Chemical Equilibria

1
Chemical Equilibria

• Dynamic Equilibria
• Acid Base Equilibria
• Presented by Simon Hung Kong Sang

2
A Dynamic Equilibrium

• Two opposing changes are continuously taking
  place on the molecular level
• Forward rate Backward rate
• Composition/bulk properties (e.g. vapor pressure)
  of the closed system remains unchanged once the
  equilibrium is reached at a constant temperature
• The equilibrium can be reached from both sides

3
Examples of equilibrium system

• Equilibria existing between the phases
  Br2(l) Br2(g)
• N2(g) 3H2(g) 2NH3(g)
• BiCl3(aq) H2O(l) BiOCl(s) 2HCl(aq)
• NH3(g) H2O(l) NH4(aq) OH-(aq)
• 2SO2(g) O2(g) 2SO3(g)
• Br2(aq) H2O(l) H(aq) Br-(aq) HOBr(aq)
  
• Cr2O72-(aq) H2O(l) 2CrO42-(aq) 2H(aq)

4
Equilibrium position shift

• Add little concentrated HCl to solid BiOCl to
  shift the equilibrium position to the left. The
  dissolution can be reversed by adding a large
  excess of water
• Orange color of Cr2O72- deepens upon addition of
  acid, yellow CrO42- is formed when Cr2O72- is
  dissolved in an alkaline medium
• Adding OH- to bromine water makes it colorless.
• Fe3(aq) NCS-(aq) FeNCS2(aq) Adding a
  soluble Fe3 salt to the equilibrium mixture
  shifts the position of equilibrium to the right,
  giving a deep red solution.

5
The Constancy of the Mass Action Expression

• If the concentration of one or more of the
  reactants in a reversible equilibrium is changed,
  then the equilibrium will shift in such a way as
  to preserve the constancy of the relative
  concentrations of the products and reactants
• For aA bB cC, Cc/AaBb is termed
  the Mass action expression. It has a constant
  value at a particular temperature. Any
  alternation in individual concentration terms
  would change the remainder of the concentration
  terms so as to keep a constant value as that
  before the change

6
Concentration changes with time
Time for equilibrium to be reached
Moles of reactants products
b
products
a
Concentrations of all substances remain constant
at equilibrium
Reactants
Time
a moles of each reactant mix to form b moles of
products
7
Equilibrium Expression

• If the reaction for the equilibrium is
  represented by the equation aA bB xX
  yY then the expression Xx Yy
  
  

Kc
Aa Bb
is known as the equilibrium expression where Kc
is a constant at a particular temperature, so
called the molar concentration equilibrium
constant.
8
Relationship of Kc to the balanced equation

• When stating Kc for a particular reaction, it is
  useful to state the equation on which the
  constant is based.
• For SO2 1/2 O2 SO3, Kc SO3/SO2O21/2
  For 2SO2 O2 2SO3, Kc
  SO32/SO22O2, Kc are stated in terms
  of the relative molar concentrations
• Kc Kc2
• If the number of particles on each side is the
  same, Kc has no unit. For others, units of Kc
  must be stated.

9
Equilibrium constants in terms of partial
pressures for gaseous systems, Kp

• Assume ideal gas equation P/RT n/V molar
  concentration P ? gas at same temperature
• Consider PCl5(g)
  PCl3(g) Cl2(g) eqm. partial pressure
  P1 P2 P3
• moles at equilibrium a - b b
  b
• Pi Xi Ptotal P1(a-b)Pt/(ab),
  P2P3bPt/(ab)
• Kp P2P3/P1 b2 Pt /(ab)(a-b)

10
Significance of the Equilibrium Constant

• Kc or Kp gives a measure of the inherent tendency
  for a reaction to occur. For a reaction with
  large equilibrium constants, the equilibrium
  mixture will have a larger proportion of products
  than reactants
• K does not give any information about the
  reaction rate
• For a certain reaction, K depends only on the
  temperature
• Cc (1/RT)cPcc,, Bb (1/RT)bPbb, Aa
  (1/RT)aPaa For aA bB cC,
  Kc Cc/AaBb Kp (1/RT)c-a-b Kc Kp
  (1/RT)/\n where /\n product - reactant
  particles

11
Different equilibrium positions at the same
temperature
12
Different equilibrium positions
te
3.0
H2
te
N2
2.0
NH3
H2
1.0
N2
1.0
NH3
Time
Time
Experiment III
Experiment II
13
Equilibrium positions

• Each set of equilibrium concentrations is called
  an equilibrium position. There are many
  equilibrium positions for the same equilibrium
  constant for the reaction N2(g) 3H2(g)
  2NH3(g) at 773 K
• There is only one equilibrium constant for a
  given reaction system at a particular
  temperature, but there are an infinite number of
  equilibrium positions
• The specific equilibrium position adopted by a
  system depends on the initial concentrations but
  Kc does not.

14
Determination of Kc

• There is no further overall change in the
  reaction mixture once equilibrium is reached. To
  confirm the attainment of equilibrium, small
  samples must be taken out of the reaction mixture
  at different times and analyzed. At equilibrium,
  the same result should be obtained for successive
  analyses.
• A steady colorimetric reading indicates the
  achievement of equilibrium. A steady pressure
  reading indicates the attainment of equilibrium
  for a gaseous reaction

15
Determine Kc in Lab.

• Reaction temperature is kept constant by a
  thermostat system. For reactions carried out at
  high temperature, a sample of the mixture can be
  taken out for anaysis by first quenching it,
  thus freezing the mixture into the same
  proportions that it had at high temperature.
• The concentrations/pressures for gases can be
  found by performing a titration, using
  colorimeters or electrochemical cells. A gaseous
  reaction mixture can be quenched and the soluble
  gases dissolved in water for titrimetric analyses.

16
Determine Kc of esterification

• Stopper a flask containing equal volumes of acid
  and ethanol for equilibration of the volatile
  reactants/products. The flask is placed in a
  water bath fitted with a thermostat. At 50oC,
  equilibrium is achieved after 2 hours. (At lower
  temperatures the flask may have to be left for
  several days before equilibrium is achieved.)
• Pipette 10 cm3 sample of the equilibrium
  mixture into a flask containing iced water to
  quench the esterification reaction. The
  equilibrium concentration of ethanoic acid is
  found by titration with say 1 M NaOH, using
  phenolphthalein as an indicator.K
• CH3COOC2H5 (l)H2O(l)

Kc
CH3COOH(l)C2H5OH(l)
17
Determine K c of Fe3 NCS- Fe(NCS)2

• Being blood red in color, the concentration of
  Fe(NCS)2 ions can be found
  colorimetrically, by measuring the absorbance of
  the solution using a colorimeter or by visual
  comparison.
• Successive dilution of standard solutions of
  Fe(NO3)3 and NaNCS are done so that several
  different initial concentrations of Fe3(aq)
  SCN-(aq) are available.
• The colorimeter is calibrated by measuring the
  absorbance of several solutions which contained
  known concentrations of Fe(NCS)2(aq) ions. The
  equilibrium concentration of the complex is found
  by comparing their absorbance with the standard
  solutions of Fe(NCS)2 by the colorimetric
  method

18
Simple calculations of Kc and Kp

• Write a balanced equation for the reaction
• Write the equilibrium constant expression for the
  equation you have written
• List the initial concentrations/number of moles
• Define the change needed to reach equilibrium,
  and define the equilibrium concentrations/number
  of moles by applying this change to the initial
  concentrations
• Substitute the equilibrium concentrations into
  the equilibrium constant expression, and solve
  for the unknown

19
Degree of dissociation

• Consider N2O4(g)
  2NO2(g)

0
Initial no. of moles before dissociation
1
2x
Final no. of moles after dissociation
(1 - x)

2x/(1x)
Mole fraction at equilibrium
(1-x)/(1x)
where x is the degree of dissociation
(1 x) total no. of moles at equilibrium
Molar mass of N2O4 and NO2 are respectively 92
46
For 20 dissociation, average molar mass 92
(0.8) 46 (2 x 0.2)/(1 0.2) 76.6
Equilibrium pressures for gases are p(NO2)
2(0.2)Pt/1.2 Pt/3,

p(N2O4) 0.8Pt/1.2 2Pt/3

Kp Pt/32 / 2Pt/3 Pt/6 1.7 x 104 N m-2
Pt 1.01 x 105 N m-2
20
Mole fraction and degree of dissociation

• In vapor phase, methanoic acid can exist as
  monomers and dimers
  (HCOOH)2 (g) 2HCOOH(g)
• (1-x)/(1x) 2x/(1x) mole
  fraction at equilibrium
• At 283 K and 1330 N m-2, (1-x)/(1x) 0.772, x
  0.157, that is, the degree of dissociation is
  0.157.

Kp Ptotal .4x2/(1-x2) 1330(4)(0.157)2/1-0.157
2 134.5 N m-2
As dissociation is endothermic, an increase in
temperature will shift the equilibrium position
to the right, resulting in a larger value of K
At 333 K 2130 N m-2, x is 0.556 Kp
2130(4)(0.556)2/(1-0.5562)3820 N m-2
21
Effect of Concentration on Equilibria

• Le Chateliers Principle if a change is imposed
  on a system at equilibrium, the equilibrium
  position will shift in a direction that tends to
  reduce that change.
• This helps predict qualitatively the effects of
  changes in temperature, concentration/pressure on
  a system at equilibrium
• Kc NH32/N2H23 0.22/(0.4)(1.2)3
  0.05 dm6 mol-2
• Upon addition of nitrogen, Le Chateliers
  Principle predicts that the system will shift in
  a direction that consumes nitrogen, thus reducing
  the effect of the addition

Calculate Kc for the Haber Process from one set
of equilibrium concentrations
22
Effect of temperature on equilibria

• Under constant temperature, any concentration
  changes on a system at equilibrium will result in
  adjustment of the system to preserve the
  constancy of K at that temperature
• For an exothermic reaction like the synthesis of
  ammonia, N2(g) 3H2(g) 2NH3(g)
  /\H -92 kJ At higher
  temperatures, Le Chateliers Principle predicts
  that the shift will be n the direction that
  consumes energy, that is to the left. This shift
  decreases NH3(g) increases N2(g) and
  H2(g), thus reducing the value of K.
• Le Chateliers Principle cannot predict the size
  of the change in K. It correctly predicts the
  direction of the change

23
Vant Hoffs Equation

• It relates K to the value of 1/T, from which we
  can have an idea of the size of the change in K
  with temperature T ln K (-/\H/R) 1/T C
  where R gas constant For an
  exothermic reaction, the term (-/\H/RT) is
  positive. As temperature rises, the value of
  (-/\H/RT) will decrease and the equilibrium
  constant, K, will decrease, resulting in less
  product being formed.
• A catalyst changes the rate of both the forward
  and reverse reactions to the same extent but it
  does not change the equilibrium position and the
  equilibrium concentrations of the various
  species. It helps to speed up the rate of
  attainment of the equilibrium without affecting
  the value of K

24
Simple calculations on equilibrium composition

• Consider the equilibrium for the synthesis of NH3
  at constant pressure
  N2(g) 3H2(g) 2NH3(g)

Initial no. of moles 2.0
6.0 0
No. of moles at equilibrium 2- 1/2(2) 1
6- 3/2(2) 3 2
Volume of container 10 dm3 NH3 0.2 M,
N2 0.1 M, H2 0.3 M At 600 K, Kc
0.22/01 (0.3)3 14.8 dm6 mol-2
At 600 K, V dm3, x 0.5 N2 (1x), H2
(33x), NH3 (2-2x)

V
V
V
14.8 (1/V)2/(1.5/V) (4.5/V)3
V 45
25
Justify the shift in equilibrium position

• H2(g) F2(g)
  2HF(g)
• initial concentration 2 mol dm-3 2 mol
  dm-3 2 mol dm-3
• At the temperature of experiment, Kc 115
• Value of mass action expression 22/22 (22)
  1 Since the value is
  much less than Kc, the system must shift to the
  right to reach equilibrium
• Equilibrium concentrations H2 2-x, F2
  2-x, HF 22x Kc 115
  (22x)2/(2-x)2, x 1.53
• Equilibrium concentrations H2 F2 0.47 M,
  HF 5.06 M

26
Using the equilibrium constant (1)

• At 720 K, H2(g) I2(g)
  2HI(g)
  Equilibrium moles
  7.83 1.68 25.54 Kc
  25.542/(7.832) (1.682) 49.6 at 720 K
• H2(g)
  I2(g) 2HI(g)
• Initial moles 0 0
  20
• Equilibrium moles 0 x 0 x
  20 -2x
• Kc 49.6 (20 -2x)2/x2 x 2.21

27
Using the equilibrium constant (2)

• At 1000 K, COCl2(g) CO(g)
  Cl2(g)
• equilibrium concentration x
  x y
• Kc 0.329 x(y)/x y, Cl2eqm 0.329
• At 1000 K, COCl2(g) CO(g)
  Cl2(g)
• initial concentration 0. 5 M
  0 0
• equilibrium concentration 0.5 - x
  x x
• Kc 0.329 x2/(0.5 - x) x 0.2732
• Fraction of COCl2 undecomposed (0.5-x)/0.5
  0.454

28
Using the equilibrium constant (3)

• 2NH3(g)
  N2(g) 3H2(g) Initial concentration
  2 M 0 0
• Equilibrium concentration 2 - 2x
  x 3x
• For 50 decomposition,
• 2 - 2x 1, x 0.5 1.0
  0.5 1.5
• Kc 1.53 (0.5)/1.0 27/16 mol dm-3
• At this temperature, 3 moles of hydrogen 2
  moles of ammonia are allowed to react in 1 dm3
  vessel. Find N2 at equilibrium.
• Kc 27/16 (33y) (y)/(2 - 2y)
• 24y2 51y -27 0, y N2eqm 0.4388 M
  

29
Using the equilibrium constant (4)

• At 25oC, Kc for CH3COOC2H5 H2O CH3COOH
  C2H5OH is 0.27. Find the mass of CH3COOC2H5
  formed at equilibrium when 3 moles of ethanol 2
  moles of ethanoic acid are mixed.
• CH3COOC2H5(l) H2O(l) CH3COOH(l)
  C2H5OH(l)
• x x
  (2 - x) (3 - x)
• 0.27 (2 - x) (3 - x)/x2 0.73x2 - 5x 6
  0
• x 1.5455 as molar mass of CH3COOC2H5 is 88,
  the mass of ethyl ethanoate at equilibrium is 88
  (1.5455) or 136 g.

30
Using the equilibrium constant (5a)

• At 565 K, 4 moles of A2(g) were mixed with 12
  moles of B2(g) in a 20 dm3 vessel. 4 moles of
  AB3(g) were formed at equilibrium.
• A2(g) 3B2(g)
  2AB3(g)
• Initial moles 4 12
  0
• Equilibrium moles 4-x 12-3x
  2x 4
• Equilibrium conc. 2/V 6/V
  4/V
• Kc (4/V)2 / (2/V) (6/V)3 V2/27 400/27
• At 565 K, 4 moles of A2(g) were mixed with 12
  moles of B2(g) in a V3 dm3 vessel. 9 moles of
  B2(g) remained at equilibrium
• A2(g)
  3B2(g) 2AB3(g)
• Initial moles 4
  12 0
• moles at new equilibrium 4-1 3
  12-3 9 2
• Kc 400/27 22V2/3 (93) V 90 dm3

31
Using the equilibrium constant (5b)

• At constant pressure, a mixture of 4 moles of
  A2(g) 12 moles of B2(g) was placed in a 20 dm3
  vessel and heated to 565 K. 4 moles of AB3(g) was
  present when the system had reached equilibrium.
• 
  A2(g) 3B2(g) 2AB3(g)
  Initial concentration 0.2 M
  0.6 M 0
• Equilibrium concentration 0.2 - x 0.6 -
  3x 2x 0.2 M
• In a 20 dm3 vessel, 0.1 M
  0.3 M 0.2 M
• Kc 0.22/0.1 (0.33) 400/27 14.82
• 
  A2(g) 3B2(g) 2AB3(g)
  moles in 20 dm3 vessel 2 6
  4
• moles in V dm3 vessel 2 1 3 (63) 9
  4- 3(2/3) 2
• Note moles B2 increase, indicating that the
  equilibrium position shifts to the left when the
  volume changes to V dm3 by Le Chateliers
  Principle, V gt 20 dm3. 400/27 22V2/3(92)V
  90 dm3 If V decreases to 3000.5dm3, equilibrium
  position will shift to the right B2 lt 6/V

32
Using the equilibrium constant (6)

• CH3COOH(l) C2H5OH(l)
  CH3COOC2H5(l) H2O(l) Initial moles
  0.0525 0.0515
  0.0314 0.0167
• moles reacted -0.0270
  -0.0270 0.0270 0.0270
  
• Equilibrium moles 0.0255
  0.0245 0.0584 0.0437
• Kc (0.0437)(0.0584)/V2 4.1
• (0.0245)(0.0255)/V2
• 4H2O(g) 3Fe(s)
  Fe3O4(s) 4H2(g) Equilibrium pressure 2.4
  kPa 3.2
  kPa
• At 873 K, Kp (3.2/2.4)4 256/81 3.16

33
Calculations using Kp (1)

• Assume ideal behavior, initial partial pressure
• P (0.05)(8.314)(298)/0.001 123850 N m-2
  123.85 kPa
• N2O4(g)
  2NO2(g)
• initial moles 0.05
  0
• equilibrium moles 0.05-x
  2x
• mole fraction at equilibrium 0.05-x/(0.05x)
  2x/(0.05x)
• equilibrium partial pressure (0.05-x)Pt/(0.05x)
  2xPt/(0.05x)
• Total pressure at equilibrium 142.5 kPa
• total moles at equilibrium/initial moles
  (0.05-x2x)/0.05
• Pt(Equilibrium)/Pt(initial) 142.5/123.85
  (0.05x)/0.05
• x 0.00753 (degree of dissociation .00753/.05
  0.1506)
• Kp 4x2Pt/(0.052 - x2) 13.14 kPa

34
Calculations using Kp (2a)

• At 673 K Ptotal 10 atm., NH3(g) is 98
  dissociated
• 2NH3(g) N2(g)
  3H2(g) Equilibrium moles 1.0-2x
  x 3x
• For 98 dissociation 0.02
  0.49 1.47
• Equilibrium mole fraction 0.02/1.98
  0.49/1.98 1.47/1.98
• Equilibrium partial pressure Ptotal /99
  49Ptotal/198 147Ptotal/198
  
• Kp 1473(49)(992)(102)/1984 99270 atm2
• PV nRT, (10)(V) 1.98 (0.08203) (673), V
  10.94 dm3
• Kc 1.473(0.49)/0.022V2 32.57 mol2 dm-6

35
Calculations using Kp (2b)

• N2(g) 3H2(g)
  2NH3(g)
• a
  b c
• a-x
  b-3x c 2x
• Kc
  (c2x)2V2/(a-x)(b-3x)3 32.57
• Assume ideal gas behavior, ni /V Pi /RT
• Kp P(NH3)2 /P(N2) P(H2)3
• Kc NH32/ N2 H23
• P(NH3)2 (RT)-1 /P(N2)P(H2)3(RT)- 4
• Kp (RT) 4 -2 Kp (RT) -/\ n
• Kp Kc (RT)/\ n

36
Calculations using Kp (3)

• 2NOBr(g) 2NO(g)
  Br2(g)
• Equilibrium moles 1.0 - 0.12 0.12
  0.06
• Mole fraction at equilibrium 0.88/1.06
  0.12/1.06 0.06/1.06
• At 300 K and 1/4 atm.
• Equilibrium partial pressure 11/53 atm
  3/106 atm. 3/212 atm.
• Kp (3/212) (3/106)2/(11/53)2 2.63 x 10-4
  atm.
• KcKp(RT)- /\n 0.000263(300x 0.08208)-1
  1.068x10-5

Mol dm-3
37
Calculations using Kp (4)

• N2O4(g)
  2NO2(g) Initial moles
  1 0
• Equilibrium moles 1-x
  2x
• Equilibrium partial pressure (1-x)Pt/(1x)
  2xPt/(1x)
• Kp 4x2Pt / (1-x2)
• At 308 K and Pt 1 atm. x 0.27,
• At this temperature, Kp 4(.27)2/(.9271) 0.3145

atm
38
Calculations using Kp (5)

• 1 mole of N2 3 moles of H2 are mixed at 600oC
  10 atm. The mole fraction of NH3 formed at
  equilibrium is 0.15. Find Kp at 600oC
• N2(g)
  3H2(g) 2NH3(g) Initial moles
  1 3
  0
• Equilibrium moles 1-x
  3-3x 2x
• Equilibrium mole fraction (1-x)/(4-2x)
  3(1-x)/(4-2x) 2x/(4-2x)
• Since 2x/(4-2x) 0.15, x 0.3/1.15 0.2609
  ntotal 3.478
• Partial pressure 2.125 atm
  6.375 atm. 1.50 atm.
• Kp 1.52/(6.3753 x 2.125) 4.09x10-3 atm-2

39
Calculations using Kp (6a)

• At a certain temperature Ptotal 1.01 x 105
  N/m2, the dissociation of PCl5 is 30. Find the
  value of Kp.
• PCl5(g)
  PCl3(g) Cl2(g) Initial moles
  1 0 0
• Equilibrium mole fraction 0.7
  0.3 0.3
• Equilibrium partial pressure 7Pt/13
  3Pt/13 3Pt/13
• Kp (9/169)(13Pt/7) 9Pt/91 9989 N m-2

40
Calculations using Kp (6b)

• 0.1 moles of PCl5 is heated at 423 K in 1 dm3
  vessel, a pressure of 4.38 x 105 N/m2 is measured
  at equilibrium.
• PCl5(g)
  PCl3(g) Cl2(g) Initial moles
  0.1 0 0
• Equilibrium mole fraction (0.1-x)/(0.1x)
  x/(0.1x) x/(0.1x)
• Assume ideal gas behavior, initial pressure
  (0.1)(8.314)(423)/0.001 3.517 x 105 N/m2 Since
  P ? n, 4.38/3.517 (0.1x)/0.1 thus x 0.0247
• Degree of dissociation 0.0247/0.1 0.247
  (24.7) lt 30 At the higher pressure of 438
  kPa, the equilibrium position shifts to the left
  as predicted by the Le Chateliers Principle.
• Kp x2 Pt/(0.12-x2) 8500 N m-2 (Pt438kPa)

41
Calculations using Kp (7)

• When 1 mole of H2 1 mole of I2 are allowed to
  reach equilibrium in 1 dm3 flask at 723 K and
  1.01 x 105 N m-2, 1.56 moles of HI are formed.
  Calculate the Kp at 723 K.
• 
  H2(g) I2(g) 2HI(g)
  Initial moles 1
  1 0
• Equilibrium moles 1-x
  1-x 2x
• Equilibrium partial pressure (1-x)Pt/2
  (1-x)Pt/2 xPt
• Since 2x 1.56, x 0.78
• Kp 4x2/(1-x)2 50.28 Kc

42
Calculations using Kp (8)

• 0.2 moles of CO2(g) was heated with excess carbon
  in a closed vessel until the equilibrium CO2(g)
  C(s) 2CO(g) was attained The average
  molecular mass of the gaseous equilibrium mixture
  was 36
• 
  CO2(g) C(s) 2CO(g) Initial
  moles 0.2
  0
• Equilibrium moles 0.2 -x
  2x
• Equilibrium mole fraction (0.2-x)/(0.2x)
  2x/(0.2x)
• (44) (0.2-x) (28) (2x) 36, x 1/15
  equilibrium mole
• (0.2x) (0.2x)
  fraction of CO 0.5
• At a total pressure of 12 atm. in equilibrium, Kp
  62/6 6 atm.
• Reducing the total pressure to 2 atm. at the same
  temperature, partial pressures of CO and CO2 are
  2XCO atm 2(1-XCO) atm
• Kp 6 4XCO2/2(1-XCO) XCO2 3XCO - 30
  XCO0.791

43
Calculations using Kp (9)

• At 1000 K, Kc for I2(g) 2I(g) is 3.76
  x 10-5 mol dm-3
  
• Kc I2/I2
• Kp Kc(RT)/\n 3.76x 10-5(0.08208 x1000)(2-1)
  3.09 x 10-3 atm
• Initial moles 1
  0
• Equilibrium moles 1-x 2x
  Pt 1 atm
• Kp 4x2Pt /(1-x2) 3.09 x 10-3 4x2
  (1)/(1-x2), x 0.0278
• At total pressure of 0.10 atm, 3.09 x 10-3
  0.4x2/(1-x2) x 0.0874
• A ten-fold rise in total pressure reduces the
  dissociation three-fold

Higher pressure shifts the equilibrium position
to the left because there is fewer of particles
on the left hand side
44
Brönsted-Lowry Concept of Acid-Base

• An acid is a proton (H) donor, and a base is a
  proton acceptor
• A HCl molecule donates a proton to a water
  molecule and so acts as a Brönsted-Lowry acid.
  Each water molecule acts as a Brönsted-Lowry base
  when it accepts a proton from the acid HCl to
  form a hydronium ion
• HCl(aq) H2O(l) H3O(aq)
  Cl-(aq) . Acid Base
  conjugate acid conjugate base
• HCl and Cl- H2O and H3O are 2 conjugate
  acid-base pairs.
• H2O(l) N(C2H5)3(l) OH-(aq)
  HN(C2H5)3(aq)
• Acid Base conjugate base
  conjugate acid

45
Acid-Base Concept (2)

• Under the Brönsted -Lowry concept, a substance
  can act both as an acid and as a base, depending
  on the demands of other substances present.
• Acid Base Conjugate
  base Conjugate acid CH3COOH(aq) H2O
  CH3COO-(aq) H3O(aq)
  H2O(l) NH3(aq) OH-(aq)
  NH4(aq) HClO4(l)
  HNO3(l) ClO4-(aq)
  H2NO3(aq) HNO3(l) CH3CO2H(l)
  NO3-(l) CH3CO2H2(l)
  H2PO4-(aq) HCO3-(aq) HPO42-(aq)
  CO2 (H2O)
• HNO3(aq) H2O(l) NO3-(aq)
  H3O(aq) HF(l)
  HNO3(l) F-(l)
  H2NO3(l)
• Nitric(V) acid is a Brönsted acid in water, but
  it acts as a Brönsted base in liquid hydrogen
  fluoride

46
Dissociation of Water and Kw

• H2O(l) H(aq) OH-(aq)
• This equilibrium represents the self-dissociation
  of water Here one water molecule acts as an acid
  by furnishing a proton, and the other acts as a
  base by accepting a proton
• KcH2O(l) H(aq)OH-(aq) Kw, the ionic
  product of water
• The temperature-dependent Kw 1.0 x 10-14 mol2
  dm-3 at 25oC
• Pure water is neutral because it has equal
  concentrations of H and OH- ions At pure water,
  H(aq) OH-(aq) 10-7 M
• In any aqueous solution at 25oC, no matter what
  it contains, the product of H(aq) OH-(aq)
  is equal to 10-14 mol2 dm-6

47
Acid, Base and pH

• An acidic solution H(aq) gt OH-(aq)
• An alkaline solution H(aq) lt OH-(aq)
• Kw increases with temperature. According to the
  Vant Hoffs equation, the self-dissociation of
  water must be endothermic. At temperatures other
  than 25oC, the pH of a neutral solution is not
  exactly 7 as Kw is not 10-14 at these
  temperatures.
• pH log10H3O(aq)-1 - log10H3O(aq)
• A tenfold rise in H3O(aq) results in a
  decrease in one pH unit.

48
pH measurement

• A combination electrode for measuring pH has
• A reference electrode whose electrode potential
  is constant Silver/silver chloride is often used
  as a reference electrode as its half potential is
  constant, irrespective of the change in pH
  outside
• An electrode responsive to H changes e.g.
  glass electrode
• A glass electrode has a glass surface on the thin
  walled bulb across which the potential difference
  varies as H outside differs. The electrode
  potential of the glass electrode is dependent on
  H outside.

49
pH measurement (2)

• The glass electrode tends to change its voltage
  or e.m.f. over a period of time. The voltage is
  also affected by temperature. Its usual practice
  to calibrate the pH meter first before use, by
  placing the electrode assembly in a buffer
  solution whose pH is accurately known. The pH
  meter reading can then be adjusted accordingly.
• To calibrate a pH meter, the electrode assembly
  is first rinsed in pure water then dipped into
  a standard buffer of known pH. The adjusting knob
  of the pH meter is then turned so that its
  reading agrees with the stated buffer pH. Repeat
  rinsing the electrode with pure water various
  buffers until the whole scale of pH is correctly
  established in the pH meter.

50
pH measurement (3)

• Care in handling the pH meter
• Keep the glass bulb immersed in solutions
  prolonged drying of the glass membrane lowers
  sensitivity to respond to changes of H Never
  touch the glass membrane with your fingers. Any
  grease contamination on the glass membrane
  affects the voltage across the membrane hence
  the pH value being read.
• Rinse thoroughly with distilled water the
  electrode assembly when it is transferred
  between solutions of different pH values

51
Strong and Weak Acids (1)

• The strength of an acid is defined by the
  equilibrium position of its dissociation
  reaction
• HA(aq) H2O(l) H3O(aq)
  A-(aq) A strong acid is one for which
  this equilibrium lies far to the right. This
  means that all the original HA is dissociated at
  equilibrium and the Heqm is comparable to the
  original concentration of HA.
• Much of a weak acid remains undissociated at
  equilibrium. Also, the increase in pH is not
  proportionate to dilution for weak acids because
  they are not completely dissociated. HA(aq)
  H2O(l) H3O(aq) A-(aq)

52
Strong and Weak Acids (2)

• As ethanoic acid is diluted, its pH gets closer
  closer to the pH of the strong acid at the same
  concentration. This shows that dilution increases
  the extent of dissociation of ethanoic acid. The
  same is true of other weak acids.
• A strong acid/base cannot be judged from the pH
  of its solution because a very dilute strong acid
  has the same pH as an aqueous solution of a weak
  acid.
• pH measurements can be used to compare the
  strength of different weak acids at the same
  concentration.
• In an oxyacid, H-O-X, the O-X bond is strong
  covalent if X is highly electronegative. Once
  dissolved in water, the O-X bond remains intact.
  With more O-atoms bonded to X, stronger electron
  withdrawing effect is exerted on the H-O-X group,
  thus weakening the H-O bond and makes the release
  of a proton easier.

53
Strong and Weak Acids (3)

• Hydrohalic acids have their acidity dependent on
  the polarity and strength of the H-X bond.
  Although the H-Cl bond is just slightly stronger
  than a C-H bond , it readily dissociates when
  dissolved in water due to its strong dipole
  moment. However, H-F is a weak acid even though
  the H-F bond is very polar, because the H-F bond
  is unusually strong.
• H3PO4, HNO2, HOCl and R-COOH are weak acids. When
  a weak acid HA is dissolved in water, an
  equilibrium is set up as some of the weak acid
  molecules dissociate into ions HA(aq)
  H(aq) A-(aq), where A- is the conjugate
  base
• The equilibrium constant Ka for the reaction is
  the acid dissociation constant of acid HA.

54
Dissociation of Weak Acids

• Ka H(aq)A-(aq) The larger
  the value of Ka,
• HA(aq) the
  stronger is the acid
• In an equimolar mixture of CH3CO2H CH3CO2-,
  HKa
• The weaker the acid, the stronger its conjugate
  base.
• HA HF HNO2 CH3CO2H HOCl HCN
  NH4
• Ka x 106 720 400 18
  0.035 6.2x10-4 5.6x10-4
• Solid AlCl3 dissolves in water to form
  Al(H2O)63, a weak acid that dissociates to form
  some H ions. The high charge on the metal ion
  polarizes the O-H bonds in the attached water
  molecules, weakening the O-H bond, forming H
  ions
• Al(H2O)63(aq) Al(OH)(H2O)52(aq)
  H(aq)

55
Systematic approach for Solving Acid-Base
Equilibria Problems

• Decide the major species present, what
  equilibrium dominates the solution and which
  reaction can be assumed to be complete
• Find the pH of a 0.10 M ethanoic acid whose Ka
  1.8 x 10-5 M
• List the major species in the solution - CH3COOH,
  H2O
• Choose the species that can produce H and write
  balanced equations for these reactions. Both
  species can produce H
• CH3COOH(aq) H(aq) CH3COO-(aq) Ka
  1.8x10-5 M
• H2O(l) H(aq) OH-(aq)
  Kw1.0x10-14 M2
• Use the Kc values to decide which equilibrium
  will dominate in producing H(aq). Since KagtgtKw
  it dominates in making H

56
Solving Solution Equilibria

• Write the equilibrium expression for the dominant
  equilibrium List the initial concentrations of
  the species participating in the dominant
  equilibrium. Defne the change needed to achieve
  equilibrium that is, define x.
• CH3COOH(aq)
  H(aq) CH3COO-(aq)
• initial concentration 0.100 M
  0 0
• equilibrium concentration (0.100 - x) M
  x M x M
• Substitute the equilibrium concentrations into
  the equilibrium expression to get what is
  required of the question.
• Ka 1.8 x 10-5 x2/(0.100 - x) at equilibrium
• Assume HXxHAo, Ka1.8x10-5 x2/0.10, x
  H 1.3 x 10-3 M Compare x to HAo to validate
  the 5 approximation rule.

57
Reactions of weak bases with water

• Ions like CN-, CH3CO2- and CO32- remove H from
  H2O to form alkaline solution by hydrolysis
• CN-(aq) H2O(l) HCN(aq) OH-(aq)
  CH3CO2-(aq) H2O(l) CH3CO2H(aq)
  OH-(aq) CO32-(aq) H2O(l)
  HCO3-(aq) OH-(aq)
• B(aq) H2O(l) BH(aq) OH-(aq)
  base acid
  conjugate acid conjugate base
• Kb BH(aq)OH-(aq)/B(aq) BH
  HOH-
• 
  B H
• (1/Ka) Kw or Kw KaKb (Kadissociation
  constant of

conjugate acid )
58
Solving Weak Base Equilibria Problems

• The pH of a 15.0 M solution of ammonia at 298 K
  is 12.2. Find the equilibrium constant Kb at 298
  K for the reaction NH3(aq)
  H2O(l) NH4(aq) OH-(aq) Kw10-14 at
  298 K H2O(l) H(aq) OH-(aq)
• As Kwlt Kb, contribution of OH- from ammonia is
  greater than that from water. The equilibrium for
  NH3 dominates in making H
• NH3(aq)
  H2O(l) NH4(aq) OH-(aq)
  initial concentration 15.0 M
  0 0
  
• equilibrium conc.(M) 15.0-x
  x x
• Kb x2/(15-x) x2/15, x OH- 10 - (14
  -12.2) 0.01585 M
• Kb 0.0162/15 1.7 x 10 -5 mol dm-3

59
Weak Base Equilibria Calculations

• Calculate the pH of a 0.10 M sodium ethanoate
  solution. The Ka value for ethanoic acid is 1.8
  x10-5 M, Kw1.0 x10-14
• The major species in solution are Na, CH3CO2-
  and H2O Since CH3CO2H is a weak acid, its
  conjugate base CH3CO2- has a significant affinity
  for H, the dominant reaction is CH3CO2-(aq)
  H2O(l) CH3CO2H(aq) OH-(aq)
• 0.10 - x
  x x
• Kb CH3CO2HOH-/CH3CO2- Kw /Ka 5.6x10-10
  M
• Kb x2/0.1, x OH- 7.48 x 10-6 M (5 appr.
  Okay)
• pOH 5.13 or pH 14 - 5.13 8.87

60
Simple Acid-Base Equilibria

• HOBr(aq) H2O(l) H3O(aq) OBr -(aq)
  0.12 M 0
  0
• 0.12-x x
  x
• Ka x2/(0.12-x) x2/0.12 2.1 x 10-9
• x 1.59 x 10-5 H3O, pH 4.80
  Common Ion Effect -
• CH3CO2H(aq) H2O(l) CH3CO2-(aq) H3O(aq)
  0.10 M
  0.05 M 0
• (0.10 - x)
  (0.05 x) x
• Ka (0.05x)x/(0.10-x) 0.05(x)/0.10 1.8 x
  10-5
• x 3.6 x 10-5 M (Validate the approximation
  x ltlt 0.05)
• Find the percent dissociation of 0.1 MCH3CO2H at
  a pH of 2

0.18
61
Simple Acid-Base Equilibria (2)

• Ka for methanoic acid at 298 K is 2.14 x 10-4
• HCOOH(aq) H2O(l) HCOO-(aq)
  H3O(aq) 0.10 -x
  x x
• Ka 2.14 x 10-4 x2/(0.1-x) x2/0.1, x
  4.63 x 10-3 M
• dissociation 4.63 x 10-3 x 100/0.10 4.63
• Ka for benzoic acid at 298 K is 6.6 x 10-5
• C6H5COOH(aq) H2O(l) C6H5COO-(aq)
  H3O(aq)
• a
  0 0
• a - x 0.9 a
  0.1 a 0.1 a
• Ka 6.6 x 10-5 0.01 a2/(0.9 a) a/90, a
  5.94 x 10-3 M
• pH -log (0.1 x 5.94 x 10-3) 3.23
  
  

62
Simple Acid-Base Equilibria (3)

• What volume of 0.1 M sodium ethanoate solution
  should be added to 1 dm3 of 0.1 M ethanoic acid
  to give a solution of pH 4?
• CH3COOH(aq) H2O(l) CH3COO-(aq)
  H3O(aq) 0.1/(1V) M
  0.1V/(1V) 0
• 0.1/(1V) - 10 -4
  0.1V/(1V) 10 -4 10 -4
• Ka 1.8 x 10-5 10-4 0.1V/(1V)(1V)/0.1
  V (10-4)
• V 0.18 dm3
• 0.1/1.18 0.018/1.18 gtgt 10-4 so that the
  approximation is valid.

63
Simple Acid-Base Equilibria (4a)

• Excess Mg(OH)2(s) is shaken with 1 dm3 of 1 M
  NH4Cl. When the reaction is complete, the
  resulting saturated solution has a pH of 9.0 at
  298 K. Calculate the concentrations of NH3 NH4
  in this solution, if Kb for NH3 is 1.8 x 10-5 mol
  dm-3 at 298 K, and Kw 1.0 x 10-14 mol2 dm-6 at
  the same temperature.
• The stoichiometry problem. Has the acid base
  reaction run to completion? The equilibrium
  problem. From what has remained determine the
  equilibrium position and the quantities
  involved..
• All NH4 ions react completely with excess
  Mg(OH)2 to give NH3. Mg(OH)2(s) 2NH4(aq)
  Mg2(aq) 2NH3(aq) 2H2O(l)
  The equilibrium problem. The ammonia so
  formed dissociates according to the equilibrium
  NH3(aq) H2O(l) NH4(aq)
  OH-(aq)
• The major species present in solution are Mg2,
  NH3, OH- H2O

64
Simple Acid-Base Equilibria (4b)

• H2O(l) H(aq) OH-(aq)
• NH3(aq) H2O(l) NH4(aq) OH-(aq)
• Mg(OH)2(s) Mg2(aq) 2OH-(aq)
• The equilibrium constants of the 3 reactions are
  small and it is not easy to judge which one is
  dominant in making OH- or perhaps none is
  dominant
• NH3(aq) H2O(l) NH4(aq) OH-(aq)
• 1.00
  0 0 initial
  concentration
• 1.00-x
  x 1 x 10-5 equilibrium
  conc.
• Kb 10-5 x/(1-x) 1.8 x 10-5 x 0.64 M
• NH4 0.64 M and NH3 0.36 M

65
Simple Acid-Base Equilibria (5a)

• 50 cm3 of 0.1 M HCN(aq) was titrated against NaOH
  at 298K. At the equivalence point of the
  titrtion, the pH was 10.96. Find the volume of
  NaOH added to reach the equivalence point.
• The stoichiometry problem. The equivalence point
  occurs when all of the HCN is stoichiometrically
  neutralized by NaOH to CN- 0.1 x 0.05 5 x 10-3
  moles of OH- is needed for neutralization.
• The equilibrium problem. At the equivalence
  point, the major species in solution are Na,
  CN-, an H2O..
• The dominant equilibrium CN-(aq) H2O(l)
  HCN(aq) OH-(aq) initial concentration
  5/(50V) 0
  0
• Equilibrium conc. 5/(50V) - x
  x x

66
Simple Acid-Base Equilibria (5b)

• Given pH 10.96, Kw 10-14 x OH-
  8.9x10-4 M
• Ka of HCN 6.2 x 10-10, Kh 10-14/6.2x10-10
  1.6x10-5
• 1.6x 10-5 (8.9x10-4)2/5x10-3/(50V) V 50
  cm3
• 50 cm3 of NaOH has been used and OH- 0.1 M

pH
10.96
5.1
Volume of NaOH added
50
67
Multiple Equilibria in Aqueous solution

• Given K14.3x10-7, K24.8x10-11 for CO2(aq) at
  298 K
• CO2(aq) H2O(l) HCO3-(aq) H(aq)
  K1 HCO3-(aq) H2O(l)
  H3O(aq) CO32-(aq) K2
• Write balanced equations for 5 equilibrium
  reactions each involving HCO3- in an aqueous
  solution of NaHCO3 at 25oC
• Acid Base conj. acid conjugate base
  
• HCO3- H2O H3O CO32- K2 4.8 x
  10-11 M
• H2O HCO3- CO2 H2O OH- K
  Kw/K12.3x10-8 H3O HCO3- CO2
  H2O K1/K12.3 x 106
• HCO3- OH- CO32- H2O KK2
  /Kw 4800
• HCO3- HCO3- CO2 H2O CO32-
  KK2/K11.12x10-4

68
Common Ion Effect on pH

• Find the mass of NH4Cl(s) to be added to 5 dm3 of
  0.1M of NH3(aq) if the solution is to have a pH
  of 10.45
• As Kb 10-5, 0.1 M NH3(aq) has a pH of 11.0.
  Adding NH4 to the ammonia solution shifts the
  equilibrium position to the left by the common
  ion effect, reducing the OH- or pH
• Suppose 5a moles of NH4Cl(s) has been added.
• NH3(aq) H2O(l) NH4(aq)
  OH-(aq) (Kb10-5) 0.1
  a M 0
• 0.1 - x a x
  x 10 -(14 -10.45) 10-3.55
• Kb (ax)(10 -3.55)/(0.1-10-3.55) 10-5, x
  is not too small
• a x 10-63.55 10-2.45 a 10-2.45
  -10-3.55 3.27 x 10-3
• Mass of NH4Cl 5(10-2.45 - 10-3.55)(53.5)
  0.875 g

69
Equilibria with very large Kc

• Sufficient KCN is added to 0.10 M AgNO3 to give
  Ag(CN)2- and an excess CN- of 0.02 M
• Ag(aq) 2CN-(aq) Ag(CN)2-(aq)
  0.10 M gt 0.02 M (0.22 M) 0
• 0.10-x 0.02 M x
  (0.1) where x has a value very
  close to 0.10 because Kc is very large
• Kc 1021 0.1/Ag(0.02)2 Ag 2.5
  x10-19 M

70
Simple Acid-Base Equilibria

• HCN(aq) H2O(l) H3O(aq)
  CN-(aq) 0.1 M
  0 0
• 0.1 - x x
  10-5.2 x
• Ka 10-10.4/ (0.1 - 10-5.2) 4 x 10-10 M
• HIn(aq) H(aq) In-(aq) Ka
  1.3 x 10-8 M Ka H In- For acid
  color, HIn/In- 10,
• HIn
  H 1.3 x 10-7 M
• For base color,
  In-/HIn 10
• 
  H 1.3 x 10-9 M
• The pH range for the indicator is 6.89 - 8.89

pH of 0.1 M HCN(aq) is 5.2
71
Calculations on Hydrolysis (1)

• C6H5COO-(aq) H2O(l) C6H5COOH(aq)
  OH-(aq) 0.1 M
  0 0
• 0.1 - x
  x x
• Kh Kw/Ka 10-14/6.6 x 10-5 1.5 x 10-10
  x2/(0.1-x)
• x 3.87 x 10-6 OH-, pH 14 - 5.41 8.59
  
• C6H5O-(aq) H2O(l) C6H5OH(aq)
  OH-(aq) 0.1 M
  0 0
• 0.1 - x
  x x
• Kh 10-14/1.4 x 10-9 7.14 x 10- 6 M x2
  7.14 x 10-6
• x 2.67 x 10-3 M, pH 14 - 2.57 11.43

Ka for benzoic acid is 6.6 x 10-5
Ka for phenol is 1.4 x 10-11 M
72
Calculations on Hydrolysis (2)

• HCO3-(aq) H2O(l) H3O(aq) CO32-(aq)
  
• CO32-(aq) H2O(l) HCO3-(aq) OH-(aq)
  0.10 M 0
  0
• 0.10 - x x
  x
• Kh Kw/Ka 2.08 x 10-4 M x2/(0.1-x)
• x 4.56 x 10-3 M pOH 2.34, pH 11.66
• Find the pH values of 0.1 M KCN and 0.1 M NH4Cl,
  given Ka for HCN is 7.2 x 10-10 and Kb for
  NH3(aq) is 2.3 x 10-5.
• 11.1 Kh Kw
  /Kb 4.35 x 10-10 H 6.6 x 10-6 M, pH
  5.18
• NH4 H2O NH3 H3O NH3 H2O NH4
  OH- Kh NH3H3O/NH4

Ka for HCO3- is 4.8 x 10-11
CO32- is conjugate base of HCO3 -
73
Action of a Buffer System

• A buffer solution is one which resists the change
  of pH on addition of a small amount of acid or
  alkali, upon dilution.
• A strong acid greatly increases H when added
  into water
• A buffer system is made of a mixture of a weak
  acid its conjugate base (or a weak base its
  conjugate acid) in similar amounts
• HA(aq) H2O(l) H3O(aq) A- (aq)
• The strong conjugate base present in the buffer
  changes the strong acid added into a slightly
  ionized weak acid HA. Upon addition of a strong
  alkali, the weak acid present in the buffer
  changes into its strong conjugate base

74
How Buffers can be made

• Buffer systems can be made
  
• At the acid pH by dissolving equal concentrations
  of a weak acid its salt with a strong base in
  the same aqueous solution
• At the alkaline pH by dissolving equal
  concentrations of a weak base its salt with a
  strong acid in the same aqueous solution
• In CH3CO2H/CH3CO2Na buffer, the large amount of
  CH3CO2- ions contributed by the fully
  dissociated salt tends to repress the
  dissociation of the weak acid, according to Le
  Chateliers Principle
• Added H ions will be mopped up by the large
  quantity of CH3CO2- ions to form relatively
  undissociated CH3CO2H.Change to HA is complete
  the equilibrium H3OCH3CO2-
  CH3CO2HH2O lies well to the right, removing all
  the H ions added restoring the pH of the
  buffer.

75
The Buffer System

• For an acid buffer, the essence of buffering
  depends on the relatively large amounts of a weak
  acid HA its conjugate base A-, as related by
  the equilibirum HA H A- Any excess
  acid/base added is being dealt with by the
  respective components of the buffer.
• H2O(l) A-(aq)
• OH-(aq)
• HA(aq) H(aq) A-(aq)
  HKaHA/A-
• 
  H(aq) the pH is found by
• 
  HA(a) HA/A- ratio

Any perturbation of the weak acid dissociation
equilibrium results in a pH change of the buffer
system
76
The pH of a Buffer Henderson-Hasselbach Equation

• The pH of a buffer is obtained from the form of
  the acid dissociation equilibrium expression
  H KaHA/A-
• pH pKa -logHA(aq)/A-(aq) or
  pKalog10Base/Acid
• This log form of the expression for Ka is called
  the Henderson-Hasselbach equation is useful for
  finding the pH of solutions when the ratio
  HA/A- is known.
• A buffered system is said to be centered about
  the pKa values. The buffer pH pKa 1
• Solid sodium carbonate is added to 0.5 dm3 of 0.4
  M NaHCO3 in order to make a buffer solution at pH
  9.5? (pKa for HCO3- 10.25)
• CO32-/HCO3- 10(9.5 - 10.25) 0.18
  m(Na2CO3)0.18x0.4x0.5x1063.82 g

77
Calculation of the pH of a Buffer

• Calculate the pH of a solution made by adding
  0.001 mole of NaOH to 100 cm3 of 0.50 M CH3CO2H
  0.50 M CH3CO2Na
• CH3CO2H(aq) NaOH(aq) CH3CO2Na(aq)
  H2O(l)
  0.5x0.10.05 0.001
  0.5x0.10.05 0
• 0.05 - 0.001 0.001-0.0010 0.05
  0.001
• CH3CO2H(aq) H2O(l) CH3CO2-(aq)
  H3O(aq)
• (0.049 - x)
  (0.051 x ) x
• Ka (0.051x)x/(0.049-x) 1.8 x 10-5 x
  1.73 x 10-5 M
• pH 4.76 (x ltlt 0.049 M)
  

78
Buffer pH Calculation

• In what proportions must 0.1M solutions of NH3
  NH4Cl be mixed, in order to obtain a buffer
  solution of pH 10 ? (Ka for NH4 is 6 x 10-10
  M at 298 K Kw 10-14 M2)
• NH3(aq) H2O(l) NH4(aq) OH-(aq)
• Kb 10-14/6x10-10 1.67x10-5, OH- 10-4
• NH4/NH3 Kb/OH- 0.166 Or
• 10 9.22 - log NH4/NH3 NH4/NH3
  0.166
• The solutions must be mixed in proportions of
  0.16 volume of 0.1 M NH4Cl(aq) to 1 volume of 0.1
  M NH3(aq)

79
Buffer pH Calculation

• 0.4 mole of HCl(g) is passed into 1 dm3 of 0.65 M
  NH3(aq), the pH of the resulting solution X is
  found to be 9.05 at 298 K.
• Stoichiometry problem. HCl(g) NH3(aq)
  NH4(aq) Cl-(aq)
• Before reaction 0.4 0.65
  0
• After reaction 00.4-0.4 0.25
  0.4 0.4
• Equilibrium problem. The major species are
  NH3, NH4, Cl-, H2O the important equilibrium is
  NH3(aq)H2O(l) NH4(aq)OH-(aq)
• 0.25
  M 0.40 M
• 
  0.25-x 0.4x x10-4.95
• Kb 0.4/(0.25) (1.1 x 10-5) 1.8 x 10-5 M

NH4/NH3
80
pH Titration Curves - Strong Acid - Strong Base
Titration

• The progress of an acid-base titration can be
  monitored by plotting the pH of the solution
  against the volume of added titrant to give a pH
  titration curve.
• The curve starts off with an almost horizontal
  section where a lot of acid is added without
  changing much the pH. This is followed by a steep
  portion of the curve, where a single drop of acid
  varies the pH rapidly from 11 to 3. This sharp
  fall of pH allows the easy detection of the
  equivalence point, which is the point at which
  the original acid (or base) in the solution has
  been exactly consumed by the titrant base (or
  acid). Since the ionization of water is the only
  significant equilibrium in the solution left
  after the strong acid-strong base titration, the
  solution is neutral, and its pH is 7.

The pH change near the end point is from 5.3 to
8.7 during a strong acid -strong alkali titration
81
Strong Base - Weak Acid Titration

• Initially the curve falls slowly, because excess
  OH- controls the pH in this region. At the
  equivalence point of a titration of a weak acid
  with a strong base, the pH is greater than 7 as
  the conjugate base of the weak acid hydrolyzes to
  give more OH- ions A-(aq) H2O(l)
  HA(aq) OH-(aq)
• The actual position of the equivalence point can
  be detected by noting where the pH changes most
  rapidly. The pH then decreases slowly levels
  off, because of the buffering action

Equivalence point
14
Buffer range
7
Buffer pH pKa
Acid added
82
Strong Acid-Weak Base Titration

• The pH at the equivalence point is always less
  than 7 as the conjugate acid hydrolyzes. Before
  the equivalence point is the buffering range of
  the system beyond the equivalence point the pH
  is determined by the excess H.

pH
NH4(aq) H2O(l) NH3(aq) H3O(aq)
Buffer range
Ka 5.6 x 10-10
Buffer pH pKa
When 12.5 cm3 2 M HCl is added to 25 cm3 2 M NH3
the pH of the mixture is equal to -log Ka or 9.25
7
Equivalence point
At equivalence point, NH4 1 M, the salt
hydrolyzes or dissociates to H3O. pH -log
Ka1/2 4.63
Acid added
83
Two-Equivalence point Titration

• For a weak base(Na2CO3), titration with a strong
  acid proceeds in two stages CO32-(aq) H(aq)
  HCO3-(aq)
• HCO3-(aq) H(aq)
  H2O(l) CO2(g) The pH titration curve has
  two kinks as the whole process is essentially
  made up of two separate acid-base reactions
  

pH
12
First rapid change in pH at pH 8.5 due to
formation of NaHCO3
7
Second rapid change in pH at pH 4 due to
formation of NaCl
Acid added
Equivalence point 1
Equivalence point 2
84
Acid-Base Indicators

• The equivalence point is defined by the reaction
  stoichiometry An indicator marks the end point of
  a titration by changing color
• Use an indicator which shows an end-point of a
  titration close enough to the equivalence point
  so that the error is negligible
• HIn(aq) H(aq) In-(aq) Ka
  H(aq)In-(aq)/HIn(aq) For most indicators,
  the HIn/In- ratio must be greater than 10
  before the acid color due to HIn is apparent to
  the human eyes. The In-/HIn ratio must be
  greater than 10 before the base color due to In-
  is apparent to the human eyes. Assuming this
  ratio and applying the Henderson-Hasselbach
  equation to the equilibrium, an indicator changes
  color at pH values of pKa1 and pKa-1 during an
  acid base titration.

85
Choice of Indicators for Acid-Base Titration

• When an acid is being titrated the color change
  occurs at pH pKa-1. When a base is titrated, the
  color change occurs at pH pKa1. The useful pH
  range for color change of an indicator is
  pKa(indicator) 1 when the indicator swings from
  being predominantly HIn to In- and vice versa.
• To determine the pH range over which an indicator
  changes color, one can add an alkali or an acid,
  1 cm3 at a time, to a solution containing this
  indicator and a buffer, and measure the pH by a
  pH meter at each addition. The pH values at which
  a color change begins and ends define the pH
  range
• An indicator whose Ka is 10-7 changes color in
  the pH range 6-8

Buffer use ensures a mild pH change upon addition
of strong acid/alkali
86
Indicators for the two-stage neutralization
between HCl Na2CO3

• With the indicator phenolphthalein, an end point
  of 25 cm3 is seen at the pH range of 8-10 with
  another indicator methyl orange another end point
  of 50 cm3 is seen at the pH range of 4-6 The
  double indicator method has applications in the
  estimation of the amount of sodium carbonate in a
  mixture of Na2CO3 NaHCO3, or the amount of
  Na2CO3 NaOH

pH
pH must correspond with the equivalence point
of the titration
10
phenolphthalein
8
6
Methyl orange
4
Volume of 0.1 M HCl added /cm3
50
25
87
Estimation of NaHCO3

• 0.282 g of a sample containin Na2CO3, NaHCO3
  inert material is titrated with 0.113 M
  hydrochloric acid. A mixture of methyl orange
  phenolphthalein is used as indicators, 21.60 cm3
  of the given acid is needed to reach the end
  point of phenolphthalein. 45.35 cm3 of the HCl is
  needed to reach the end point of methyl orange.
  Calculate the of Na2CO3 NaHCO3 in the
  mixture.
• Phenolphthalein indicates the stage H CO32-
  HCO3- as NaHCO3 formed is a weak base that
  forms an alkaline solution.
• Mass of Na2CO3 0.113 x 0.0216 x 106 0.258 g
  
• Methyl orange indicates both CO32- 2H
  CO2 H2O and HCO3- H CO2 H2O.
  Number of moles of NaHCO3 in the sample 0.113
  (0.04535 - 2 x 0.0216) 2.42 x 10-4
• Mass of NaHCO3 2.42 x 10-4 x 84 0.02 g,
  of NaHCO3 0.02 x
  100/0.282 7.1

88
Neutralization of triprotic phosphoric acid

• When H3PO4 is titrated with NaoH, the pH
  titration curve has two kinks only. In the
  neutralization essentially all the H3PO4
  molecules are first changed to NaH2PO4, one mole
  of NaOH being required for each mole of H3PO4 to
  reach this first equivalence point. At this
  equivalence point, the solution is essentially
  NaH2PO4. This is an acidic solution because K2 gt