11

Buffers & TitrationsBuffers & Titrations

Chapter 18Chapter 18

22

Buffers Buffers

• A soln that resists change in pH when A soln that resists change in pH when strong acid or strong base is addedstrong acid or strong base is added

• Made from Made from weak acid/conjugate-baseweak acid/conjugate-base• OrOr• Made from Made from weak base/conjugate-acidweak base/conjugate-acid• Strong acids and strong bases don’t Strong acids and strong bases don’t

make buffers!make buffers!• So how does it work?So how does it work?

33

How buffers workHow buffers work

• Acetic acid/acetate buffer system:Acetic acid/acetate buffer system:

CC22HH33OO22HH(aq) (aq) + H+ H22OO(l)(l) C C22HH33OO22--(aq) (aq) + H+ H33OO++

(l)(l)

• Add base: Add base:

CC22HH33OO22HH(aq)(aq) + OH + OH--(aq)(aq) C C22HH33OO22

--(aq) (aq) + H+ H22OO(l)(l)

• Add acid:Add acid:

HH33OO++(aq)(aq) + C + C22HH33OO22

--(aq) (aq) H H22OO(l) (l) +C+C22HH33OO22HH(aq)(aq)

44

CCommon ommon IIon on EEffect: CIEffect: CIE

• The ionization of an acid or a base is The ionization of an acid or a base is limited by the presence of its conjugate limited by the presence of its conjugate base or acid.base or acid.

HAcHAc(aq)(aq) + H + H22OO(l)(l) Ac Ac--(aq)(aq) + H + H33OO++

(aq)(aq)

• Acetate ion is added in form of NaAcAcetate ion is added in form of NaAc– Which way will this shift the rxn?Which way will this shift the rxn?– Would you expect a greater or lesser acidity if Would you expect a greater or lesser acidity if

the CIE was lacking?the CIE was lacking?

• Let’s look at the next problemLet’s look at the next problem

55

Calculating pH of a buffer Calculating pH of a buffer solnsoln

• You have an acetic acid/acetate You have an acetic acid/acetate buffer with a 0.700 M conc of acetic buffer with a 0.700 M conc of acetic acid and a 0.600 M conc of acetate acid and a 0.600 M conc of acetate ion. What’s the pH of the buffer? (Kion. What’s the pH of the buffer? (Kaa = 1.8 x 10= 1.8 x 10-5-5))

66

Solution Solution

- +(aq) 2 (l) (aq) 3 (aq)

-5a

HAc + H O Ac + H O

I 0.700M -- 0.600 0

C -x -- +x +x

E 0.700 - x -- 0.600+x x

[0.600+x][x]K = 1.8 x 10

[0

-5

-5

.700 x]

CIE: addition or subtraction of "x" from original concentrations very small

[0.600][x]

[0.700]

x=2.1 10 M

pH=-log(2.1 10 )=4.68

77

Let’s work on thisLet’s work on this

•Consider 100.0 mL of a buffer solution that is 1.00M in HAc and 1.00M in NaAc. What is the pH after addition of 25.0 mL of 1.00M NaOH?

88

Solution Solution - +

(aq) 2 (l) (aq) 3 (aq)

-5a

HAc + H O Ac + H O

I 1.00M -- 1.00 0

C -x -- +x +x

E 1.00 - x -- 1.00 + x x

[1.00+x][x] [1.00][x]K = 1.8 x 10

[1.00 x] [1

-5

-5

- -5

- -(aq) (aq) (aq) 2 (l)

.00]

x=1.8 x 10 M

Thus, [HAc] 1.00M 1.8 x 10 M =1.00M

And [Ac ]=1.00M +1.8 x 10 M=1.00M

(By the way, the pH=4.74)

HAc OH Ac + H O

mol 1acid: 1.00 0.1000L acid 0.800M

L acid 0.1250L total

bas

a

mol 1e: 1.00 0.0250L 0.200M

L 0.1250Lmol 1

acetate ion: 1.00 0.1000L 0.800ML 0.1250L

I 0.800M 0.200 0.800 --

C -0.200 -0.200 +0.200 --

E 0.600 0 1.000

So K

-5 + -53

-5

[1.000][x]= 1.8 x 10 , x = [H O ] 1.1 x 10

[0.600]

pH = -log(1.1 x 10 ) 4.96

99

Henderson-Hasselbalch Henderson-Hasselbalch equationequation

• Useful for previous problemUseful for previous problem– Let’s take a lookLet’s take a look

+ -3

a

+3 a-

-

[H O ][A ]K =

[HA]

[HA]Re-arrange: [H O ]= K

[A ]

Take the negative log

[A ]pH = pKa + log( )

[HA]

1010

TitrationsTitrations

• Used to determine quantity of acid or base Used to determine quantity of acid or base in unknown (analyte)in unknown (analyte)

1111

• When sample is neutralized (HWhen sample is neutralized (H33OO++ = = OHOH--))– Equivalence pointEquivalence point

• Determined by use of Determined by use of pH indicatorpH indicator

1212

Different titration typesDifferent titration types

• Strong acid-strong base: equivalence Strong acid-strong base: equivalence pt = 7.0 (contains neutral salt)pt = 7.0 (contains neutral salt)

1313

Different titration typesDifferent titration types

• Weak acid-strong base: equivalence Weak acid-strong base: equivalence pt is greater than 7 (basic salt)pt is greater than 7 (basic salt)

• pH @ half-equivalence pt (halfway pH @ half-equivalence pt (halfway pt) = pKpt) = pKaa

1414

ProblemProblem

•We start with 50.0 mL of 0.100 M HAc. 25.0 mL of 0.100 M NaOH is then added. What is the pH of the resulting solution?

1515

Solution Solution - -

(aq) (aq) (aq) 2 (l)HAc OH Ac + H O

mol 1acid: 0.100 0.0500L 0.0667M

L 0.0750Lmol 1

base: 0.100 0.0250L 0.0333ML 0.0750L

I 0.0667 0.0333 0 --

C -0.0333 -0.0333 +0.0333 --

E 0.0334 0

--5

0.0333 --

[A ] [0.0333]pH = pKa + log( )= -log(1.8 x 10 ) log( ) 4.74

[HA] [0.0334]