1

Buffer Solutions

pH of solution adding 0.10 M HCl to 100 mL water

HCl added pH

0 mL 7.00

2 mL 2.71

5 mL 2.32

10 mL 2.04

20 mL 1.78

50 mL 1.48

mL of 0.10 M HCl added

0 10 20 30 40 501

2

3

4

5

6

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Buffer Solutions

A buffer helps a solution maintain its pH whenacid or base is added

A buffer must contain two components to work

a weak acid that reacts with added base

a weak base that reacts with added acid

Buffers usually contain approximately equalamounts of a weak acid and its conjugate base

2

Buffer Solutions

Solution that is 0.100 M CH3COOH (acetic acid)and 0.100 M NaCH3COO (sodium acetate)

Find pH of buffer solution:CH3COOH(aq) + H2O CH3COO-(aq) + H3O

+(aq)

[CH3COOH] [CH3COO-] [H3O+]

initial 0.100 0.100 0

-x x x

equil 0.100 – x 0.100 + x x

Buffer Solutions

Find pH of buffer solution:CH3COOH(aq) + H2O CH3COO-(aq) + H3O

+(aq)

Ka = [CH3COO-][H3O

+]

[CH3COOH] =

(.100 + x)x

(.100 - x) = 1.8 x 10-5

x = 1.80 x 10-5 M

pH = 4.74

assume x is negligiblecompared to .100 M

3

Buffer Solutions

Add 5 mL .10 M HCl

Find pH of resulting solution

Assume all acid added reacts with acetate ion to formacetic acid (remember that acids react with bases)

CH3COOH(aq) + H2O CH3COO-(aq) + H3O+(aq)

[H3O+] added = (5 mL)(.10 M)/(105 mL) = 4.76x10-3 M

[CH3COOH] = (0.100)(100/105) + 0.005 = 0.100 M

[CH3COO-] = (0.100)(100/105) - 0.005 = 0.090 M

(100/105)=dilution factor for addition of HCl

Buffer Solutions

Now let solution come to equilibrium

[CH3COOH] [CH3COO-] [H3O+]

initial .100 .090 0

-x x x

equil .100 – x .090 + x x

Ka = [CH3COO-][H3O

+]

[CH3COOH] =

(.090 + x)x

(.100 - x) = 1.8 x 10-5

x = 2.00 x 10-5 M

pH = 4.70

4

Buffer Solutions

pH of buffered solution adding 0.10 M HCl to 100mL soln

HCl added pH

0 mL 4.74

5 mL 4.70

10 mL 4.66

15 mL 4.61

25 mL 4.52

50 mL 4.28mL of 0.10 M HCl added

0 10 20 30 40 501

2

3

4

5

6

7

Buffered solution

Buffer Solutions

Henderson-Hasselbach Equation

Allows calculation of pH of a buffer ifconcentrations of conjugate acid and conjugatebase are known

HA(aq) + H2O H3O+(aq) + A-(aq)

Ka = [H3O

+][A-][HA]

[H3O+] =

Ka[HA]

[A-]

5

Buffer Solutions

Take -log of both sides

log[H3O+] = - log

Ka[HA]

[A-]

= - log Ka( ) - log

[HA]

[A-]

-log(Ka) = pKa log[HA]

[A ]

= log

[A-]

[HA]

pH = pKa + log[A-]

[HA]

Henderson-Hasselbach Eqn

Buffer Solutions

Using the Henderson-Hasselbach Eqn, we can:

Determine pH of a solution

Determine ratio of conjugate base toconjugate acid to achieve specific pH

pH = pKa + log[A-]

[HA]

6

Buffer Solutions

Let’s go back to problem of adding HCl to buffersolution:

We can use H-H eqn. to make the calculationsmuch easier

[CH3COOH] = 0.100 + [HCl]added

[CH3COO-] = 0.100 – [HCl]added

pH = pKa + log[A-]

[HA]

Buffer Solutions

VHCl [HCl] [CH3COOH] [CH3COO-] pH

5mL (.1)(5mL)/105mL .095+.005 .095-.005

= .00476 M =.100 M =.090 M 4.70

10mL (.1)(10)/110 .091+.009 .091-.009

= .00909 M =.100 M =.082 M 4.66

25mL (.1)(25)/125 .080+.020 .080-.020

= .0200 M =.100 M =.060 M 4.52

50mL (.1)(50)/150 .067+.033 .067-.033

= .0333 M =.100 M =.034 M 4.28

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Buffer Solutions

Buffer Capacity—the amount of acid or base thatcan be added to a buffer without the pHsignificantly changing

Suppose we acid to a buffer solution:

The acid will react with the conjugate base until it isdepleted

Past this point, the solution behaves as if no bufferwere present

Acid-Base Titrations

A titration is a method used to determine theconcentration of an unknown species

Add a measured amount of a known reactant

Determine when the reaction has gone tocompletion

[unknown] + [known] products

At the equivalence point

moles unknown = moles known added

CunknownVunknown = CknownVknown

8

Acid-Base Titrations

Titrate 50.00mL unknownHCl soln.with 0.2137M NaOH

mL NaOH added

0 5 10 15 20 25 30 35 40

1

2

3

4

5

6

7

8

9

10

equivalence point

Acid-Base Titrations

At equivalence point, VNaOH

= 23.96 mL

mol(NaOH) =

(.2137 M)(.02396 L)

= 5.120 x 10-3 mol

mol(HCl) = 5.120 x 10-3 mol

(mol known = mol unknown)

[HCl] =

(5.120x10-3 mol)/(.05000 L)

= 0.1024 M

mL NaOH added

0 5 10 15 20 25 30 35 40

1

2

3

4

5

6

7

8

9

10

equivalence point

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Indicators

An indicator is a chemical species that changescolor depending on the pH of the solution

An indicator is a conjugate acid-conjugate basepair in which the acid and base forms of thecompound have different colors

HIn(aq) + H2O In-(aq) + H3O+(aq)

color 1 color 2

Indicators are used to determine the endpointof a titration

Indicators

The pKa of the indicator determines the pHrange over which the color changes

[HIn]/[In-] 10 acid color

[HIn]/[In-] 0.1 base color

[HIn]/[In-] 1 intermediate color

Remember: pH = pKa + log{[In-]/[HIn]}

If [HIn]/[In-] = 1, log{[HIn]/[In-]} = 0

pH = pKa at point when indicator is

changing color

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Indicators

Indicator pKa pH range color change

Methyl orange 3.7 3.1 – 4.4 red to yellow

Bromophenol blue 4.0 3.0 – 4.6 yellow to blue

Methyl red 5.1 4.2 – 6.3 red to yellow

Bromothymol blue 7.0 6.0 – 7.6 yellow to blue

Phenol red 7.9 6.8 – 8.4 yellow to red

Phenolphthalein 9.3 8.2 – 10.0 clear to pink

Indicators

Figure 17.5: pH curve fortitration of 0.100 M HCl with0.100 M NaOH

change before endpoint

change after endpoint

changearound

endpoint

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Indicators

Titration of weak acid with strong base

HA(aq) + OH-(aq) A-(aq) + H2O

At equivalence point

A-(aq) + H2O HA(aq) + OH-(aq)

the solution is basic because conjugate base ofweak acid reacts with water to form OH-(aq)

Indicators

Titrate 25.00 mL 0.100 M formic acid (HCOOH) with0.100 M NaOH

Ka = 1.8 x 10-4

Find pH at equivalence point and select appropriateindicator

At equivalence point, mol(HCOOH) = mol(OH-)

mol(fa) = (0.100 M fa)(0.02500 L) = 2.5 x 10-3 mol

fa = formic acid

VNaOH added = (2.5 x 10-3 mol)/(0.100 M) = 25.0 mL

Vtotal = 50.0 mL

12

Indicators

Assume HCOOH + OH- reaction goes to completion:

[HCOO-] = (2.5 x 10-3 mol)/(0.0500 L) = 0.0500 M

Determine Keq for reaction of formate ion:

HCOO-(aq) + H2O HCOOH(aq) + OH-(aq)

Keq = [HCOOH][OH-]

[HCOO ] =

Kw

Ka

= 6.67 x 10-11

Indicators

[HCOO-] [HCOOH] [OH-]

initial .0500 0 0

-x x x

equil .0500 – x x x

Keq = [HCOOH][OH-]

[HCOO ] =

Kw

Ka

= 6.67 x 10-11

Keq = [HCOOH][OH-]

[HCOO-] =

x2

0.0500 x = 6.67 x 10-11

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Indicators

x = 1.83 x 10-6 M = [OH-]

pOH = -log(1.83 x 10-6) =5.74

pH = 14.00 – 5.74 = 8.26

Phenol red (6.8 – 8.4) or phenolphthalein (8.2 – 10.0)would be appropriate indicators

Keq = [HCOOH][OH-]

[HCOO-] =

x2

0.0500 x = 6.67 x 10-11

Indicators

Titration of weak base with strong acid

B(aq) + H3O+(aq) BH+(aq) + H2O

At equivalence point

BH+(aq) + H2O B(aq) + H3O+(aq)

the solution is acidic because conjugate acid ofweak base reacts with water to form H3O

+(aq)

14

Indicators

Figure 17.8: titration of 0.100 MNH3 with 0.100 M HCl

Acid Rain

Carbon dioxide in the air is in equilibrium withH2O in atmospheric water droplets (clouds &fog):

CO2(aq) + H2O H2CO3(aq)

carbonic acid Ka = 4.2 x 10-7

H2CO3(aq) + H2O H3O+(aq) + HCO3

-(aq)

Natural rain water has pH = 5.6

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Acid Rain

Emitted pollutants can form additional acidsources in clouds:

NO2:

2 NO2(aq) + H2O HNO3(aq) + HNO2(aq)

nitric acid nitrous acid

strong Ka = 4.5 x 10-4

Acid Rain

Emitted pollutants can form additional acidsources in clouds:

SO2:

SO2(aq) + H2O H2SO3(aq)

sulfurous acid Ka = 1.2 x 10-2

2 SO2(g) + O2(g) 2 SO3(g)

SO3(aq) + H2O H2SO4(aq)

sulfuric acid

strong

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Acid Rain

Solubility Products

Many salts are only slightly soluble

The solubility product is a measure of theconcentration of ions in a solution saturated withthe saltMA(s) M+(aq) + A-(aq) Ksp = [M+][A-]

Examples

AgCl(s) Ag+(aq) + Cl-(aq) Ksp=[Ag+][Cl-]=1.8x10-10

PbCl2(s) Pb2+(aq) + 2 Cl-(aq) Ksp=[Pb2+][Cl-]2=1.7x10-5

AuBr3(s) Au3+(aq) + 3 Br-(aq) Ksp=[Au3+][Br-]3=4.0x10-36

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Solubility Products

Knowing the Ksp, we can calculate theconcentration of ions in solution

Examples

AgCl(s) Ag+(aq) + Cl-(aq) Ksp=[Ag+][Cl-]=1.8x10-10

-x x x

x2 = 1.8 x 10-10 x = 1.3 x 10-5 M = [Ag+] = [Cl-]

Solubility Products

Examples

PbCl2(s) Pb2+(aq) + 2 Cl-(aq) Ksp=[Pb2+][Cl-]2=1.7x10-5

-x x 2x

x(2x)2 = 1.7 x 10-5

4x3 = 1.7 x 10-5 x = 1.6 x 10-2 M

[Pb2+] = 1.6 x 10-2 M

[Cl-] = 3.2 x 10-2 M

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Solubility Products

Examples

AuBr3(s) Au3+(aq) + 3 Br-(aq) Ksp=[Au3+][Br-]3=4.0x10-36

-x x 3x

x(3x)3 = 4.0 x 10-36

27x4 = 4.0 x 10-36 x = 6.2 x 10-10 M

[Au3+] = 6.2 x 10-10 M

[Br-] = 1.9 x 10-9 M

Solubility Products

Examples—Common ion effect

How much PbI2 will dissolve in a 0.0100 M solution of NaI?

PbI2(s) Pb2+(aq) + 2 I-(aq) Ksp = 8.7 x 10-9

-x x 2x + .0100

x(2x+.0100)2 = 8.7 x 10-9

x(4x2 + 0.0200x + 1.0x10-4) = 8.7 x 10-9

4x3 + .0200x2 + 1.0x10-4x – 8.7x10-9 = 0

x = 8.6 x 10-5 M

vs 1.3 x 10-3 M if no I-(aq) were present initially

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Factors Affecting Solubility

Salts that are slightly soluble in water can bemuch more soluble in acid if one or both of itsions are moderately basic:CaCO3(s) Ca2+(aq) + CO3

2-(aq) Ksp = 8.7 x 10-9

But CO32-(aq) is the conjugate base of HCO3

-(aq)CO3

2-(aq) + H3O+(aq) HCO3

-(aq) + H2O Kb = 2.1x10-4

HCO3-(aq) + H3O

+(aq) H2CO3(aq) + H2O Kb = 2.4x10-8

H2CO3(aq) H2O + CO2(g) Keq 105

Works for carbonates, some sulfides, phosphates,

etc. (species that behave as bases [no too weak])

Precipitation

Define ion quotient, Q, as:

MxAy(s) x My+(aq) + y Ax-(aq)

Q = [My+]x[Ax-]y

A precipitate will form only when Q exceeds Ksp

Q < Ksp: solution is unsaturated—no precipitate

Q > Ksp: solution is saturated—precipitate forms

Q = Ksp: solution at saturation point