Tro, Chemistry: A Molecular Approach 2
added 30.0 mL NaOH0.00050 mol NaOH xspH = 11.96
added 35.0 mL NaOH0.00100 mol NaOH xspH = 12.22
Adding NaOH to HCHO2
added 12.5 mL NaOH0.00125 mol HCHO2
pH = 3.74 = pKa
half-neutralization
initial HCHO2 solution0.00250 mol HCHO2
pH = 2.37
added 5.0 mL NaOH0.00200 mol HCHO2
pH = 3.14
added 10.0 mL NaOH0.00150 mol HCHO2
pH = 3.56
added 15.0 mL NaOH0.00100 mol HCHO2
pH = 3.92
added 20.0 mL NaOH0.00050 mol HCHO2
pH = 4.34
added 40.0 mL NaOH0.00150 mol NaOH xspH = 12.36
added 25.0 mL NaOHequivalence point0.00250 mol CHO2
−
[CHO2−]init = 0.0500 M
[OH−]eq = 1.7 x 10-6
pH = 8.23
added 50.0 mL NaOH0.00250 mol NaOH xspH = 12.52
Tro, Chemistry: A Molecular Approach 6
Titrating Weak Acid with a Strong Base• the initial pH is that of the weak acid solution– calculate like a weak acid equilibrium problem • e.g., 15.5 and 15.6
• before the equivalence point, the solution becomes a buffer– calculate mol HAinit and mol A−
init using reaction stoichiometry
– calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−
init
• half-neutralization pH = pKa
Tro, Chemistry: A Molecular Approach 7
Titrating Weak Acid with a Strong Base• at the equivalence point, the mole HA = mol Base, so
the resulting solution has only the conjugate base anion in it before equilibrium is established– mol A− = original mole HA
• calculate the volume of added base like Ex 4.8– [A−]init = mol A−/total liters– calculate like a weak base equilibrium problem
• e.g., 15.14• beyond equivalence point, the OH is in excess– [OH−] = mol MOH xs/total liters– [H3O+][OH−]=1 x 10-14
Tro, Chemistry: A Molecular Approach 8
Ex 16.7a – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the volume of KOH at the
equivalence point
Write an equation for the reaction for B with HA.
Use Stoichiometry to determine the volume of added B
HNO2 + KOH NO2 + H2O
KOH L .02000
KOH mol 0.200
KOH L 1
NO mol 1
KOH mol 1
NO L 1
NO mol 0.100NO L .04000
22
22
L 0400.0mL 1
L 0.001mL 0.40
mL 0.20L 0.001
mL 1L 0200.0
Tro, Chemistry: A Molecular Approach 9
Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH
Write an equation for the reaction for B with HA.
Determine the moles of HAbefore & moles of added B
Make a stoichiometry table and determine the moles of HA in excess and moles A made
HNO2 + KOH NO2 + H2O
KOH mol 00100.0L 1
KOH mol 200.0
mL 1
L 0.001mL 00.5
HNO2 NO2- OH−
mols Before 0.00400 0 ≈ 0
mols added - - 0.00100
mols After ≈ 0
22 HNO mol 00400.0
L 1
HNO mol 100.0
mL 1
L 0.001mL 0.40
0.00300 0.00100
Tro, Chemistry: A Molecular Approach 10
Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after adding 5.00 mL KOH.
Write an equation for the reaction of HA with H2O
Determine Ka and pKa for HA
Use the Henderson-Hasselbalch Equation to determine the pH
HNO2 + H2O NO2 + H3O+
HNO2 NO2- OH−
mols Before 0.00400 0 ≈ 0
mols added - - 0.00100
mols After 0.00300 0.00100 ≈ 0
Table 15.5 Ka = 4.6 x 10-4
15.3106.4loglogp 4 aa KK
2
2
HNO
NOlogppH aK
67.20.00300
00100.0log15.3pH
Tro, Chemistry: A Molecular Approach 11
Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at
the half-equivalence pointWrite an equation for the reaction for B with HA.
Determine the moles of HAbefore & moles of added B
Make a stoichiometry table and determine the moles of HA in excess and moles A made
HNO2 + KOH NO2 + H2O
HNO2 NO2- OH−
mols Before 0.00400 0 ≈ 0
mols added - - 0.00200
mols After ≈ 0
22 HNO mol 00400.0
L 1
HNO mol 100.0
mL 1
L 0.001mL 0.40
0.00200 0.00200
at half-equivalence, moles KOH = ½ mole HNO2
Tro, Chemistry: A Molecular Approach 12
Ex 16.7b – A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at
the half-equivalence point.
Write an equation for the reaction of HA with H2O
Determine Ka and pKa for HA
Use the Henderson-Hasselbalch Equation to determine the pH
HNO2 + H2O NO2 + H3O+
HNO2 NO2- OH−
mols Before 0.00400 0 ≈ 0
mols added - - 0.00200
mols After 0.00200 0.00200 ≈ 0
Table 15.5 Ka = 4.6 x 10-4
15.3106.4loglogp 4 aa KK
2
2
HNO
NOlogppH aK
15.30.00200
00200.0log15.3pH
Tro, Chemistry: A Molecular Approach 13
Titration Curve of a Weak Base with a Strong Acid
Tro, Chemistry: A Molecular Approach 14
Titration of a Polyprotic Acid• if Ka1 >> Ka2, there will be two equivalence
points in the titration– the closer the Ka’s are to each other, the less
distinguishable the equivalence points are
titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH
Tro, Chemistry: A Molecular Approach 15
Monitoring pH During a Titration• the general method for monitoring the pH during the
course of a titration is to measure the conductivity of the solution due to the [H3O+]– using a probe that specifically measures just H3O+
• the endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve
• if you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator
Tro, Chemistry: A Molecular Approach 16
Monitoring pH During a Titration
Tro, Chemistry: A Molecular Approach 17
Indicators• many dyes change color depending on the pH of the solution• these dyes are weak acids, establishing an equilibrium with the
H2O and H3O+ in the solutionHInd(aq) + H2O(l) Ind
(aq) + H3O+(aq)
• the color of the solution depends on the relative concentrations of Ind:HInd– when Ind:HInd ≈ 1, the color will be mix of the colors of Ind and
HInd – when Ind:HInd > 10, the color will be mix of the colors of Ind
– when Ind:HInd < 0.1, the color will be mix of the colors of HInd
18
Phenolphthalein
Tro, Chemistry: A Molecular Approach 19
Methyl Red
C
C CH
CH
CH
CH
C
CH
CH
C
CH
CH
(CH3)2N N N NH
NaOOC
C
C CH
CH
CH
CH
C
CH
CH
C
CH
CH
(CH3)2N N N N
NaOOC
H3O+ OH-
Tro, Chemistry: A Molecular Approach 20
Monitoring a Titration with an Indicator
• for most titrations, the titration curve shows a very large change in pH for very small additions of base near the equivalence point
• an indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH– pKa of HInd ≈ pH at equivalence point
21
Acid-Base Indicators
Tro, Chemistry: A Molecular Approach 22
Solubility Equilibria
• all ionic compounds dissolve in water to some degree – however, many compounds have such low
solubility in water that we classify them as insoluble
• we can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water
Tro, Chemistry: A Molecular Approach 23
Solubility Product• the equilibrium constant for the dissociation of a solid
salt into its aqueous ions is called the solubility product, Ksp
• for an ionic solid MnXm, the dissociation reaction is:MnXm(s) nMm+(aq) + mXn−(aq)
• the solubility product would be Ksp = [Mm+]n[Xn−]m
• for example, the dissociation reaction for PbCl2 isPbCl2(s) Pb2+(aq) + 2 Cl−(aq)
• and its equilibrium constant is Ksp = [Pb2+][Cl−]2
Tro, Chemistry: A Molecular Approach 24
Tro, Chemistry: A Molecular Approach 25
Molar Solubility• solubility is the amount of solute that will dissolve in a
given amount of solution– at a particular temperature
• the molar solubility is the number of moles of solute that will dissolve in a liter of solution– the molarity of the dissolved solute in a saturated solution
• for the general reaction MnXm(s) nMm+(aq) + mXn−(aq)
mnmn
sp
mn
K
solubilitymolar
Tro, Chemistry: A Molecular Approach 26
Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
[Pb2+] [Cl−]
Initial 0 0
Change +S +2S
Equilibrium S 2S
PbCl2(s) Pb2+(aq) + 2 Cl−(aq)
Ksp = [Pb2+][Cl−]2
Tro, Chemistry: A Molecular Approach 27
Ex 16.8 – Calculate the molar solubility of PbCl2 in pure water at 25C
Substitute into the Ksp expression
Find the value of Ksp from Table 16.2, plug into the equation and solve for S
[Pb2+] [Cl−]
Initial 0 0
Change +S +2S
Equilibrium S 2S
Ksp = [Pb2+][Cl−]2
Ksp = (S)(2S)2
M 1043.14
1017.1
4
4
2
35
3
3
S
SK
SK
sp
sp
Tro, Chemistry: A Molecular Approach 28
Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M
Tro, Chemistry: A Molecular Approach 29
Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
[Pb2+] [Br−]
Initial 0 0
Change +(1.05 x 10-2) +2(1.05 x 10-2)
Equilibrium (1.05 x 10-2) (2.10 x 10-2)
PbBr2(s) Pb2+(aq) + 2 Br−(aq)
Ksp = [Pb2+][Br−]2
Tro, Chemistry: A Molecular Approach 30
Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25C is 1.05 x 10-2 M
Substitute into the Ksp expression
plug into the equation and solve
Ksp = [Pb2+][Br−]2
Ksp = (1.05 x 10-2)(2.10 x 10-2)2
[Pb2+] [Br−]
Initial 0 0
Change +(1.05 x 10-2) +2(1.05 x 10-2)
Equilibrium (1.05 x 10-2) (2.10 x 10-2)
6
222
1063.4
1010.21005.1
sp
sp
K
K
Tro, Chemistry: A Molecular Approach 31
Ksp and Relative Solubility
• molar solubility is related to Ksp
• but you cannot always compare solubilities of compounds by comparing their Ksps
• in order to compare Ksps, the compounds must have the same dissociation stoichiometry
Tro, Chemistry: A Molecular Approach 32
The Effect of Common Ion on Solubility• addition of a soluble salt that contains one of
the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt
• for example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2
PbCl2(s) Pb2+(aq) + 2 Cl−(aq)
addition of Cl− shifts the equilibrium to the left
Tro, Chemistry: A Molecular Approach 33
Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C
Write the dissociation reaction and Ksp expression
Create an ICE table defining the change in terms of the solubility of the solid
[Ca2+] [F−]
Initial 0 0.100
Change +S +2S
Equilibrium S 0.100 + 2S
CaF2(s) Ca2+(aq) + 2 F−(aq)
Ksp = [Ca2+][F−]2
Tro, Chemistry: A Molecular Approach 34
Ex 16.10 – Calculate the molar solubility of CaF2 in 0.100 M NaF at 25C
Substitute into the Ksp expression
assume S is small
Find the value of Ksp from Table 16.2, plug into the equation and solve for S
[Ca2+] [F−]
Initial 0 0.100
Change +S +2S
Equilibrium S 0.100 + 2S
Ksp = [Ca2+][F−]2
Ksp = (S)(0.100 + 2S)2
Ksp = (S)(0.100)2
M 1046.1
100.0
1046.1
100.0
8
2
10
2
S
S
SKsp
Tro, Chemistry: A Molecular Approach 35
The Effect of pH on Solubility• for insoluble ionic hydroxides, the higher the pH, the
lower the solubility of the ionic hydroxide– and the lower the pH, the higher the solubility– higher pH = increased [OH−]
M(OH)n(s) Mn+(aq) + nOH−(aq)
• for insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility
M2(CO3)n(s) 2 Mn+(aq) + nCO32−(aq)
H3O+(aq) + CO32− (aq) HCO3
− (aq) + H2O(l)
Tro, Chemistry: A Molecular Approach 36
Precipitation• precipitation will occur when the concentrations of the
ions exceed the solubility of the ionic compound• if we compare the reaction quotient, Q, for the current
solution concentrations to the value of Ksp, we can determine if precipitation will occur– Q = Ksp, the solution is saturated, no precipitation– Q < Ksp, the solution is unsaturated, no precipitation– Q > Ksp, the solution would be above saturation, the salt
above saturation will precipitate
• some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions
Tro, Chemistry: A Molecular Approach 37
precipitation occurs if Q > Ksp
a supersaturated solution will precipitate if a seed crystal is added
Tro, Chemistry: A Molecular Approach 38
Selective Precipitation
• a solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others
• a successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different
Tro, Chemistry: A Molecular Approach 39
Ex 16.13 What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater?
precipitating may just occur when Q = Ksp
22 ]OH][Mg[ Q
6
13
132
109.1059.0
1006.2]OH[
1006.2]OH][059.0[
spKQ
Tro, Chemistry: A Molecular Approach 40
Ex 16.14 What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from
seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M 22 ]OH][Ca[ Q
2
6
62
1060.2011.0
1068.4]OH[
1068.4]OH][011.0[
spKQ
Tro, Chemistry: A Molecular Approach 41
Ex 16.14 What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from
seawater?
precipitating Mg2+ begins when [OH−] = 1.9 x 10-6 M
M 108.4
1060.2
1006.2]Mg[
1006.2]1060.2][Mg[
when
1022
132
13222
spKQ
precipitating Ca2+ begins when [OH−] = 2.06 x 10-2 M 22 ]OH][Mg[ Q when Ca2+ just begins to
precipitate out, the [Mg2+] has dropped from 0.059 M to 4.8 x 10-10 M
Tro, Chemistry: A Molecular Approach 42
Qualitative Analysis• an analytical scheme that utilizes selective
precipitation to identify the ions present in a solution is called a qualitative analysis scheme– wet chemistry
• a sample containing several ions is subjected to the addition of several precipitating agents
• addition of each reagent causes one of the ions present to precipitate out
Tro, Chemistry: A Molecular Approach 43
Qualitative Analysis
44
Tro, Chemistry: A Molecular Approach 45
Group 1
• group one cations are Ag+, Pb2+, and Hg22+
• all these cations form compounds with Cl− that are insoluble in water– as long as the concentration is large enough– PbCl2 may be borderline• molar solubility of PbCl2 = 1.43 x 10-2 M
• precipitated by the addition of HCl
46
Group 2• group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+,
Pb2+, Sb3+, and Hg2+
• all these cations form compounds with HS− and S2− that are insoluble in water at low pH
• precipitated by the addition of H2S in HCl
47
Group 3
• group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+ precipitated as hydroxides
• all these cations form compounds with S2− that are insoluble in water at high pH
• precipitated by the addition of H2S in NaOH
Tro, Chemistry: A Molecular Approach 48
Group 4
• group four cations are Mg2+, Ca2+, Ba2+ • all these cations form compounds with PO4
3− that are insoluble in water at high pH
• precipitated by the addition of (NH4)2HPO4
49
Group 5• group five cations are Na+, K+, NH4
+
• all these cations form compounds that are soluble in water – they do not precipitate
• identified by the color of their flame
Tro, Chemistry: A Molecular Approach 50
Complex Ion Formation
• transition metals tend to be good Lewis acids• they often bond to one or more H2O molecules to form
a hydrated ion– H2O is the Lewis base, donating electron pairs to form
coordinate covalent bondsAg+(aq) + 2 H2O(l) Ag(H2O)2
+(aq)• ions that form by combining a cation with several
anions or neutral molecules are called complex ions– e.g., Ag(H2O)2
+
• the attached ions or molecules are called ligands– e.g., H2O
Tro, Chemistry: A Molecular Approach 51
Complex Ion Equilibria
• if a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand
Ag(H2O)2+
(aq) + 2 NH3(aq) Ag(NH3)2+
(aq) + 2 H2O(l) – generally H2O is not included, since its complex ion is
always present in aqueous solution
Ag+(aq) + 2 NH3(aq) Ag(NH3)2
+(aq)
Tro, Chemistry: A Molecular Approach 52
Formation Constant
• the reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction
Ag+(aq) + 2 NH3(aq) Ag(NH3)2
+(aq)
• the equilibrium constant for the formation reaction is called the formation constant, Kf
23
23
]NH][[Ag
])[Ag(NH
fK
Tro, Chemistry: A Molecular Approach 53
Formation Constants
Tro, Chemistry: A Molecular Approach 54
Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the
[Cu2+] at equilibrium?
Write the formation reaction and Kf expression.
Look up Kf value
Determine the concentration of ions in the diluted solutions
Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)
134
32
243 107.1
]NH][Cu[
])Cu(NH[
fK
M 107.6L 0.250 L 200.0
L 1mol 101.5
L 200.0]Cu[ 4
-3
2
M 101.1L 0.250 L 200.0
L 1mol 100.2
L 250.0]NH[ 1
-1
3
55
Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the
[Cu2+] at equilibrium?
Create an ICE table. Since Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium
[Cu2+] [NH3] [Cu(NH3)22+]
Initial 6.7E-4 0.11 0
Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4
Equilibrium x 0.11 6.7E-4
Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)
132
43
43
2
107.1])Cu(NH[
]NH][Cu[
fK
Tro, Chemistry: A Molecular Approach 56
Ex 16.15 – 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3. What is the
[Cu2+] at equilibrium?
Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)
134
32
243 107.1
]NH][Cu[
])Cu(NH[
fK
Substitute in and solve for x
confirm the “x is small” approximation [Cu2+] [NH3] [Cu(NH3)2
2+]
Initial 6.7E-4 0.11 0
Change -≈6.7E-4 -4(6.7E-4) + 6.7E-4
Equilibrium x 0.11 6.7E-4
13413
4
4
413
107.211.0107.1
107.6
11.0
107.6107.1
x
x
since 2.7 x 10-13 << 6.7 x 10-4, the approximation is valid
Tro, Chemistry: A Molecular Approach 57
The Effect of Complex Ion Formation on Solubility
• the solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands
AgCl(s) Ag+(aq) + Cl−
(aq) Ksp = 1.77 x 10-10
Ag+(aq) + 2 NH3(aq) Ag(NH3)2
+(aq) Kf = 1.7 x 107
• adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+
58
Tro, Chemistry: A Molecular Approach 59
Solubility of Amphoteric Metal Hydroxides
• many metal hydroxides are insoluble• all metal hydroxides become more soluble in acidic
solution– shifting the equilibrium to the right by removing OH−
• some metal hydroxides also become more soluble in basic solution– acting as a Lewis base forming a complex ion
• substances that behave as both an acid and base are said to be amphoteric
• some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+
Tro, Chemistry: A Molecular Approach 60
Al3+
• Al3+ is hydrated in water to form an acidic solutionAl(H2O)6
3+(aq) + H2O(l) Al(H2O)5(OH)2+
(aq) + H3O+(aq)
• addition of OH− drives the equilibrium to the right and continues to remove H from the molecules
Al(H2O)5(OH)2+(aq) + OH−
(aq) Al(H2O)4(OH)2+
(aq) + H2O (l)
Al(H2O)4(OH)2+
(aq) + OH−(aq) Al(H2O)3(OH)3(s) + H2O
(l)
Tro, Chemistry: A Molecular Approach 61