Class notes

9
%ct,ss: q,z×ÿO-ÿxÿOO = 3,ÿoxÿo;
O,?AO b-ÿ
Applications of Aqueous Equilibria
I. Common Ion Effect
A. Objectives:
1. Perform calculations on acidic solutions involving a common ion.
B. The addition of a "common ion" to a solution can 5ÿ#A0f'ÿ the dissociation
ofanacid. This based on ÿ ÿigdÿ5--ÿ principle.
C. Consider acetic acid:
HC2H302 (aq) ÿ-"-) a+ (aq) + C2H302 (aq)
1. Add sodium acetate, }4 &ÿ-Z. ÿ A 0 ÿ. , which completely dissociates to
form ÿ#L ffÿ and ÿ.ÿ0ÿ0 ÿ . The acetate ion is common to the
solution. It is called the COMMON ION.
2. The common ion imposes a Sntr!gg$ on the system.
3. The system responds by shifting __] eÿ+ therefore, there is much
, less dissociation than before.
4. The general effect of a common ion is to depress the
D. Because dissociation is depressed, we can assume that the equilibrium
concentration of acetic acid equals the ]lfi/ÿ0kÿ ÿÿ 0ÿ . The
same is true of the acetate ion.
E. Example: Common ion effect
1. Calculate the pH and the percent dissociation of the acid in each of the
following situations:
a. 0.200 M HCzH302 (Ka = 1.8 X 10-5)
b. 0.200 M HCaH302 in the presence of 0.500 M NaC2H302
2. Solutions:
Iÿ+( )tCz.%o:2 ( )
a.. % _ "q
Cÿ4 e7.ÿ%o.ÿ3
7-= ITS+3 =[ezP,ÿoZ3 : I"ff':al°-sN , pie: z,7
o!o &ssodoÿ4noÿ =- l,q × 10-3 X t00% : ooqS%
[ ÿt4ÿoÿyJ : o,zoom
: 0,5oo Ui
!,9× ÿo-ÿ: ÿH+3f o,s-oo1
D,ZOO

©
Buffered Solutions // ¢tA'ÿCÿ "I+S 5,'4"@- ÿ)

go+,t h0,4 wgll
i ;
. j
D.
1.
Example: pH Changes in buffered solutions
Calculate the change in pH that occurs when 0.010 mol solid NaOH is
added to 1.0L of the buffered solution in example C-1 above. Compare
this pH change with that which occurs when 0.010mol solid NaOH is .... o
added to 1.0L of water.
2. Solution:
V' ,5'
' -3
iÿ6,< I0
k L ' '!
\
,, '-.\
beÿ4_ÿarce
loÿs 1+1
€ ,OL)'-O,50ÿ O,OÿOÿ° i ÿOLx O,SOÿ --*
= o,s-oÿo I ÿ _- 0°b-owd
0,ÿ50- 0,010 Omo/ 0ÿ$o+ 6,010a
3.
¢
g,
ezl4ÿOÿ ÿ ÿ4+ +
o.ÿtq, o
,qq-ÿ x
e %o/
O,S'ÿ
+X
,Sl+X
I,ÿX tO-S" =_
oÿ- 40 44ÿm baÿa4 sotÿn
,q(o- ÿ,qV -- + o,os

®
g.
Calculation of buffered solution and introduction of
Henderson/Hasselbalch equation.
1. Calculate the pH of a solution containing 0.75 M lactic acid (gO =
1.4 X 10-4 ) and 0.25 M sodium lactate. Lactic acid (HCgH503) is a
common constituent of biological systems. It is present in milk and
human muscle tissue during exercise.
2. Solution:
eÿt4goÿÿ" oM ÿ.o
I: ,qÿ o
E: ÿlS-'g X
4×
,ÿ+X
['xlg zÿ¢xa {xJ(,zs)
° oTN ÿY ,7bÿ
q-°Z:ÿ Io-¢ )ÿ
f, er: fÿKo. +
I
1,ÿ'¢ ÿd¢ =. 3,{35)

F. Another buffer solution example: -ÿ
1. A buffered solution contains 0.25 M NH3 (Kb = 1.8 X 105) and
0.40 M NI-I4C1. Calculate the pH of this solution.
2. Solution:
+Lc)
- o,qo o
+x 4-X
! ,% g tO"¢ÿ Ap, o+x3
O,'Zg- X
y
6bÿ

G. Adding strong acids and bases to buffers
1. Real utility of buffers comes when we add a strong acid or base.
2. Buffered solutions should 1Pÿ5ÿ$ÿ pH change when a substantial
amount of ÿ:t"@ or 0 ÿ ÿ ions are added. (Substantial amount
will be discussed later- will vary from solution to solution based on
"buffer capacity."
3. If we add a strong base to a solution that contains a weak acid, the
reaction is C.I)¢ÿ#ÿ]'ÿ.:..
HA(aq) + OH (aq) :-> A (aq) + H20 (1)
. Similarly, adding a strong acid to a weak base gives a complete
reaction:
A- (aq) + H+(aq) ÷ HA(aq)
. Notice that in these equations, unless ÿJlÿ%5 strong acid or base
is added, NO STRONG ACID OR BASE BUILDS UP. Only o
=llCÿ species remain. This is why a buffer works so ÿ!,ÿ
H. Example: Addition of a strong acid to a buffer
OÿF ÿ
1. Calculate the pH of the solution that results when 0.10 mole gaseous
HC1 is added to 1.0L of the buffered solution from the example in II F. 4ÿ prÿ¢0n5
1ÿ!tÿ ÿ 1ÿtÿÿ: , e.ÿ btÿ: aÿ± ÿ4ÿo
4o4" +
bÿ:( I,0c)l 0.ÿN)
= 0,ÿ.$'ÿo\
vÿiH rÿ+'. .
lqÿÿ = O,qOÿol
0 O,q-O+ 0°i0
= 0,ÿ"0¢ÿol

III.
Buffer Capacity
A. Objectives:
1. Calculate the pH of a solution where the buffer capacity has been
exceeded.
2. Choose among alternatives for the best buffer system for a given pH. n ÿ ÿ ,ÿ
Q
2V_a t+ÿl fÿ f4
pH:
B. Two important points: z ÿ===ÿ-'--- ÿ C,/ÿ .- ÿ kÿJ ]
1. The pH is determined by the ratio oÿ,[HA]/[A- ] 2ÿ ÿ[-ÿÿ !/k,ÿ r ÿ ÿ,Aÿ'
2. The buffer capacity is detennined by t'h-ÿnSaÿnltuÿ des of [I-IA] and [A-]. ÿZÿ
The more you have, the more strong acid or strong base that can be 0
neutralized.
C. Example: Adding a strong acid to a buffered solution
1. Calculate the change in pH that occurs when 0.010 mole gaseous HC1
is added to 1.0L of each of the following solutions: ÿÿ dkÿl'ÿ
Solution A: 5.00 M HC2H302 and 5.00 M NaC2H302 I, ÿ N t ;- ÿ
Solution B: 0.050 M HC2H302 and 0.050 M NaC2H302
2. Solution:
# ÿ + io9 ( rÿ h zn ÿaÿ, cÿ, Lo aÿ3czJ=
£r4 g.zÿ430ÿ-ÿJ . , crNCx1430zÿ
: - Ioq ÿ¢, ÿ ×/0-s') :
-ÿ eÿ43 oÿ_ eÿ) eoÿplÿbÿ
if, Oopl
5ÿ,OIM
4b,- son ÿ ÿÿ
vSoOÿ /" , 0017=
q, vg

4- HCÿ ÿ4a; 0ÿ_
®
o, 0£oÿ
0, oÿoN
qnÿ+ lo5 lo,°ÿ° )
\ O.¢too
o , OSo H
o , o (ÿ o ?q
o
or
,%'
"D. The example above compares the pH change brought about by adding a
strong acid to solutions with different buffering capacities. Let's try a
problem where we exceed the buffering cÿpacity of a solution.
1. Example: Calculate the pH ofa 0.50(!ÿolution that contains 0.15 M
HCOOH (Ka = 1.8 X 10-4) and 0.20 M HCOONa. Then calculate the
pH of the solution after the addition of 10.0(ÿ of 12.0 M NaOH.
HeÿoH(=!ÿ Hÿ ÿaI3 + ÿoo- c@
\\ ,ÿ.G ÿ[' ÿ?ÿ.ÿ'ÿ' '
/' "ÿ" r '- , , C)
.ÿ - t ,o ' ,,[; , y ÿr 32\
- LHq-3EOcoo-3 : ÿ,ll×I0-q = C A*% o,ÿ.o"
g hÿco o ÿ3
gHtÿ = ! q-×¢o"qN J
¢eooC-¢ ,%..0 (.0
L.¢.. ÿ mmoles
lq IooH [nhhM =
ffeAgO o =
. .---oÿLa.qgmmol
O, 15mmoÿmL g,,,bo ,. o
/ 0o mmo m L % 10-o ÿL = IZo °

U)
g. Preparing a Buffer
1. pH of a buffered solution depends on the ratio of the concentrations.._.,..........,,._ÿof
the buffering components
a. table:
b. Becauseÿchanges in the ratio [A-]/[HA] will produce large
changes in pH, we want to avoid this situation for the most
effective buffering>> so optimal buffering occurs when [HA]
=[A].
c. When choosing the buffering components for a specified
application, we want [A']/[HA] to equal 1ÿ
d. It follows the pKa of the weak acid to used in the buffer should be
as close as possible to the desired pH.
e. If you desire a buffer with a pH of 4, choose a weak acid with a
pKa of 4, or a Ka of 1.0 X 10-4.
2. Example: A chemist needs a solution buffered at pH 4.30 and can
choose from the following acids and their sodium salts:
a. Chloroacetic acid (Ka = 1.35 X 10-3)
b. Propanoic acid (Ka = 1.3 X 10-5)
c. Benzoic acid (Ka = 6.4 X 104)
d. Hypochlorous acid (Ka = 3.5 X 10-8)
Calculate the ratio [HA]/[A] required for each system to yield a pH of
4.30. Which system will work best? ÿal ÿ.
Eÿ-3 ° , ÿ0 eJoSeST
O ,V-g
I, ,?x ioÿ
laaÿ.r.oÿc ÿ.,u ,-, ° 5 ÿ ha.o & ÿ add

o_oÿ
@
IV. Titrations and pH Curves
A. Objectives:
1. You should be able to calculate the pH at any point along a curve for the
following titrations:
a. Strong acid-strong base
b. Weak acid-ÿe,ÿaWÿ ÿ0¢a$©
c. Weak base-strong acid
2. You should be able to define the terms:
Titranta.
b. Buretc.
Indicator
d. Titration curve-=
, , nÿcÿ Oxz.wJ
e. Equivalence point- 5%1 ¢Jÿt oÿ,,ÿLÿr,c ÿe I I .
f. Endpoint-ÿhÿ*wÿe-ÿ4-ocÿCktÿvtÿ Oÿt
3. In order for a titration to be feasible, it must be complete and fast. In order to
be complete, the reaction should have a value ofK > 107 or so.
B. Review Neutralization Reactions
. Example: Neutralization reactions I
a. What volume of 0.100 M HC1 solution is needed to neutralize 25.0 mL of
0.350 M NaOH?
b. Solution: ÿ C.15 ÿdÿt, 0ÿ
Z)ÿ01ÿ+ cxn will oÿxF? Nÿ4-4c1ÿ ÿ NaCl(.s)-ÿSo[ÿI¢ÿgÿ4ÿsJ go rxv
4-) eaÿIe.Mÿ raoles rÿcÿn4g.
cÿoa
ÿ.vxx roÿ-3 mol R÷asÿi

@
. Example: Neutralization Reactions II
a. In a certain experiment, 28.0 mL of 0.250 M HNO3 and 53.0 mL of 0.320
M KOH are mixed. Calculate the amount of water formed in the resulting
reaction. What is the concentration of I-1+ and OH- ions in excess after the
reaction goes to completion?
- t_/4u0
§5,oraL Kot-ll It... 01t'=
L Ko 4
. Example: Neutralization Reactions III
a. Problem: A student carries out an experiment to standardize (determine
the exact concentration of) a sodium hydroxide solution. To do this, the
student weighs out a 1.3009g sample of potassium hydrogen phthalate
(KHC8H404, often abbreviated KHP; molar mass 204.22 g/mol) KHP has
one acidic hydrogen. The student dissolves the KHP in distilled water,
adds phenolphthalein as an indicator, and titrates the resulting solution
with the sodium hydroxide solution to the phenolphthalein endpoint. The
difference between the final and initial buret readings indicates that 41.20
ml of the sodium hydroxide solution is required to react exactly with the
1.3009g KHP. Calculate the concentration of the sodium hydroxide
solution.

H opt N oH =
L
C. Strong Acid-Strong Base Titrations
. Will use a new unit which is more convenient for titrations
a. Burets are typically in mL
b. Titrations involve small quantities
c. Mole is inconveniently large.
d. Will use millimole (mmol) which is a thousandths of a mole:
IO00
e. Will look at molarity as: ÿ,ÿ= p¢ÿOlÿ Solctÿ
. Strong Acid-Strong Base Titration- Step by step.
a. We will illustrate the calculations involved in a strong acid-strong base
titration by considering the titration of 50.0 mL of 0.200M HNOs with
0.100 M NaOH. We will calculate the pH of the solution at selected points
during the course of the titration, where specific volumes of 0. ! 00 M
NaOH have been added.
Graph of progress:

@
Point A: No NaOH has been added. 0
14ÿ4o3 ÿmptÿ+ÿtÿ &ÿocio.ÿ SCer4'ÿ:
pH ÿeÿcÿlnÿd ÿom H+ ,
lncÿ 0,Z0OM HNOg eorrgams O,zooÿ
gt4+3 = O, aoo M, PN = 0,6q9
Point B: 10.0 mL of 0.100 M NaOH has been added.
.. H+ 4 OH- --ÿ Hÿo
rWl. ÿ-Omlÿ.ÿZM Io,OmlX ,IH .
= IO,Ommol = l, oommol
rrm /Oÿl = ÿ retools 1,0o-/.oo=
o (nÿo le3
VolOmÿ ÿ - (SN,OelO,O)mc
,ÿ. ÿ0ÿ ÿ" vol, ÿoÿ4 Mdÿd
Point C: 20.0 mL (total) of 0.100 M NaOH has been added.__
H÷ + off - -ÿ Hÿ.o
=lOmrno]
=ÿmcÿo[s
(ÿo, o ÷ao, owc)
Point D: 50.0 mL (total) of 0.100MNaOH has been added.ÿ ÿY*ÿ(ÿ ÿ #ÿOV'ÿ 3 fÿ - I ÿ-ÿ-
Point E: 100.0 mL (total) of0.100M NaOH has been aÿ.0
IOOmL x O, loom= lo, ommÿIs
0rÿtngl ÿvmÿc ÿ05 cOÿ; EOoOmL "aO, ZOO¢¢l- ;Ommo/5

®
f.ÿ,k[oHÿj= JXro
3. Strong base with strong acid will produce this type of graph.
D°
gO ,00, t
Titrations of Weak acids with strong bases.
1. Strong acids/bases completely dissociate, so calculations are straightforward.
2. When weak acid is being titrated, to calculate ItS] after a certain amount of
strong base has been added, we must deal with the weak acid dissociation
equilibrium.
3. Really a series of buffer problems.
4. Remember that even though the acid is weak, it reacts essentially to
completion with hydroxide ion, a very strong base.
5. Calculating the pH curve for a weak acid is a two step procedure:
a. A stoichiometry problem
b. An equilibrium problem
c. Do these steps separately! You already know how to do them.
6. Example: Weak acid- Strong base titration
a. Consider the titration of 50.0 mL of 0.10 ÿ acetic acid (HC2H3 02, Ka=
1.58 X 104) with 0.10 M NaOH.

Graph of Progess:
Point A: No NaOH has been added.
Point B: 10.0 mL of 0.10 MNaOH has been added.
bÿ ÿfÿdÿ4: {+cÿHsoÿ, oH7 t4ÿ,% nÿo
OH- ÿ
HgzHÿOÿ
+x
: #. o/eo = x ×
= i,S-x ÿo
__ =
=ÿZs'X
Point C: 25.0 mL (total) of 0.10 M NaOH has been added.
OH- + BetH,0z
%,OÿL'x,/M
.o5= g,5-ÿ 0

÷Y
[Hca Hÿ 0ÿ3 a,::ÿ _ )

Point C: 25.0 mL (total) of 0.10 M NaOH has been added.
Point D: 40.0 mL (total) of 0.10M NaOH has been added.
Point E: 50.0 mL (total) of 0.10M NaOH has been added.
to=- 5, (0X" t 0ÿ1°
Z O, o5o ÿ o 0
---- ÷ÿ +Y
C" -X
C'. o, oÿo-7:
.ÿ y. f
5,L+y.IO-ÿo V._b ÿ- gHCÿ1430-z]£°H-3- (XÿX)
O, 05-o
[ÿ÷3Co ÿ-2 : i.o ÿ \o-'ÿ

Point F: ÿ'- ÿ.
oH- -ÿ Hcÿ_Hsoÿ. CÿHsOÿ- gÿo
Soÿo,ÿk x, I h-I
=. %0 ,'r, mo I
5

Point G:
7. Compare two graphs
a. Shapes axe quite different before equivalence point although very similar
after that point (because excess OH" controls the pH in the later region in
both cases.)
b. Near the beginning of the titration of the weak acid, the pH increases more
rapidly that it does in the strong acid. It levels off near the halfway point
and then increases rapidly again. The leveling offnear the halfway point is
caused by buffering effects. Optimal buffering occurs when [HA] = [A- ].
This is exactly the case at the halfway point of the titration, pH changes
least rapidly in this region of titration.
c. For the titration of a strong acid, the equivalence point occurs at pH "7
d. For the titration of a weak acid, the pH at the equivalence point is
than 7 because of the basicity of the conjugate base of the weak acid.
e. Remember, the equivalence point in an acid-base titration is defined by the
5ÿK4.(ÿÿCÿ, not by the ÿ.
E. Titrations of Weak Bases with Strong Acids
1. Problem: 100.0 mL of 0.050 M NH3 is titrated with 0.10 M HC1.
Graph and Results:
Before the addition of any HCI:
1. Major species: ÿJ ÿ and ÿ._O
NH3 is a base and will seek a source of protons. In this case, water is the only
available source.
2. No reactions occur that go to completion since NH3 cannot readily take protons
from water. This is evidenced by the small Kb value for NH3.

. The equilibrium that controls the pH involves the reactions of ammonia with
water:
Use Kb to calculate [OH-]. Although NH3 is a eak base (compared with OH-), it produces
much more OH- in this reaction than is produced from the autoionization of water.
Before the equivalence point:
1. Major species (before any reaction occurs):
Rt O__ t- kilo
2. The NH3 will react with I-I+ from the added HCI:
This reaction proceeds essentially to completion because the NH3 readily reacts with a
free proton. This case is much different than the previous case, where water was the only
source of protons. The stoichiometric calculations are then carried out using the known
volume of 0.10 M HC1 added.
° After the reaction of NH3 with H÷ is run to completion, the solution contains the
following major species: uz -C-o,ÿ ÿn 4-dnca.'kÿvÿ
, , t4.. o
Note that the solution contains NH3 and NH4 +, and the equilibria involving these species
will determine [H+]. You can use either the dissociation reaction ofNI-I4+
Or the reaction of NH3 with water:
At the equivalence point:
. By definition, the equivalence point occurs when all the original NH3 is converted
to NH4+. Thus the major species in solution are:
.
3.
No reactions go to completion.
The dominant equilibrium (the one that controls the [I-l+]) will be dissociation of
the weak acid NH4+ for which

Beyond the equivalence point:
1. Excess HC1 has been added, and the major species are:
ff _.o
) } I
2. No reaction occÿs that goes to completion.
3. Although NI-I4 + will dissociate, it is such a weak acid that [H+] will be
determined simply the excess H+:
4. The results of the calculation are shown below: 09. ÿ@(ÿ ÿ-ÿd.ÿ_ÿ / ÿ-,
go Acid - Base Indicators
A. Two common ways to determine the equivalence point of an acid-base
titration:
1. Use a pH meter.
2. Use an acid-base indicator
a. Marks the end point of titration by turning color
b. Most are weak acids that exhibit one color when the Woton is
attached to the molecule and another when the proton is absent.
Ex. Phenolphthalein is pink in base and colorless in acid.j

1. Example: HC1 + NaOH could use BTB with a pH color change
range of 6.2-7.6, but not methyl yellow which has a pH range
of 2.9-4.0
D. How much indicator should be used?
1. Usually only a very small amount needed
2. Are very weak acids and are not the dominant source of protons in an
acid base reaction.
E. Color chart of indicators is in textbook on pg.

VI. Solubility Equilibria
A. Objectives: In this section, we consider equilibria associated with solids
dissolving to form aqueous solutions, interconvert between solubility and
Ksp, solve problems relating to the common ion effect.
B. Solubility is an important phenomenon.
1. Example: tooth decay involves solubility- When food is lodged
between teeth, acids form containing hydroxapatite, Cas(PO4)3OH.
If treated with fluoride, F replaces the OH- to produce fluorapatite,
Ca5 (PO4)3F and CaC12. both of these are less soluble in acids than the
original and thus deter tooth decay.
C. How it works:
1. Take for example, calcium fluoride dissolving in water.
a. When the solid salt is first added to water, no calcium and fluorine
ions are present. However, as dissolution proceeds, the
concentrations of Ca+2 and F" increase, making it more and more
likely that these ions will collide and re-form the solid- so two
competing processes are occurring- the dissolution reaction and the
reverse.
b. Ultimately, dynamic equilibrium is reached:
At this point, no more solid dissolves and the solution is said to be
saturated. The equilibrium expression for this process according to
law of mass action is written as:
c. Ksp is called the solubility product constant or the solubility
• product for the equilibrium expression.
1. The amount of excess solid present does not affect the position
of the solubility equilibrium
a. When ions in solution reform the solid, they do so on the
surface of the solid, so doubling the surface area of the
solid not only doubles the rate of dissolving, but also
doubles the rate of re-formation of the solid. The amount of
excess solid present has no effect on the equilibrium
position.
b. Increasing the surface area by grinding up the solid or
stirring the solution speeds up the attainment of
equilibrium, but does not change the amount of solid
dissolved at equilibrium. Neither the amount of excess
solid nor the size of the particles present will shift the
position of the solubility equilibrium.
c. Remember, the solubility product is an equilibrium
constant and has one ÿvalue for a given solid at a given

temperature. Solubility, on the other hand, is an equilibrium
position. The Ksp expressions of many solids are listed in
the table attached. Units are customarily omitted.

-ÿ- q--,.o x t o-ÿ
D° Example: Calculating Ksp from solubility
1. Copper (I) bromide has a measured solubility of 2.0 X 10-4 moliL at

.
E. Example: Calculating solubility from Ksp
The Ksp value for copper (II) iodate is 1.4 X 10-7 at 25°C. Calculate its solubility
at 25°C.
O 0 0
L_.,
"ÿkuz) /
L
F. Relative Solubilities
1. A salt'sÿsp value gives us information about its solubility. However,
we mum be careful in using Ksp values to predict the relative
solubilities of a group of salts. There are two possible cases:
a. The salts being compared produce the same number of ions:
x /v

Each of these solids dissolves to oroduce two io1ÿ'.
Ifx is the solubility of the salt in mol/L, then at equilibrium,
Therefore, in this case we can compare the solubilities for these solids by comparing the
Ksp values:
t ÿ¢g-Sg b. The salts being compared produce different numbers of ions. For
o
example, consider: . ,
8tzSCis) ÿs(p = g, $-x lO-qÿ
Jtÿ St (s.) /

H. pH and solubility
1. The pH of a solution can greatly affect a salt's solubility.
2. Consider the dissolution of Magnesium hydroxide:
a.
b.
C°
d.
Addition of OH- ion (an i P/Aa.stÿ n ÿ ÿ in pH) will, by the common ion
effect, foÿtÿ'he equilibriumÿeasing the ÿof
Mg(OH)2. ÿ o'
On the other hand, if I-1+ ions (a Jÿjlÿ.dÿ3.xÿin pH ) are added, the solubility
j_]_ÿÿ because OH ions are removed from solution by reacting
with the H+ ions. In response to the lower concentration of OH-, the equilibrium
moves to the ÿ .
This is why asusp£'tasion of solid milk of magnesia dissolves as required in the
stomach to combat acidity.
General Rule: If the anion X- is an effective base, that is ifHX is a weak acid- the
salt MX will show increased solubility in an acidic solution.
1. Common ions that are effective bases are: OH-, S-2, CO32, C20412, and CrO4-2
2. The above salts are much more soluble in acidic solution than in pure water.
3. Is the reason for the formation of huge limestone caves- carbon dioxide in
groundwater makes it acidic increasing the acidity of the calcium carbonate.
As the carbon dioxide escapes to the air, the pH of the dripping water goes up
and the calcium carbonate precipitates forming stalactites and stalagmkes.
VII.
Precipitation and Qualitative Analysis
A. Let's consider the formation of a solid from solution.
B. We will predict whether a precipitate will form when two solutions are
mixed using the l 0 tQ /3g'oÿ oÿl=- which is defined just like
Ksp except, that I ¢71'ÿ ÿ I concentrations are used instead of
i#ÿt"l ÿ concentrations. For solid CaF2' the expression
forCthe ion product is written:
C. To predict whether precipitation will occur, we consider the relationship
between Q and Ksp.
1. If Q is greater than Ksp, precipitation OCpÿLO(-.$ and will
continue until the concentrations are reduced to the point that they
satisfy Ksp.
2. If Q is less than Ksp, no precipitation occurs.
D. Example:

1. A solution is prepared by adding 750.0 mL of 4.00 X 10-3 M Ce(NO3)3
to 300.0 mL of 2.00 X 10.2 M KIO3. Will Ce(IO3)3 (Ksp = 1.9 X 10"1°)
precipitate from this solution?
rÿn ÿÿoÿ
/it)ÿ
, (-pS-o oÿc)_( ¢,o0/ 1o-3 mmo _
o = (-Tgo.o +-ÿoo,o )too
1 %-7o (3oo, o c) ( ooj
g.
Sometimes we want to do more than simply predict whether precipitation
will occur; we may want to calculate the equilibrium concentrations in the
solution after precipitation occurs.
1. Calculate the equilibrium concentrations of Pb+2 and I- ions inÿ--/'ÿ3ÿ
solution formed by mixing 100.0 mL of 0.0500 M Pb(NOg)ÿaÿ-'and 200.0 ÿ.ÿ
2. Solution:
SOO,O

1. Mixtures of metal ions in aqueous solution are often separated by
5jÿ -ÿ j,qY(0- ÿ q-2 ff)ÿ ÿLfiÿ - ÿelective precipitate precipitation, with only that one is, or a by few using ofthe a reagent metal ions whose in the anion mixture. forms For a
@(ÿ, 3"5g t0-z H- ÿ ×)SL example, suppose we have a solution containing both Ba+2and ag+'.
"- X ÿ ÿ, 3 "5 )ÿ I oÿz)'2-- ions. If NaCI is added to the solution, AgC1 precipitates as a white
[9b °z4ÿ ÿ X - iÿ ?:x ÿs- solid, but since BaC12 is soluble, the Ba+2 remains in solution.
2. Example: A solution contains 1.0 X 10.4 M Cu+ and 2.0 X 10-3 M Pb+2.
If a source of 1-is added gradually to this solution, will PbI2 (Ksp= ÿ I.
3"b/t 0=
x l0-8 ) or CuI (Ksp = 5.3 X 10-12) precipitate first? Specify the
concentration of I- necessary to begin precipitation of each salt.

G. Qualitative Analysis (chart)
VIII.
Complex ion equilibria
A. Complex ions are charged species consisting of a t4/ÿ
surrounded by i ÿÿoÿ . A ligand is a Lewis bcÿoeÿ
ion
-a

molecule or ion having a lone electron pair that can be donated to an
empty orbital on the metal ion to form a G0xt-0.ÿz4-,ÿ bond.
g.
C.
Common ligands:
Number of ligands attached to the metal ion is called the
@_2xÿ . The most common coordination
numbers are : ÿl ÿ t (o ..
D° Metal ions add ligands one at a time in steps characterized by equilibrium
constants called formation constants or stability constants.
1. Consider when solutions ofAg÷ and NH3 molecules are mixed, the
following reactions take place
° Calculating the concentrations of all these components can be
complicated, however usually the total concentration of the ligand is
much larger than the total concentration of the metal ion, and
approximations can greatly simplify problems.
. Example:
a. Consider a solution prepared by mixing 100.0 mL of 2.0 M NH3
with 100.0 mL of 1.0 X 10-3 AgNO3. Before any reactions occurs,
the mixed solution contains the major species:
We know that the reaction that will occur is:
However, we are interested in the reaction between NH3 and Ag+
to form complex ions and since the position of the preceding
equilibrium lies fax to the left (Kb for NH3 is 1.8 X 10-5), we can
neglect the amount of NH3 used up in the reaction with water.

f,ÿlJ
¢
/,ÿÿ
2, S4