Aqueous Ionic Equilibrium

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


[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


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


[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


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


[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


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


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


= 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


addedpH increasesas more base

is added

Ch 16.36 of 77


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


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


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


M HCH3CO2 (Ka=1.8x10-5) with 0.100 M KOH1. Before addition of any base:


• [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


M HCH3CO2 (Ka=1.8x10-5) with 0.100 M KOH2. After addition of 10 mL of base:


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


= 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


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


= 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


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


= 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


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

Aqueous Ionic Equilibrium

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