Buffers, Titrations, and Buffers, Titrations, and Aqueous EquilibriaAqueous Equilibria

Common Ion EffectCommon Ion Effect

The shift in equilibrium that occurs because The shift in equilibrium that occurs because of the addition of an ion already involved in of the addition of an ion already involved in the equilibrium reaction. (Le Chatelier’s the equilibrium reaction. (Le Chatelier’s principle)principle)

AgCl(AgCl(ss) ) Ag Ag++((aqaq) + Cl) + Cl((aqaq))

This affects the concentrations of other ions, This affects the concentrations of other ions, notably Hnotably H++

adding NaCl( ) shifts equilibrium positionaq

Calculations with ICECalculations with ICEWhat is the [H+] and % ionization of 1.0 M HF mixed with 1.0 M NaF. (1.0 M HF alone, [H+]=2.7 x 10-2, %ion=2.7%)

HF(aq) <====> H+ (aq) + F- (aq)

I 1.0 0 1.0

C -x +x +x

E 1.0-x x 1.0 + x

Ka = 7.2 x 10-4 = x (1.0+x) / (1.0 - x) = x/1

x=[H+]= 7.2 x 10-4 % ion = 7.2 x 10-4 /1.0 = .072%

A Buffered SolutionA Buffered Solution

. . . resists change in its pH when either H. . . resists change in its pH when either H++ or OH or OH are added.are added.

1.0 L of 0.50 M H1.0 L of 0.50 M H33CCOOH CCOOH

+ 0.50 M H+ 0.50 M H33CCOONaCCOONa

pH = 4.74pH = 4.74

Adding 0.010 mol Adding 0.010 mol solid NaOH solid NaOH raises the raises the pHpH of of the solution to the solution to 4.764.76, a very minor change., a very minor change.

Key Points on Buffered Key Points on Buffered SolutionsSolutions

1.1. They are weak acids or bases containing a They are weak acids or bases containing a common ioncommon ion..

2.2. After addition of strong acid or base, deal After addition of strong acid or base, deal with with stoichiometry firststoichiometry first, then equilibrium., then equilibrium.

Demonstration of Buffer Demonstration of Buffer ActionAction

A buffered solution of 1.0 L of 0.5 M HC2H3O2 Ka=1.8 x 10-5 and 0.5 M NaC2H3O2 has 0.01 mol of solid NaOH added. What is the new pH?

HC2H3O2 (aq)<====> C2H3O2-(aq) + H+ (aq)

I 0.5 0.5 0

C -x +x +x

E 0.5 -x 0.5 +x x

Ka=1.8 x 10-5 = 0.5(x) / 0.5 x= 1.8 x 10-5 pH= 4.74

When OH- is added, it takes away an equal amount of H+ ions, and will affect also the acid concentration by the same amount. It will also add to the common ion.

HC2H3O2 (aq)<====> C2H3O2-(aq) + H+ (aq)

I 0.5 0.5 1.8 x 10-5

S -0.01 +0.01 - 1.8 x 10-5 due to adding base

I 0.49 0.51 0

C -x +x +x

E 0.49-x 0.51 +x x

Ka= 1.8 x 10-5 = x (.51)/ .49 x= 1.73 x 10-5 pH= 4.76

Practically no change in pH after base is added.

Henderson-Hasselbalch Henderson-Hasselbalch EquationEquation

- Useful for calculating pH when the Useful for calculating pH when the [A[A]/[HA] ratios are known.]/[HA] ratios are known.

pH p log( A HA

p log( base acid

a

a

K

K

/ )

/ )

Base Buffers With H/H Base Buffers With H/H EquationEquation

pOH = pKb + log ( [HB+] / [B] ) =

pKb + log ([acid]/ [base])

Base buffers can be calculated in similar fashion to acid buffers. pH = 14 - pOH

Buffered Solution CharacteristicsBuffered Solution Characteristics

- Buffers contain relatively large amounts of weak Buffers contain relatively large amounts of weak acid and corresponding base.acid and corresponding base.

- Added HAdded H++ reacts to completion with the weak reacts to completion with the weak base.base.

- Added OHAdded OH reacts to completion with the weak reacts to completion with the weak acid.acid.

- The pH is determined by the ratio of the The pH is determined by the ratio of the concentrations of the weak acid and weak base.concentrations of the weak acid and weak base.

H/H CalculationsH/H CalculationsFrom the previous example, [HC2H3O2] = .49 M after the addition of base, and [C2H3O2

-] = .51 M. Using the H/H equation,

pH= pKa + log (.51)/(.49)

= 4.74 + 0.02 = 4.76 same as using ICE, but easier

A weak base buffer is made with 0.25 M NH3 and 0.40 M NH4Cl. What is the pH of this buffer? Kb = 1.8 x 10-5

NH3 (aq) + H2O (l) <===> NH4+ (aq) + OH- (aq)

pOH = pKb +log ([NH4+] / [NH3]) = 4.74 +log (.40/.25)= 4.94

pH = 14- 4.94 = 9.06

H/H CalculationsH/H CalculationsThe weak base buffer with pH of 9.06 from the previous example has 0.10 mol of HCl added to it. What is the new pH?

NH3 (aq) + H2O (l) <===> NH4+ (aq) + OH- (aq)

I .25 .40 1.15 x 10-5

S - .10 + .10 - 1.15 x 10-5 due to acid

I .15 .50 0

Using H/H, pOH = 4.74 + log (.5)/(.15) = 5.26

pH= 14 - 5.26 = 8.74

Buffering CapacityBuffering Capacity

. . . represents the amount of H. . . represents the amount of H++ or OHor OH the buffer can absorb the buffer can absorb without a significant change without a significant change in pH.in pH.

Titration (pH) CurveTitration (pH) Curve

A plot of pH of the solution being analyzed as A plot of pH of the solution being analyzed as a function of the amount of titrant added.a function of the amount of titrant added.

Equivalence (stoichiometric) pointEquivalence (stoichiometric) point: Enough : Enough titrant has been added to titrant has been added to react exactly react exactly with the with the solution being analyzed.solution being analyzed.

15_327

01.0

Vol NaOH added (mL)

50.0

7.0

13.0pH

100.0

Equivalencepoint

Strong Acid/Base TitrationStrong Acid/Base TitrationIn titrations, it is easier to calculate millimoles(mmol), which would be Molarity x mL, in stoichiometry.

What would be the pH of 50.0 mL of a 0.2 M solution of HCl after 20.0 mL of 0.1 M NaOH have been added?

HCl---50.0 x 0.2 =10.0 mmol NaOH--- 20.0 x 0.1= 2.0 mmol

H+ + OH- ----> H2O volumes must be added

I 10mmol 2 mmol 50 + 20 = 70

S -2 mmol -2mmol [H+] = 8 mmol/ 70 mL = 0.11 M

End 8mmol 0 mmol pH= 0.95

Weak Acid - Strong Base Weak Acid - Strong Base TitrationTitration

Step 1 -Step 1 - A stoichiometry problem A stoichiometry problem - reaction is - reaction is assumed to run to completion - then assumed to run to completion - then determine remaining species.determine remaining species.

Step 2 -Step 2 - An equilibrium problem An equilibrium problem - determine - determine position of weak acid equilibrium and position of weak acid equilibrium and calculate pH.calculate pH.

15_329

Vol NaOH added (mL)

25 50

3.0

9.0

12.0 Equivalencepoint

pH

Weak Acid/Strong BaseWeak Acid/Strong BaseFirst do a stoichiometry, then equilibrium using H/H equation

What is the pH of 50.0 mL of a 0.100M HCN solution after 8.00 mL of 0.100 M NaOH has been added? Ka= 6.2 x 10-10

HCN + OH- -----> H2O + CN- Volume

I 5 mmol 0.8 mmol 0 mmol 50 + 8 = 58

S - 0.8 mmol -0.8 mmol +0.8 mmol

I 4.2 mmol /58mL 0.8 mmol /58 mL

0.072M 0.0138 M

pH = pKa + log (0.0138/0.072) = 8.49

15_330

Vol NaOH

Strong acid

pH

Weak acid

15_331

Vol 0.10 M NaOH added (mL)

10 20 30 40 50 60

2.0

4.0

6.0

8.0

10.0

12.0

0

Ka = 10–2

Ka = 10–4

Ka = 10–6

Ka = 10–8

Ka = 10–10

Strong acid

pH

15_328

Vol 1.0 M HCl added

7.0

14.0

50.0 mL

Equivalencepoint

pH

15_332

0Vol 0.10 M HCl (mL)

10

2

20 30 40 50 60 70

4

6

8

10

12

EquivalencepointpH

Acid-Base IndicatorAcid-Base Indicator

. . . marks the . . . marks the end point end point of a titration by of a titration by changing color.changing color.

The The equivalence point equivalence point is not necessarily the is not necessarily the same as the same as the end pointend point..

15_333

– O

C

O

C O–

O

(Pink base form, In– )

HO

COH

C O–

O

(Colorless acid form, HIn)

OH

15_3340 1 2 3 4 5 6 7 8 9 10 11 12 13

The pH ranges shown are approximate. Specific transition ranges depend on the indicator solvent chosen.

pH

Crystal Violet

Cresol Red

Thymol Blue

Erythrosin B

2,4-Dinitrophenol

Bromphenol Blue

Methyl Orange

Bromcresol Green

Methyl Red

Eriochrome* Black T

Bromcresol Purple

Alizarin

Bromthymol Blue

Phenol Red

m - Nitrophenol

o-Cresolphthalein

Phenolphthalein

Thymolphthalein

Alizarin Yellow R

* Trademark CIBA GEIGY CORP.

15_335AB

pH

00

Vol 0.10 M NaOH added (mL)

2

4

6

8

10

12

14

20 40 60 80 100 120

Equivalencepoint pH

00

Vol 0.10 M NaOH added (mL)

2

4

6

8

10

12

14

20 40 60 80 100 120

Equivalencepoint

Phenolphthalein

Methyl red

Phenolphthalein

Methyl red

Homework !!Homework !!

p. 761 33, 35, 45 (for 35 only)p. 761 33, 35, 45 (for 35 only)

Solubility ProductSolubility Product

For solids dissolving to form aqueous solutions.For solids dissolving to form aqueous solutions.

BiBi22SS33((ss) ) 2Bi 2Bi3+3+((aqaq) + 3S) + 3S22((aqaq))

KKspsp = solubility product constant = solubility product constant

andand KKspsp = [Bi = [Bi3+3+]]22[S[S22]]33

Solubility ProductSolubility Product““Solubility” = Solubility” = ss = concentration of Bi = concentration of Bi22SS33 that that

dissolves, which equals 1/2[Bidissolves, which equals 1/2[Bi3+3+] and 1/3[S] and 1/3[S22].].

BiBi22SS3 3 (s) <==> 2 Bi(s) <==> 2 Bi3+3+ (aq)+ 3 S (aq)+ 3 S22aq)aq)

2 2 ss 3 3 ss

Note:Note: KKspsp is constant (at a given temperature) is constant (at a given temperature)

ss is variable (especially with a common is variable (especially with a common ion present)ion present)

Relation of KRelation of Kspsp to to ss

Ionic compounds will dissociate according to Ionic compounds will dissociate according to set equations, which will give set Kset equations, which will give set Kspsp to to ss

relation. Use ICE to find it.relation. Use ICE to find it.

AA22B <==> 2A + BB <==> 2A + B

22ss ss K Kspsp = (2 = (2ss))22((ss)=4s)=4s33

ABAB33 <==> A + 3B <==> A + 3B

ss 3 3ss K Kspsp= (= (ss)(3)(3ss))33=27=27ss44

KKspsp from from ssIf 4.8 x 10-5 mol of CaC2O4 dissolve in 1.0 L of solution, what is its Ksp?

[CaC2O4]= mol / L = 4.8 x 10-5 / 1.0 L = 4.8 x 10-5 M

CaC2O4 (s) <==> Ca+2 (aq) + C2O4-2 (aq)

s s

Ksp= (s)(s)= s2 = (4.8 x 10-5 )2 = 2.3 x 10-9

If the molar solubility (s) of BiI3 is 1.32 x 10-5, find its Ksp.

BiI3 (s) <==> Bi+3 + 3I-

s 3s Ksp= (s)(3s)3= 27 s4

Ksp = 27(1.32 x 10-5)4= 8.20 x 10-19

ss from K from Kspsp

If Ksp for Ag2CO3 is 8.1 x 10-12, find its molar solubility.

Ag2CO3 (s) <==> 2 Ag+ (aq) + CO3-2 (aq)

2 s s

Ksp = (2s)2(s) = 4 s3 = 8.1 x 10-12 s= 1.27 x 10-4

If Ksp for Al(OH)3 is 2 x 10-32, find its molar solubility.

Al(OH)3 (s) <==> Al+3 (aq) + 3 OH- (aq)

s 3 s

Ksp = (s)(3s)3 = 27s4 = 2 x 10-32 s = 5.22 x 10-9

Equilibria Involving Complex Equilibria Involving Complex IonsIons

Complex IonComplex Ion: A charged species consisting of a : A charged species consisting of a metal ion surrounded by metal ion surrounded by ligandsligands (Lewis bases). (Lewis bases).

Coordination NumberCoordination Number: Number of ligands attached : Number of ligands attached to a metal ion. (Most common are 6 and 4.)to a metal ion. (Most common are 6 and 4.)

Formation (Stability) ConstantsFormation (Stability) Constants: The equilibrium : The equilibrium constants characterizing the stepwise addition of constants characterizing the stepwise addition of ligands to metal ions.ligands to metal ions.

Complex Ion ExamplesComplex Ion Examples

Coordination number=2 Coordination number=2

Ag(NHAg(NH33))22++

Coordination number=4Coordination number=4

Al(OH)Al(OH)44-1-1 Cu(NHCu(NH33))44

+2+2

Coordination number=6Coordination number=6

Fe(CN)Fe(CN)66-4-4 Fe(SCN)Fe(SCN)66

-3-3

Homework and XCRHomework and XCR

p. 762ff 50, 51, 59, 60p. 762ff 50, 51, 59, 60

XCR 72, 76, 86XCR 72, 76, 86

Super XCR 90Super XCR 90