NaOH?

What is the pH at each of the What is the pH at each of the following points in the following points in the titration of 25.00 mL of 0.100 titration of 25.00 mL of 0.100 M HCM HC22HH33OO22 by 0.100 M NaOH? by 0.100 M NaOH?

1. Before addition of any NaOH? 1. Before addition of any NaOH?

2. After addition of 10.00 mL of 2. After addition of 10.00 mL of 0.100 M NaOH? 0.100 M NaOH?

3. After addition of 12.5 mL of 3. After addition of 12.5 mL of 0.100 M NaOH?0.100 M NaOH?

4. At the equivalence point. 4. At the equivalence point.

5. After the addition of 26.00 mL 5. After the addition of 26.00 mL of of

0.100 M NaOH?0.100 M NaOH?

Exercise 9Exercise 9 Titration of a Weak Titration of a Weak AcidAcidHydrogen cyanide gas (HCN), a Hydrogen cyanide gas (HCN), a

powerful respiratory inhibitor, is powerful respiratory inhibitor, is

highly toxic. highly toxic.

It is a very weak acid (KIt is a very weak acid (Kaa = 6.2 X = 6.2 X

1010-10-10) when dissolved in water.) when dissolved in water.

If a 50.0-mL sample of 0.100 If a 50.0-mL sample of 0.100 MM

HCN is titrated with 0.100 HCN is titrated with 0.100 MM NaOH, NaOH,

calculate the pH of the solution:calculate the pH of the solution:

a. after 8.00 mL of 0.100 M NaOH a. after 8.00 mL of 0.100 M NaOH

has been added.has been added.

b. at the halfway point of the b. at the halfway point of the

titration.titration.

c. at the equivalence point of the c. at the equivalence point of the

titration.titration.

SolutionSolution

A: pH = 8.49A: pH = 8.49

B: pH = 9.21B: pH = 9.21

C: pH = 10.96C: pH = 10.96

Exercise 10Exercise 10 Calculating K Calculating Kaa

A chemist has synthesized a A chemist has synthesized a

monoprotic weak acid and monoprotic weak acid and wants to wants to

determine its Kdetermine its Kaa value. value.

To do so, the chemist dissolves To do so, the chemist dissolves

2.00 mmol of the solid acid in 2.00 mmol of the solid acid in 100.0 100.0

mL water and titrates the mL water and titrates the resulting resulting

solution with 0.0500 solution with 0.0500 MM NaOH. NaOH.

After 20.0 mL NaOH has been After 20.0 mL NaOH has been

added, the pH is 6.00. added, the pH is 6.00.

What is the KWhat is the Kaa value for the value for the acid?acid?

SolutionSolution

[H[H++] = K] = Kaa = 1.0 X 10 = 1.0 X 10-6-6

ACID-BASE ACID-BASE INDICATORSINDICATORS

IndicatorIndicator

a substance a substance

that changes that changes

color in color in some some

known pH known pH

range.range.

HIn + HHIn + H22O O H H33OO++ + In + In--

Indicators are usually weak acids, Indicators are usually weak acids,

HIn. HIn.

They have one color in their acidic They have one color in their acidic

(HIn) form and another color in (HIn) form and another color in

their basic (Intheir basic (In--) form. ) form.

Usually 1/10 of the initial form of Usually 1/10 of the initial form of

the indicator must be changed to the indicator must be changed to

the other form before a new color the other form before a new color is is

apparent.apparent.

……very little effect on overall pH of very little effect on overall pH of

reaction.reaction.

End PointEnd Point

point in titration where indicator point in titration where indicator

changes colorchanges color

When choosing an indicator, we When choosing an indicator, we

want the indicator end point and want the indicator end point and

the titration equivalence point to the titration equivalence point to

be as close as possible.be as close as possible.

Since strong acid-strong base Since strong acid-strong base

titrations have a large vertical titrations have a large vertical area, area,

color changes will be sharp and color changes will be sharp and a a

wide range of indicators can be wide range of indicators can be

used. used.

For titrations involving weak For titrations involving weak

acids or weak bases, we acids or weak bases, we must must

be more careful in our be more careful in our choice of choice of

indicator.indicator.

A very common indicator, A very common indicator,

phenolphthalein, is colorless in its phenolphthalein, is colorless in its

HIn form and pink in its InHIn form and pink in its In-- form. form.

It changes color in the range of It changes color in the range of pH pH

8-10.8-10.

The following equations can The following equations can be be

used to determine the pH at used to determine the pH at

which an indicator will which an indicator will change change

color:color:

For Titration of an Acid:For Titration of an Acid:

pH = pKpH = pKaa + log 1/10 = pK + log 1/10 = pKaa--11

For Titration of a Base:For Titration of a Base:

pH = pKpH = pKaa + log 10/1 = pK + log 10/1 = pKaa+1+1

The useful range of an indicator is The useful range of an indicator is

usually its pKusually its pKaa ±1. ±1.

When choosing an indicator, When choosing an indicator,

determine the pH at the equivalence determine the pH at the equivalence

point of the titration and then choose point of the titration and then choose

an indicator with a pKan indicator with a pKaa close to that. close to that.

Exercise 11Exercise 11 Indicator Color ChangeIndicator Color Change

Bromthymol blue, an indicator Bromthymol blue, an indicator with with

a Ka Kaa value of 1.0 X 10 value of 1.0 X 10-7-7, is , is yellow yellow

in its HIn form and blue in its Inin its HIn form and blue in its In-- form. form.

Suppose we put a few drops of this Suppose we put a few drops of this

indicator in a strongly acidic indicator in a strongly acidic

solution. If the solution is then solution. If the solution is then

titrated with NaOH, at what pH will titrated with NaOH, at what pH will

the indicator color change first be the indicator color change first be

visible?visible?

SolutionSolution

pH = 6.00pH = 6.00

Solubility Euilibria Solubility Euilibria (The Solubility Product)(The Solubility Product)

Saturated solutions of salts are Saturated solutions of salts are

another type of chemical another type of chemical equilibria.equilibria.

Slightly soluble salts establish Slightly soluble salts establish a a

dynamic equilibrium with the dynamic equilibrium with the

hydrated cations and anions hydrated cations and anions in in

solution. solution.

When the solid is first added to When the solid is first added to

water, no ions are initially water, no ions are initially present. present.

As dissolution proceeds, the As dissolution proceeds, the

concentration of ions increases concentration of ions increases

until equilibrium is established. until equilibrium is established.

This occurs when the solution This occurs when the solution is is

saturated.saturated.

The equilibrium constant, the The equilibrium constant, the

KKspsp, is no more than the product , is no more than the product of of

the ions in solution. the ions in solution.

(Remember, solids do not (Remember, solids do not

appear in equilibrium appear in equilibrium expressions.)expressions.)

For a saturated solution of For a saturated solution of

AgCl, the equation would AgCl, the equation would be: be:

AgCl AgCl (s)(s) Ag Ag+ + (aq)(aq) + Cl + Cl- - (aq)(aq)

The solubility product The solubility product expression expression

would be:would be:

KKspsp = [Ag = [Ag++] [Cl] [Cl--]]

The AgClThe AgCl(s)(s) is left out since is left out since

solids are left out of solids are left out of equilibrium equilibrium

expressions (constant expressions (constant

concentrations).concentrations).

You can find loads of You can find loads of KKspsp’s on tables.’s on tables.

Find the KFind the Kspsp values & write values & write the the

KKspsp expression for the expression for the following:following:

CaFCaF2(s)2(s) Ca Ca+2+2 + 2 F + 2 F- - KKspsp = =

AgAg22SOSO4(s)4(s) 2 Ag 2 Ag++ + SO + SO44-2 -2 KKspsp = =

BiBi22SS3(s)3(s) 2 Bi 2 Bi+3+3 + 3 S + 3 S-2 -2 KKspsp = =

Determining KDetermining Kspsp From From Experimental Experimental MeasurementsMeasurements

In practice, KIn practice, Kspsp’s are determined ’s are determined

by careful laboratory by careful laboratory

measurements using various measurements using various

spectroscopic methods.spectroscopic methods.

Remember STOICHIOMETRY!!Remember STOICHIOMETRY!!

ExampleExample

Lead (II) chloride dissolves to a Lead (II) chloride dissolves to a

slight extent in water according slight extent in water according

to the equation:to the equation:

PbClPbCl22 Pb Pb+2+2 + 2Cl + 2Cl--

Calculate the KCalculate the Kspsp if the lead if the lead ion ion

concentration has been found concentration has been found to to

be 1.62 x 10be 1.62 x 10-2-2M.M.

SolutionSolution

If lead’s concentration is “x” ,If lead’s concentration is “x” ,then chloride’s concentration is then chloride’s concentration is ““2x”. 2x”.

So. . . . So. . . .

KKspsp = (1.62 x 10 = (1.62 x 10-2-2)(3.24 x 10)(3.24 x 10-2-2))22 = 1.70 x 10= 1.70 x 10-5-5

Exercise 12 Exercise 12 Calculating KCalculating Kspsp from from Solubility ISolubility ICopper(I) bromide has a Copper(I) bromide has a

measured measured

solubility of 2.0 X 10solubility of 2.0 X 10-4-4 mol/L at mol/L at

25°C. 25°C.

Calculate its KCalculate its Kspsp value. value.

SolutionSolution

KKspsp = 4.0 X 10 = 4.0 X 10-8-8

Exercise 13Exercise 13 Calculating Calculating Ksp from Solubility IIKsp from Solubility II

Calculate the KCalculate the Kspsp

value for bismuth value for bismuth

sulfide (Bisulfide (Bi22SS33), ), which which

has a solubility of has a solubility of

1.0 X 101.0 X 10-15-15 mol/L at mol/L at

25°C.25°C.

SolutionSolution

KKspsp = 1.1 X 10 = 1.1 X 10-73-73

ESTIMATING SALTESTIMATING SALT

SOLUBILITY FROM KSOLUBILITY FROM Kspsp

ExampleExample

The KThe Kspsp for CaCO for CaCO33 is 3.8 x 10 is 3.8 x 10-9-9 @ @

25°C. 25°C.

Calculate the solubility of calcium Calculate the solubility of calcium

carbonate in pure water in carbonate in pure water in

a) moles per liter a) moles per liter

b) grams per literb) grams per liter

The relative solubilities can be The relative solubilities can be deduced by comparing values of deduced by comparing values of KKspsp..

BUT, BE CAREFUL!BUT, BE CAREFUL!

These comparisons can only be These comparisons can only be made for salts having the same made for salts having the same ION:ION ratio.ION:ION ratio.

Please don’t forget solubility Please don’t forget solubility changes with temperature! changes with temperature!

Some substances become less Some substances become less soluble in cold while some soluble in cold while some become become

more soluble! more soluble!

Aragonite.Aragonite.

Exercise 14Exercise 14 Calculating Calculating Solubility from KSolubility from Kspsp

The KThe Kspsp value for copper(II) value for copper(II) iodate, iodate,

Cu(IOCu(IO33))22, is 1.4 X 10, is 1.4 X 10-7-7 at 25°C. at 25°C.

Calculate its solubility at 25°C.Calculate its solubility at 25°C.

SolutionSolution

= 3.3 X 10= 3.3 X 10-3-3 mol/L mol/L

Exercise 15Exercise 15 Solubility Solubility and Common Ionsand Common Ions

Calculate the solubility of solid Calculate the solubility of solid CaFCaF22

(K(Kspsp = 4.0 X 10 = 4.0 X 10-11-11) )

in a 0.025 in a 0.025 MM NaF solution. NaF solution.

SolutionSolution

= 6.4 X 10= 6.4 X 10-8-8 mol/L mol/L

KKspsp and the Reaction and the Reaction Quotient, QQuotient, Q

With some knowledge of the With some knowledge of the

reaction quotient, we can decidereaction quotient, we can decide

1) whether a ppt will form, AND 1) whether a ppt will form, AND

2) what concentrations of ions 2) what concentrations of ions

are required to begin the ppt. of are required to begin the ppt. of

an insoluble salt.an insoluble salt.

1. Q < K1. Q < Kspsp, the system , the system is notis not at at equil. (equil. (ununsaturated)saturated)

2. Q = K2. Q = Kspsp, the system , the system isis at at equil. (saturated)equil. (saturated)

3. Q > K3. Q > Kspsp, the system , the system is notis not at at equil. (equil. (supersupersaturatedsaturated))

Precipitates form when the Precipitates form when the

solution is supersaturated!!!solution is supersaturated!!!

Precipitation of Insoluble Precipitation of Insoluble SaltsSalts

Metal-bearing ores often contain Metal-bearing ores often contain

the metal in the form of an the metal in the form of an

insoluble salt, and, to complicate insoluble salt, and, to complicate

matters, the ores often contain matters, the ores often contain

several such metal salts.several such metal salts.

Dissolve the metal salts to obtain Dissolve the metal salts to obtain

the metal ion, concentrate in some the metal ion, concentrate in some

manner, and ppt. selectively only manner, and ppt. selectively only

one type of metal ion as an one type of metal ion as an

insoluble salt.insoluble salt.

Exercise 16Exercise 16 Determining Determining Precipitation ConditionsPrecipitation Conditions

A solution is prepared by adding A solution is prepared by adding

750.0 mL of 4.00 X 10750.0 mL of 4.00 X 10-3-3 MM Ce(NO Ce(NO33))33

to 300.0 mL of 2.00 X 10to 300.0 mL of 2.00 X 10-2-2 MM KIO KIO33. .

Will Ce(IOWill Ce(IO33))33 (K (Kspsp = 1.9 X 10 = 1.9 X 10-10-10) )

precipitate from this solution?precipitate from this solution?

SolutionSolution

yesyes

Exercise 17Exercise 17 PrecipitationPrecipitation

A solution is prepared by mixing A solution is prepared by mixing

150.0 mL of 1.00 X 10150.0 mL of 1.00 X 10-2-2 M M Mg(NO Mg(NO33))22

and 250.0 mL of 1.00 X 10and 250.0 mL of 1.00 X 10-1-1 MM NaF. NaF.

Calculate the concentrations of Calculate the concentrations of MgMg2+2+

and Fand F-- at equilibrium with solid MgF at equilibrium with solid MgF22

(K(Kspsp = 6.4 X 10 = 6.4 X 10-9-9).).

SolutionSolution

[Mg[Mg2+2+] = 2.1 X 10] = 2.1 X 10-6-6 MM

[F[F--] = 5.50 X 10] = 5.50 X 10-2-2 MM

SOLUBILITY AND THE SOLUBILITY AND THE COMMON ION EFFECTCOMMON ION EFFECT

Experiment shows that the Experiment shows that the

solubility of any salt is always solubility of any salt is always less less

in the presence of a in the presence of a “common “common

ion”.ion”.

LeChatelier’s Principle, that’s LeChatelier’s Principle, that’s why! why!

Be reasonable and use Be reasonable and use

approximations when you can!!approximations when you can!!

Just remember what Just remember what happened happened

earlier with acetic acid and earlier with acetic acid and

sodium acetate. sodium acetate.

The same idea here!The same idea here!

pH can also affect solubility. pH can also affect solubility.

Evaluate the equation to see Evaluate the equation to see who who

would want to “react” with the would want to “react” with the

addition of acid or base. addition of acid or base.

Would magnesium hydroxide be Would magnesium hydroxide be

more soluble in an acid or a more soluble in an acid or a base? base?

Why? Why?

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

(aq)(aq)

(milk of magnesia)(milk of magnesia)

Why Would I Ever Care Why Would I Ever Care About KAbout Kspsp ??? ???

Keep reading to find out ! Keep reading to find out !

Actually, very useful stuff!Actually, very useful stuff!

Solubility, Solubility, Ion Separations, and Ion Separations, and Qualitative AnalysisQualitative Analysis

……introduce you to some basic introduce you to some basic

chemistry of various ions.chemistry of various ions.

……illustrate how the principles illustrate how the principles of of

chemical equilibria can be chemical equilibria can be applied.applied.

Objective:Objective:

Separate the Separate the

following following

metal ions: metal ions:

silver, silver,

lead, lead,

cadmium and cadmium and

nickelnickel

From solubility rules, lead and From solubility rules, lead and silver silver

chloride will ppt, so add dilute chloride will ppt, so add dilute HCl. HCl.

Nickel and cadmium will stay in Nickel and cadmium will stay in

solution.solution.

Separate by filtration: Separate by filtration:

Lead chloride will dissolve in HOT Lead chloride will dissolve in HOT

water… water…

filter while HOT and those two will filter while HOT and those two will

be separate.be separate.

Cadmium and nickel are more Cadmium and nickel are more

subtle. subtle.

Use their KUse their Kspsp’s with sulfide ion. ’s with sulfide ion.

Who ppt’s first???Who ppt’s first???

Exercise 18Exercise 18 Selective Precipitation Selective Precipitation

A solution contains 1.0 X 10A solution contains 1.0 X 10-4-4 MM Cu Cu++

and 2.0 X 10and 2.0 X 10-3-3 MM Pb Pb2+2+. .

If a source of IIf a source of I-- is added gradually to is added gradually to

this solution, will PbIthis solution, will PbI22 (K (Kspsp = 1.4 X = 1.4 X

1010-8-8) or CuI (K) or CuI (Kspsp = 5.3 X 10 = 5.3 X 10-12-12) )

precipitate first? precipitate first?

Specify the concentration of ISpecify the concentration of I--

necessary to begin necessary to begin precipitation of precipitation of

each salt.each salt.

SolutionSolution

CuI will precipitate first.CuI will precipitate first.

Concentration in excess of Concentration in excess of

5.3 X 105.3 X 10-8-8 MM required. required.

**If this gets you interested, lots **If this gets you interested, lots

more information on this topic in more information on this topic in

the chapter. the chapter.

Good bedtime reading for Good bedtime reading for

descriptive chemistry! descriptive chemistry!

THE EXTENT OF LEWIS THE EXTENT OF LEWIS ACID-BASE ACID-BASE

REACTIONS: REACTIONS:

FORMATION CONSTANTSFORMATION CONSTANTS

When a metal ion (a Lewis acid) When a metal ion (a Lewis acid)

reacts with a Lewis base, a reacts with a Lewis base, a

complex ion can form. complex ion can form.

The formation of complex ions The formation of complex ions

represents a reversible equilibria represents a reversible equilibria situation. situation.

A complex ion is a charged A complex ion is a charged

species consisting of a metal species consisting of a metal ion ion

surrounded by ligands.surrounded by ligands.

A ligand is typically an anion or A ligand is typically an anion or neutral molecule that has an neutral molecule that has an unshared electron pair that can unshared electron pair that can be shared with an empty metal be shared with an empty metal ion orbital to form a metal-ligand ion orbital to form a metal-ligand bond. bond.

Some common ligands are Some common ligands are

HH22O, NHO, NH33, Cl, Cl--, and CN, and CN--. .

The number of ligands attached The number of ligands attached to to

the metal ion is the coordination the metal ion is the coordination

number. number.

The most common coordination The most common coordination

numbers are: 6, 4, 2 numbers are: 6, 4, 2

Metal ions add ligands one at a Metal ions add ligands one at a

time in steps characterized by time in steps characterized by

equilibrium constants called equilibrium constants called

formation constantsformation constants..

AgAg++ + 2NH + 2NH33 [Ag(NH [Ag(NH33))22]]+2+2

acid baseacid base

Stepwise Reactions:Stepwise Reactions:

AgAg++((aqaq)) + NH + NH3(3(aqaq)) Ag(NH Ag(NH33))++

((aqaq))

KKf1f1 = 2.1 x 10 = 2.1 x 1033

[Ag(NH[Ag(NH33))++] ] = 2.1 x 10 = 2.1 x 1033

[Ag[Ag++][NH][NH33] ]

Ag(NHAg(NH33))++ +NH +NH3(3(aqaq)) Ag(NH Ag(NH33))22++

((aqaq))

KKf2f2 = 8.2 x 10 = 8.2 x 1033

[Ag(NH[Ag(NH33))22++] ] = 8.2 x 10 = 8.2 x 1033

[Ag(NH[Ag(NH33))++][NH][NH33]]

In a solution containing Ag and In a solution containing Ag and

NHNH33, all of the species NH, all of the species NH33, Ag, Ag++, ,

Ag(NHAg(NH33))++, and Ag(NH, and Ag(NH33))22++ exist at exist at

equilibrium. equilibrium.

Actually, metal ions in aqueous Actually, metal ions in aqueous

solution are hydrated.solution are hydrated.

More accurate representations would More accurate representations would

be be

Ag(HAg(H22O)O)22++ instead of Ag instead of Ag++, and , and

Ag(HAg(H22O)(NHO)(NH33))++ instead of Ag(NH instead of Ag(NH33))++. .

The equations would be:The equations would be:

Ag(HAg(H22O)O)22++

(aq) (aq) + NH+ NH33(aq) (aq)

Ag(HAg(H22O)(NHO)(NH33))++(aq)(aq) + H+ H22OO(l)(l)

KKf1f1 = 2.1 x 10 = 2.1 x 1033

[Ag(H[Ag(H22O)(NHO)(NH33))++] ] = 2.1 x 10 = 2.1 x 1033

[Ag(H[Ag(H22O)O)22++][NH][NH33]]

Ag(HAg(H22O)(NHO)(NH33))++(aq)(aq) + NH + NH33(aq)(aq)

Ag(NHAg(NH33))22++

(aq)(aq) + 2H + 2H22OO(l)(l)

KKf2f2 = 8.2 x 10 = 8.2 x 1033

[Ag(NH[Ag(NH33))22++] ] = 8.2 x 10 = 8.2 x 1033

[Ag(H [Ag(H22O)(NHO)(NH33))++][NH][NH33]]

The sum of the equations gives The sum of the equations gives the the

overall equation, so the product overall equation, so the product of of

the individual formation constants the individual formation constants

gives the overall formation gives the overall formation constant: constant:

AgAg++ + 2NH + 2NH3 3 Ag(NH Ag(NH33))22++

or or Ag(HAg(H22O)O)22

++ + 2NH + 2NH33 Ag(NH Ag(NH33))22++ + +

2H2H22OO

KKf1 f1 x K x Kf2f2 = K = Kff

(2.1 x 10(2.1 x 1033) x (8.2 x 10) x (8.2 x 1033) = 1.7 x 10) = 1.7 x 1077

Exercise 19Exercise 19

Calculate the equilibrium Calculate the equilibrium

concentrations of Cuconcentrations of Cu+2+2, NH, NH33, and , and

[Cu(NH[Cu(NH33))44]]+2+2 when 500. mL of 3.00 M when 500. mL of 3.00 M

NHNH33 are mixed with 500. mL of 2.00 x are mixed with 500. mL of 2.00 x

1010-3-3 M Cu(NO M Cu(NO33))22. .

KKformationformation = 6.8 x 10 = 6.8 x 101212..

Solubility Solubility and and Complex Complex IonsIons

Complex ions are often insoluble Complex ions are often insoluble in in

water. water.

Their formation can be used to Their formation can be used to

dissolve otherwise insoluble salts. dissolve otherwise insoluble salts.

Often as the complex ion forms, Often as the complex ion forms,

the equilibrium shifts to the right the equilibrium shifts to the right

and causes the insoluble salt to and causes the insoluble salt to

become more soluble. become more soluble.

If sufficient aqueous ammonia If sufficient aqueous ammonia is is

added to silver chloride, the added to silver chloride, the latter latter

can be dissolved in the form of can be dissolved in the form of

[Ag(NH[Ag(NH33))22]]++..

AgClAgCl(s)(s) Ag Ag++(aq)(aq) + Cl + Cl--(aq)(aq)

KKspsp = 1.8 x 10 = 1.8 x 10-10-10

AgAg++(aq)(aq) + 2 NH + 2 NH3(aq)3(aq) [Ag(NH [Ag(NH33))22]]++

(aq)(aq)

KKformationformation = 1.6 x 10 = 1.6 x 1077

Sum:Sum:

K = KK = Kspsp x K x Kformation formation = 2.0 x 10 = 2.0 x 10-3-3 = =

{[Ag(NH{[Ag(NH33))22++}[Cl}[Cl--] ]

[NH[NH33]]22

The equilibrium constant for The equilibrium constant for

dissolving silver chloride in dissolving silver chloride in ammonia ammonia

is not large; however, if the is not large; however, if the

concentration of ammonia is concentration of ammonia is

sufficiently high, the complex ion sufficiently high, the complex ion

and chloride ion must also be high, and chloride ion must also be high,

and silver chloride will dissolve.and silver chloride will dissolve.

Exercise 20Exercise 20 Complex IonsComplex Ions

Calculate the concentrations of Calculate the concentrations of

AgAg++, Ag(S, Ag(S22OO33))--, and Ag(S, and Ag(S22OO33))223-3- in a in a

solution prepared by mixing 150.0 solution prepared by mixing 150.0

mL of 1.00 X 10mL of 1.00 X 10-3-3 MM AgNO AgNO33 with with

200.0 mL of 5.00 200.0 mL of 5.00 MM Na Na22SS22OO33..

The stepwise formation The stepwise formation equilibria are:equilibria are:

AgAg++ + S + S22OO332-2- Ag(S Ag(S22OO33))--

KK11 = 7.4 X 10 = 7.4 X 1088

Ag(SAg(S22OO33))-- + S + S22OO332- 2- Ag(S Ag(S22OO33))22

3-3-

KK22 = 3.9 X 10 = 3.9 X 1044

SolutionSolution

[Ag[Ag++] = 1.8 X 10] = 1.8 X 10-18-18 MM

[Ag(S[Ag(S22OO33))--] = 3.8 X 10] = 3.8 X 10-9-9 MM

ACID-BASE AND PPT ACID-BASE AND PPT EQUILIBRIA OF PRACTICAL EQUILIBRIA OF PRACTICAL

SIGNIFICANCESIGNIFICANCE

SOLUBILITY OF SALTS IN WATER SOLUBILITY OF SALTS IN WATER

AND ACIDSAND ACIDS

The solubility of PbS in The solubility of PbS in water:water:

PbS PbS (s)(s) Pb Pb+2+2 + S + S-2-2

KKspsp = 8.4 x 10 = 8.4 x 10-28-28

The Hydrolysis of the SThe Hydrolysis of the S-2-2 ion in Waterion in Water

SS-2-2 + H + H22O O HS HS-- + OH + OH--

KKbb = 0.077 = 0.077

Overall Process:Overall Process:

PbS + HPbS + H22O O Pb Pb+2+2 + HS + HS-- + OH + OH--

KKtotaltotal = K = Kspsp x K x Kbb = 6.5 x 10 = 6.5 x 10-29-29

May not seem like much, but it can May not seem like much, but it can

increase the environmental lead increase the environmental lead

concentration by a factor of about concentration by a factor of about

10,000 over the solubility of PbS 10,000 over the solubility of PbS

calculated from simply Kcalculated from simply Kspsp!!

Any salt containing an anion Any salt containing an anion that is that is

the conjugate base of a weak the conjugate base of a weak acid acid

will dissolve in water to a will dissolve in water to a greater greater

extent than given by the Kextent than given by the Kspsp. .

This means salts of sulfate, This means salts of sulfate,

phosphate, acetate, carbonate, phosphate, acetate, carbonate, and and

cyanide, as well as sulfide can be cyanide, as well as sulfide can be

affected. affected.

If a strong acid is added to If a strong acid is added to water-insoluble salts such as ZnS water-insoluble salts such as ZnS

or CaCOor CaCO33, then hydroxide ions from , then hydroxide ions from the anion hydrolysis is removed by the anion hydrolysis is removed by the formation of water. the formation of water.

This shifts the anion hydrolysis This shifts the anion hydrolysis further to the right; the weak acid further to the right; the weak acid is is

formed and the salt dissolves.formed and the salt dissolves.

Carbonates and many metal Carbonates and many metal sulfides along with metal sulfides along with metal hydroxides are generally soluble hydroxides are generally soluble in strong acids. in strong acids.

The only exceptions are sulfides The only exceptions are sulfides of mercury, copper, cadmium and of mercury, copper, cadmium and a few others.a few others.

Insoluble inorganic salts containing Insoluble inorganic salts containing

anions derived from weak acids anions derived from weak acids

tend to be soluble in solutions of tend to be soluble in solutions of

strong acids. strong acids.

Salts are not soluble in strong acid Salts are not soluble in strong acid

if the anion is the conjugate base of if the anion is the conjugate base of

a strong acid!!a strong acid!!