aqueous ionic equilibrium
TRANSCRIPT
Ch 16.1 of 77
Chapter 16: Aqueous Ionic Equilibrium
Ch 16.2 of 77
16.1 Introduction• Aqueous solutions are extremely important
– Acids and bases are aqueous solutions• Equilibria influence aqueous solutions
– For example, we can describe the strength of an acid orbase in terms of equilibrium
• How does equilibrium affect mixtures of acids andbases?
• How does equilibrium affect the solubility ofcompounds in water?
Ch 16.3 of 77
Buffers• The pH of blood must be maintained between
about 7.3 and 7.5 to avoid death• Aqueous equilibria in the blood control pH• A buffer is a substance that resists changes in
pH– A buffer reacts with added acid or base with little change
in overall pH• In blood, the major components of the buffer are
H2CO3 and HCO3-
How are these relatedto each other?
Ch 16.4 of 77
Buffer Composition• Buffers contain significant quantities of an acid
and its conjugate base (or a base and itsconjugate acid)– Carbonic acid (H2CO3) + bicarbonate (HCO3
-)– Ammonia (NH3) + ammonium ions (NH4
+)• Consider an acetic acid (HCH3CO2) and (its
conjugate base) acetate (CH3CO2-) buffer:
1. Added base is neutralized by the acid HCH3CO22. Added acid is neutralized by the conjugate base
CH3CO2-
AcidicBasic
Ch 16.5 of 77
How Buffers Work1. Added base is neutralized by the acid HCH3CO2
OH-(aq) + HCH3CO2(aq) ⇌ CH3CO2-(aq) + H2O(l)
• Addition of OH- shifts the equilibrium right– As long as the amount of OH- added is small compared
with the amount of HCH3CO2, it is almost completelyneutralized
– CH3CO2- is weakly basic so pH would rise slightly
Ch 16.6 of 77
How Buffers Work2. Added acid is neutralized by the conjugate base
CH3CO2-
H+(aq) + CH3CO2-(aq) ⇌ HCH3CO2(aq)
• Addition of H+ shifts the equilibrium right– As long as the amount of H+ added is small compared
with the amount of CH3CO2-, it is almost completely
neutralized– HCH3CO2 is weakly acidic so pH would drop slightly– Acetic acid by itself is not a good buffer because the
amount of CH3CO2- available is very small
Ch 16.7 of 77
How Buffers Work
Addition of acid:HCl(aq) + NaCH3CO2(aq) ⇌HCH3CO2(aq) + NaCl(aq)
Addition of base:NaOH(aq) + HCH3CO2(aq)⇌ NaCH3CO2(aq) + H2O(l)
Buffer made by mixingacetic acid and sodium
acetate
Any acetate works
Ch 16.8 of 77
Calculating Buffer pH• At first sight, calculating the pH of a mixture of
acid and base might seem difficult• But the common ion effect simplifies the problem
HCH3CO2(aq) ⇌ H+(aq) + CH3CO2-(aq)
• Adding extra conjugate base (the common ion)suppresses ionization of the acid (Le Châtelier)
• The initial concentrations of the acid and addedconjugate base donʼt change significantly
Ch 16.9 of 77
Calculating Buffer pHHCH3CO2(aq) ⇌ H+(aq) + CH3CO2
-(aq)• To find pH we need [H+]eqm
Ka =[CH3CO2
!] " [H+ ]
[HCH3CO2 ]=[A! ] " [H+ ]
[HA]
[H+ ] =Ka " [HA]
[A! ]
#Ka " [HA]initial
[A! ]initial
Because ofcommon ion
effect,concentrations
of acid andconjugate basedonʼt changemuch upon
reachingequilibrium!
Ch 16.10 of 77
Calculating Buffer pH
[H+ ] =Ka ! [HA]initial
[A" ]initial
log[H+ ] = logKa ! [HA]initial[A" ]initial
#
$%&
'(
log[H+ ] = logKa + log[HA]initial
[A! ]initial
"
#$%
&'
! log[H+ ] = ! logKa ! log[HA]initial
[A! ]initial
"
#$%
&'
Take logs
Separateterms
Multiply by-1
Ch 16.11 of 77
Simplifying Calculating Buffer pH
! log[H+ ] = ! logKa ! log[HA]initial
[A! ]initial
"
#$%
&'
pH = pKa + log[A! ]initial
[HA]initial
"
#$%
&'
• We have arrived at the Henderson-Hasselbalchequation
• It allows us to calculate pH from initialconcentrations of acid and conjugate base
• It should only be used for buffers
Ch 16.12 of 77
The Henderson-Hasselbalch Equation• Calculate the pH of a solution of 0.34 M lactic acid
(Ka = 1.4x10-4) and 0.22 M potassium lactate• We must recognize this as a buffer solution with
[HA] = 0.34 M and [A-] = 0.22 M• Using the Henderson-Hasselbalch equation
pH = pKa + log[A! ]initial
[HA]initial
"
#$%
&'
pH = pKa + log0.22
0.34
!"#
$%&
How do wefind this?
Ch 16.13 of 77
The Henderson-Hasselbalch Equation• Remember pKa = -log(Ka) = -log(1.4x10-4) = 3.85
• The Henderson-Hasselbalch equation works bestwhere ʻx is smallʼ is appropriate
pH = 3.85 + log0.22
0.34
!"#
$%&
= 3.85 + log(0.647)
= 3.85 ' 0.19= 3.66
When [HA]= [A-] thenpH = pKabecauselog(1) = 0
Check thatʻx is smallʼ?
Canʼtbecause
assumptionis built in
Ch 16.14 of 77
Calculating Changes in Buffer pH• While buffers resist changes in pH upon addition
of strong acid or base, the pH does change slighty• We find out how much by breaking the pH
calculation into two parts:1. Stoichiometry calculation
– Use stoichiometry to find out how much conjugate baseis neutralized by addition of strong acid or
– Use stoichiometry to find out how much acid isneutralized by addition of strong base
2. Equilibrium calculation– Use Henderson-Hasselbalch calculation
Ch 16.15 of 77
Calculating Changes in Buffer pH (1)• Calculate the pH of a solution of 1.00 L of 0.34 M
lactic acid (Ka = 1.4x10-4) and 0.22 M potassiumlactate after addition of 0.05 L of 1.0 M HCl
1. Stoichiometry calculation• Addition of strong acid uses up some of the
conjugate base and makes more lactic acid
H+(aq) + A-(aq) ⇌ HA(aq)– Remember HA is a weak acid so equilibrium lies mostly
on the right
Ch 16.16 of 77
Calculating Changes in Buffer pH (1)• Itʼs easiest to work in terms of moles rather than
concentration– Moles [A-] = 1.00 L x 0.22 M = 0.22 moles– Moles [HA] = 1.00 L x 0.34 M = 0.34 moles– Moles [H+] added = 0.05 L x 1.00 M = 0.05 moles
H+(aq) + A-(aq) → HA(aq)Initial 0.00 mol 0.22 mol 0.34 mol
Addition 0.05 mol Decreases IncreasesFinal 0.00 mol 0.17 mol 0.39 mol
Ch 16.17 of 77
Calculating Changes in Buffer pH (1) H+(aq) + A-(aq) → HA(aq)
• How do you know whether the [HA] concentrationincreases or decreases?
• The addition of strong acid increases [HA]• The addition of strong base increases [A-]
Initial 0.00 mol 0.22 mol 0.34 mol
Addition 0.05 mol Decreases IncreasesFinal 0.00 mol 0.17 mol 0.39 mol
Ch 16.18 of 77
Calculating Changes in Buffer pH (1)2. Equilibrium calculation• Weʼll convert the moles back into concentration
– [HA] = 0.39 mols / 1.05 L = 0.37 M– [A-] = 0.17 mols / 1.05 L = 0.16 Mand use the Henderson-Hasselbalch equation
pH = pKa + log[A! ]initial
[HA]initial
"
#$%
&'
pH = 3.85 + log0.16
0.37
!"#
$%&= 3.49
Small 4.6%change
from 3.66
Ch 16.19 of 77
Calculating Changes in Buffer pH (2)• Calculate the pH of a solution of 1.00 L of 0.34 M
lactic acid (Ka = 1.4x10-4) and 0.22 M potassiumlactate after addition of 0.05 L of 1.0 M NaOH
1. Stoichiometry calculation• Addition of strong base uses up some of the acid
and makes more lactate
OH-(aq) + HA(aq) ⇌ A-(aq) + H2O(l)– Addition of OH- pushes the equilibrium to the right
Ch 16.20 of 77
Calculating Changes in Buffer pH (2)• Itʼs easiest to work in terms of moles rather than
concentration– Moles [A-] = 1.00 L x 0.22 M = 0.22 moles– Moles [HA] = 1.00 L x 0.34 M = 0.34 moles– Moles [OH-] added = 0.05 L x 1.00 M = 0.05 moles
OH-(aq) + HA(aq) → A-(aq)Initial 0.00 mol 0.34 mol 0.22 mol
Addition 0.05 mol Decreases IncreasesFinal 0.00 mol 0.29 mol 0.27 mol
Ch 16.21 of 77
Calculating Changes in Buffer pH (2)2. Equilibrium calculation• Weʼll convert the moles back into concentration
– [HA] = 0.29 mols / 1.05 L = 0.28 M– [A-] = 0.27 mols / 1.05 L = 0.26 Mand use the Henderson-Hasselbalch equation
pH = pKa + log[A! ]initial
[HA]initial
"
#$%
&'
pH = 3.85 + log0.26
0.28
!"#
$%&= 3.82
Small 4.4%change
from 3.66
Ch 16.22 of 77
16.3 Buffer Range• Changes in buffer pH upon addition of a strong
acid or base are smallest when:1. The concentrations of HA and A- are equal
– The buffer has reserves of both acid and base withwhich to neutralize additions
– HA and A- concentrations should be within a factor of 10– Buffers work best when the pH = pKa of the acid
2. The concentrations of HA and A- are large– The concentrations of HA and A- donʼt change much
upon additions of strong acid and base
Ch 16.23 of 77
Buffer Range• HA and A- concentrations should be within a factor
of 10 for a reasonably good buffer
• The buffer range (the range of pH over which thebuffer is effective) is pKa±1
pH = pKa + log[A! ]
[HA]
"#$
%&'
pH = pKa + log[10]
[1]
!"#
$%&
pH = pKa +1
pH = pKa + log[1]
[10]
!"#
$%&
pH = pKa '1
How does pHvary if we
change ratio?
(1) (2)
Ch 16.24 of 77
Buffer Capacity• The buffer capacity is the amount of strong
acid or base that can be added to the bufferbefore it is destroyed– This means either the acid or conjugate base is
completely used up• Buffer capacity increases as [HA] and [A-]
increase– A 0.5 M HA + 0.5 M A- buffer has a higher buffer
capacity than a 0.1 M HA + 0.1 M A- buffer– A 0.5 M HA + 0.5 M A- buffer has a higher buffer
capacity than a 0.5 M HA + 0.1 M A- buffer
Ch 16.25 of 77
Buffer Capacity• A buffer is comprised of 0.10 mols HCH3CO2 and
0.05 mols NaCH3CO2. Which of the followingadditions will destroy the buffer?
• 0.05 mols of NaCH3CO2?– No - this will simply increase [A-]
• 0.05 mols of NaOH?– No - this will use up half of the acid but leave half
remaining• 0.05 mols of HBr?
– Yes - this will use up all of the conjugate base leavingonly the weak acid (not a buffer)
Ch 16.26 of 77
16.4 Titrations and pH Curves• In an acid-base titration, an acid of known
quantity and concentration is reacted with a baseof unknown concentration
• The endpoint or equivalence point is reachedwhen an added indicator just changes color
• Care must be taken not to exceed the endpoint– The acid is exactly neutralized by the base
• The information about the substance added canbe used to determine the concentration of thetitrant (the substance added)
Ch 16.27 of 77
Titrations• At the endpoint,
enough base hasbeen added toexactly neutralize theacidH+(aq) + OH-(aq) →
H2O(l)• At the endpoint the
moles H+(aq) andmoles OH-(aq) areequal
Ch 16.28 of 77
Titrations and pH Curves
Before theendpoint, [H+]is in excess so
pH is low
After theendpoint, [OH-]is in excess so
pH is high
At theendpoint, [H+]= [OH-] so pH
is neutral
Ch 16.29 of 77
Titration and pH Curves• We can calculate the pH curve by knowing how
many moles of acid has been neutralized aftereach addition of base
• We must work in moles (not concentration)
• Which can be rearranged to
Mols = Concentration (mols/L)!Volume (L)
Volume (L) =Mols
Concentration (mols/L)
Usefulequations
Ch 16.30 of 77
Titration and pH Curves• Derive the pH curve for titration of 50 mL of 0.200
M HCl with 0.100 M KOHMols H+ = 0.200 M × 0.050 L = 0.010 mols
• At the endpoint mols H+ = mols OH- so
• The endpoint occurs after addition of 100 mL ofKOH
Volume OH!
=0.010mols OH
!
0.100mols/L= 0.100 L
Ch 16.31 of 77
Strong Acid-Strong Base pH Curves• Derive the pH curve for titration of 50 mL of 0.200
M HCl with 0.100 M KOH1. Before addition of any base:
HCl(aq) → H+(aq) + Cl-(aq)• [H+] = 0.200 M
pH = !log[H+ ]
= !log(0.200)
= 0.70
pH of astrongacid
Ch 16.32 of 77
Strong Acid-Strong Base pH Curves• Derive the pH curve for titration of 50 mL of 0.200
M HCl with 0.100 M KOH2. After addition of 10.00 mL base:
Some H+ is neutralizedH+(aq) + OH-(aq) → H2O(l)
• The moles of H+ neutralized is the same as themoles of OH- added
Mols OH! added = Concentration (M)"Volume (L)
= 0.100M"0.010 L = 0.001
Ch 16.33 of 77
Strong Acid-Strong Base pH Curves• The moles of H+ left is the original amount (0.010
mols) minus the amount that was neutralized(0.001 mols) = 0.009 mols
• The [H+] is
• And pH is
[H+ ] =0.009mols
0.050 L + 0.010 L= 0.150M
pH = !log[H+ ]
= !log(0.150)
= 0.82
50 mL of HCl +10 mL KOH
addedpH increases
as base isadded
Ch 16.34 of 77
Strong Acid-Strong Base pH Curves• Derive the pH curve for titration of 50 mL of 0.200
M HCl with 0.100 M KOH3. After addition of 20.00 mL base:
Some H+ is neutralizedH+(aq) + OH-(aq) → H2O(l)
• The moles of H+ neutralized is the same as themoles of OH- added
Mols OH! added = Concentration (M)"Volume (L)
= 0.100M"0.020 L = 0.002
Ch 16.35 of 77
Strong Acid-Strong Base pH Curves• The moles of H+ left is the original amount (0.010
mols) minus the amount that was neutralized(0.002 mols) = 0.008 mols
• The [H+] is
• And pH is
[H+ ] =0.008mols
0.050 L + 0.020 L= 0.114 M
pH = !log[H+ ]
= !log(0.114)
= 0.94
50 mL of HCl +20 mL KOH
addedpH increasesas more base
is added
Ch 16.36 of 77
Strong Acid-Strong Base pH Curves• Derive the pH curve for titration of 50 mL of 0.200
M HCl with 0.100 M KOH4. After addition of 100 mL base (the endpoint):
All H+ is neutralizedH+(aq) + OH-(aq) → H2O(l)
• The only source of H+ is from the ionization ofwater
• [H+] = 1.00x10-7 M so pH = 7.00
Ch 16.37 of 77
Strong Acid-Strong Base pH Curves• Derive the pH curve for titration of 50 mL of 0.200
M HCl with 0.100 M KOH5. After addition of 110 mL base (10 mL past the
endpoint):All H+ is neutralized but there is an excess of OH-
• 100 mL base is used to neutralize the acid so 10mL excess has been added
Mols excess OH!= Concentration (M)"Volume (L)
= 0.100M"0.010 L = 0.001
Ch 16.38 of 77
Strong Acid-Strong Base pH Curves• The [OH-] is
• pOH is
• And pH = 14.00 - pOH = 11.80
[OH! ] =0.001mols
0.050 L + 0.110 L= 0.006M
pOH = !log[OH! ]
= !log(0.006)
= 2.20
50 mL of HCl +110 mL KOH
added
pH is basic pastequivalence
point
Ch 16.39 of 77
Strong Acid-Strong Base pH Curves
12.0712011.801107.001002.15901.81801.60701.44601.30501.18401.06300.94200.82100.700pHVol KOH added (mL) 14
12
10
8
6
4
2
00 10 20 100
Volume KOH (mL)
pH
EndpointpH = 7.00
pH alwaysincreases
Ch 16.40 of 77
Strong Acid-Strong Base pH Curves1. At start, calculate the pH of a strong acid2. Before the endpoint, H+ is in excess
– Calculate [H+] from initial moles minus molesneutralized (moles OH-) divided by total volume
3. At the endpoint, H+ comes from water only– pH = 7.00
4. After the endpoint, OH- is in excess– Calculate [OH-] from moles added minus moles
neutralized (initial moles H+) divided by total volume
Ch 16.41 of 77
Weak Acid-Strong Base pH Curves• Derive the pH curve for titration of 50 mL of 0.200
M HCH3CO2 (Ka=1.8x10-5) with 0.100 M KOHTotal mols H+ = 0.200 M × 0.050 L = 0.010 mols
• At the endpoint mols H+ = mols OH- so
• The endpoint occurs after addition of 100 mL ofKOH
Volume OH!
=0.010mols OH
!
0.100mols/L= 0.100 L
Ch 16.42 of 77
Weak Acid-Strong Base pH Curves• Derive the pH curve for titration of 50 mL of 0.200
M HCH3CO2 (Ka=1.8x10-5) with 0.100 M KOH1. Before addition of any base:
HCH3CO2(aq) ⇌ H+(aq) + CH3CO2-(aq)
• [H+] = ?• This is a weak acid problem so we need to solve
with an ICE table for 0.200 M HCH3CO2
Ch 16.43 of 77
Weak Acid-Strong Base pH Curves
xx0.200 - xEquilibrium+x+x-xChange000.200Initial
CH3CO2-H+HCH3CO2
Ka =[H+ ] ! [CH3CO2
"]
[HCH3CO2 ]
1.8x10"5 =x2
0.200" x#
x2
0.200
x = [H+ ] =1.90x10"3 pH = 2.72
x is small
Ch 16.44 of 77
Weak Acid-Strong Base pH Curves• Derive the pH curve for titration of 50 mL of 0.200
M HCH3CO2 (Ka=1.8x10-5) with 0.100 M KOH2. After addition of 10 mL of base:
HCH3CO2(aq) ⇌ H+(aq) + CH3CO2-(aq)
HCH3CO2(aq) + OH-(aq) → H2O(l) + CH3CO2-(aq)
• Addition of base always converts a stoichiometricamount of the acid to its conjugate base– We ignore the small amounts H+ and CH3CO2
- made bythe equilibrium
Ch 16.45 of 77
Weak Acid-Strong Base pH Curves• The moles of HCH3CO2 used up and the moles of
CH3CO2- made is the same as the moles of OH-
added
• Moles HCH3CO2 is initial (0.010 mols) minus thatused up (0.001 mols) = 0.009 mols
• Moles CH3CO2- made is 0.001 mols
Mols OH! added = Concentration (M)"Volume (L)
= 0.100M"0.010 L = 0.001
Ch 16.46 of 77
Weak Acid-Strong Base pH Curves• We have some acid and some conjugate base• This is a buffer!
pH = pKa + log[A! ]
[HA]
"#$
%&'
= 4.74 + log0.001mols / 0.060 L
0.009 mols / 0.060 L
"#$
%&'
= 3.79
Henderson-Hasselbalch
equation
pKa = -log(1.8x10-5) = 4.74
pH increasesbecause
weʼve addedbase Total volume =
50 mL + 10 mL
Ch 16.47 of 77
Weak Acid-Strong Base pH Curves• Derive the pH curve for titration of 50 mL of 0.200
M HCH3CO2 (Ka=1.8x10-5) with 0.100 M KOH3. After addition of 20 mL of base:
HCH3CO2(aq) + OH-(aq) → H2O(l) + CH3CO2-(aq)
• Addition of base converts a stoichiometric amountof the acid to its conjugate base– [HCH3CO2] decreases and [CH3CO2] increases
according to amount of OH- added
Ch 16.48 of 77
Weak Acid-Strong Base pH Curves• The moles of HCH3CO2 used up and the moles of
CH3CO2- made is the same as the moles of OH-
added
• Moles HCH3CO2 is initial (0.010 mols) minus thatused up (0.002 mols) = 0.008 mols
• Moles CH3CO2- made is 0.002 mols
Mols OH! added = Concentration (M)"Volume (L)
= 0.100M"0.020 L = 0.002
Ch 16.49 of 77
Weak Acid-Strong Base pH Curves• We have some acid and some conjugate base• This is a buffer
pH = pKa + log[A! ]
[HA]
"#$
%&'
= 4.74 + log0.002 mols / 0.070 L
0.008 mols / 0.070 L
"#$
%&'
= 4.14
Total volume =50 mL + 20 mL
Ch 16.50 of 77
Weak Acid-Strong Base pH Curves• Derive the pH curve for titration of 50 mL of 0.200
M HCH3CO2 (Ka=1.8x10-5) with 0.100 M KOH4. After addition of 100 mL of base (at the
endpoint):HCH3CO2(aq) + OH-(aq) → H2O(l) + CH3CO2
-(aq)• At the endpoint, all the weak acid has been
converted to conjugate base• The solution is a salt in water
Ch 16.51 of 77
Weak Acid-Strong Base pH Curves• The salt is KCH3CO2 in water
– K+ is neutral (from a strong base)– CH3CO2
- is basic (from a weak acid)• The moles of CH3CO2
- made are the same as themoles of OH- added
Mols OH! added = Concentration (M)"Volume (L)
= 0.100M"0.100 L = 0.010
[CH3CO2!] =
0.010mols CH3CO2!
0.050 L + 0.100 L= 0.0667M
Ch 16.52 of 77
Weak Acid-Strong Base pH Curves• We have a weak base dissolved in water
CH3CO2-(aq) + H2O(l) ⇌ HCH3CO2(aq) + OH-(aq)
• We can solve this equilibrium with an ICE table
• Note, CH3CO2- acts as a base; we need Kb
xx0.0667 - xEquilibrium+x+x-xChange000.0667Initial
OH-HCH3CO2CH3CO2-
Ch 16.53 of 77
Weak Acid-Strong Base pH CurvesCH3CO2
-(aq) + H2O(l) ⇌ HCH3CO2(aq) + OH-(aq)• CH3CO2
- is the conjugate base of HCH3CO2 (Ka =1.8x10-5) so
• And we can insert values from the ICE table
Kb =KW
Ka
=1.0x10
!14
1.8x10!5
= 5.6x10!10
Kb =[OH! ] " [CH3CO2H]
[CH3CO2!]
Ch 16.54 of 77
Weak Acid-Strong Base pH Curves• Inserting values and solving for x (= [OH-])
• pOH = -log(6.11x10-6) = 5.21• pH = 14.00 - pOH = 8.79
Kb =[OH! ] " [CH3CO2H]
[CH3CO2!]
5.6x10!10 =x2
0.0667! x#
x2
0.0667
x = 6.11x10!6
Ch 16.55 of 77
Weak Acid-Strong Base pH Curves• Derive the pH curve for titration of 50 mL of 0.200
M HCH3CO2 (Ka = 1.8x10-5) with 0.100 M KOH5. After addition of 110 mL base (10 mL past the
endpoint):Weak acid is completely neutralized but there isan excess of OH-
• 100 mL base is used to neutralize the acid so 10mL excess has been added
Mols excess OH!= Concentration (M)"Volume (L)
= 0.100M"0.010 L = 0.001
Ch 16.56 of 77
Weak Acid-Strong Base pH Curves• The [OH-] is
• pOH is
• And pH = 14.00 - pOH = 11.80
[OH! ] =0.001mols
0.050 L + 0.110 L= 0.006M
pOH = !log[OH! ]
= !log(0.006)
= 2.20
50 mL ofHCH3CO2 +110 mL KOH
added
pH is very basicpast endpoint
Same as for strong acid-strongbase since KOH now dominates
Ch 16.57 of 77
Weak Acid-Strong Base pH Curves
12.0712011.801108.791005.70905.35805.11704.92604.74504.57404.38304.14203.79102.720pHVol KOH added (mL) 14
12
10
8
6
4
2
00 10 20 50 100
Volume KOH (mL)
pH
EndpointpH > 7.00
pKa = 4.74
Buffer range
Ch 16.58 of 77
Weak Acid-Strong Base pH Curves1. At start, calculate the pH of a weak acid
– ICE table for weak acid using Ka
2. Before the endpoint, a buffer solution exists– Use Henderson-Hasselbalch equation– Halfway to endpoint, pH = pKa
3. At the endpoint, HA has been converted to itsconjugate base– Use ICE table for salt in water using Kb
4. After the endpoint, OH- is in excess– Calculate [OH-] from moles added minus moles
neutralized (initial moles H+) divided by total volume
Ch 16.59 of 77
Polyprotic Acid-Strong Base pHCurves
• A polyprotic acid produces multiple endpoints intitrations, each endpoint corresponding to theneutralization of each proton
• Consider sulfurous acid:
H2SO3(aq) ⇌ HSO3-(aq) + H+(aq) Ka1 = 1.6x10-2
HSO3-(aq) ⇌ SO3
2-(aq) + H+(aq) Ka2 = 6.4x10-8
• In the first reaction, the first proton in H2SO3 isneutralized
• In the second reaction, HSO3- is neutralized
Ch 16.60 of 77
14
12
10
8
6
4
2
00 10 20 30 40 50 60
Volume KOH (mL)
pH
Buffer range
1st endpoint
pKa1 = 1.80 pKa2 = ~7.19
Buffer range
2nd endpoint
• 25 mL 0.100 M H2SO3 titrated with 0.100 M NaOH
Ch 16.61 of 77
Indicators• The endpoint of an acid-base titration is always
signaled by a rapid change in pH• Sometimes it is more convenient to use an
indicator, a molecule with an acid and conjugatebase that are different colors
Colorless Pink
Ch 16.62 of 77
Indicators• Indicators are weak acids represented by the
generic formula HIn
HIn(aq) + H2O(l) ⇌ In-(aq) + H3O+(aq)
• In acidic conditions, [H3O+] is high and equilibriumshifts left– At low pH the indicator is color 1
• In basic conditions, [H3O+] is small and equilibriumshift right– At high pH the indicator is color 2
Color 1 Color 2
Ch 16.63 of 77
Indicators• Remember, an equilibrium is present so both HIn
and In- will be present all the time but in differentconcentrations
HIn(aq) + H2O(l) ⇌ In-(aq) + H3O+(aq)
• When [In-]/[HIn] ≥10 the solution appears color 2• When [In-]/[HIn] ≤0.1 the solution appears color 1• When [In-]/[HIn] = 1 the solution appears
intermediate
Color 1 Color 2
Ch 16.64 of 77
Indicators• According to the Henderson Hasselbalch equation
• So indicators change color over about 2 pH unitscentered at the pKa of the indicator
pH = pKa + log[In! ]
[HIn]
"#$
%&'
= pKa + log10
1
"#$
%&'
= pKa +1
pH = pKa + log[In! ]
[HIn]
"#$
%&'
= pKa + log1
10
"#$
%&'
= pKa !1Color 2 Color 1
Ch 16.65 of 77
Indicators• What is the pH range over
which methyl red indicator(Ka = 7.9x10-6) changesfrom pink (acid form) toyellow (basic form)?
pKa = -log(7.9x10-6) = 5.1• Color changes at pKa±1• It will be pink at pH = 4.1
(and below) and yellow atpH 6.1 (and above)
pH = pKa when[In-] = [HIn]
Ch 16.66 of 77
Choice of Indicator
Indicator shouldbe chosen so itchanges color
exactly atendpoint (whereindicator pKa =pH at endpoint)
Best for strongacid-strong
base titration
Ch 16.67 of 77
16.5 Solubility Equilibria• A substance is soluble if it dissolves and
insoluble if it does not• But some substances are very soluble and others
slightly soluble• Equilibrium constants can quantitate solubility
PbBr2(s) ⇌ Pb2+(aq) + 2 Br-(aq)Ksp = [Pb2+]·[Br-]2
where Ksp is the solubility product constant forPbBr2
Ch 16.68 of 77
Solubility Product Constants• A larger Ksp means more soluble
Slightly soluble
Very insoluble
Ch 16.69 of 77
Ksp and Molar Solubility• Ksp values can be used to calculate molar
solubility (the solubility in mols/L)• What is the molar solubility of lead chloride PbCl2
(Ksp = 1.17x10-5)?
PbCl2(s) ⇌ Pb2+(aq) + 2 Cl-(aq)• Since this is an equilibrium problem, weʼll write an
ICE table– Donʼt include PbCl2 in the table because it is a solid and
wonʼt enter into expression for Ksp
Ch 16.70 of 77
Ksp and Molar SolubilityPbCl2(s) ⇌ Pb2+(aq) + 2 Cl-(aq)
• Writing an ICE table
Cl-(aq)Pb2+(aq)
2·SSEquilibrium
+2·S+SChange
0.000.00Initial
Weʼll use S (for molarsolubility) rather than x
Ch 16.71 of 77
Ksp and Molar SolubilityPbCl2(s) ⇌ Pb2+(aq) + 2 Cl-(aq)
• Insert into expression for Ksp
Ksp = [Pb2+ ] ! [Cl" ]2
1.17x10"5 = S ! (2 !S)2
= S ! (4 !S2 )
= 4 !S3
S =1.17x10"5
43 =1.43x10"2
Careful!
Molarsolubility =1.43x10-2
mols/L
Ch 16.72 of 77
Ksp and Relative Molar Solubility• Care must be taken when comparing Ksp values
• The relationship between Ksp and molar solubilitydepends on the number of moles of ions made– Ksps can only be compared if the same number of moles
of ions are made otherwise compare molar solubility
Molar solubilityKsp
5.54x10-6 M3.07x10-11FeCO3
3.72x10-5 M2.06x10-13Mg(OH)23 mols
2 mols
Ch 16.73 of 77
16.6 Precipitation Reactions• A precipitate forms when two solutions of soluble
ions are mixed and at least one of the ionicproducts is insoluble
Na2CrO4(s) → 2 Na+(aq) + CrO42-(aq)
AgNO3(s) → Ag+(aq) + NO3-(aq)
• Both solids are very soluble on their own• If mixed, insoluble Ag2CrO4(s) forms
– Really Ag2CrO4 is slightly soluble (Ksp = 1.12x10-12)– NaNO3 is soluble (Ksp large) and remains as Na+ and
NO3- ions
Ch 16.74 of 77
Precipitation Reactions• Remember, tiny quantities
of Ag2CrO4 are soluble• As long as Q < Ksp, no
precipitate forms (anunsaturated solution)
• When Q = Ksp, theprecipitate just begins toform (a saturatedsolution)
Ch 16.75 of 77
Precipitation Reactions• Under some circumstances, a supersaturated
solution can form where Q > Ksp
Supersaturatedsolutions areunstable andusually form a
precipitate,sometimessuddenly!
Ch 16.76 of 77
Precipitation Reactions• Does a solution of 0.020 M Ca(CH3CO2)2(aq)
mixed with 0.004 M Na2SO4(aq) produce aprecipitate of CaSO4(s) (Ksp = 7.10x10-5)?
• We need to compare Q with Ksp
Ca(CH3CO2)2(s) → Ca2+(aq) + 2 CH3CO2-(aq)
• [Ca2+] = 0.020 MNa2SO4(s) → 2 Na+(aq) + SO4
2-(aq)• [SO4
2-] = 0.004 M
Ch 16.77 of 77
Precipitation Reactions• The reaction relating to Ksp is
CaSO4(s) → Ca2+(aq) + SO42-(aq)
• And
• Since Q > Ksp, we have exceeded the solubilityand a small amount of CaSO4(s) shouldprecipitate
Q = [Ca2+ ] ! [SO42"]
= 0.020#0.004
= 8x10"5 Ksp = 7.10x10-5