Upload
saima
View
86
Download
0
Embed Size (px)
DESCRIPTION
APPLICATIONS OF AQUEOUS EQUILIBRIA. REACTIONS AND EQUILIBRIA INVOLVING ACIDS, BASES, AND SALTS. Common Ions. Common ion effect - The addition of an ion already present (common) in a system causes equilibrium to shift away from the common ion. General Idea. A + B C + D - PowerPoint PPT Presentation
Citation preview
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 that Any salt containing an anion that is 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!!