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From KaKb to pH of a weak acidalkali

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Title: From KaKb to pH of a weak acidalkali

1
From Ka/Kb to pH of a weak acid/alkali

Ka for CH3(CH2)2COOH 1.5 x 10-5 mol dm-3 at 298
K. Calculate the pH of an aqueous solution of
0.10 M CH3(CH2)2COOH.
CH3(CH2)2CO2H H2O CH3(CH2)2CO2- H3O
0.10 - x
x x
Ka 1.5 x 10-5 x2/(0.10 - x)
x H3O 1.23 x 10-3 mol dm-3
pH - log 1.22 x 10-3 2.91

2
From pH of a weak acid/alkali to Ka/Kb

The pH of a 0.5M solution of ammonia at 298 K is
11.48. Calculate the equilibrium constant at 298
K for the reaction NH3(aq) H2O(l)
NH4(aq) OH-(aq)
(At 298 K, Kw 1.0 x 10-14 mol2 dm-6)
89I2b 3
0.5-x
x x
Kb x2/(0.5-x) x2/0.5, where x
10-(14.0-11.48) 10-2.52
Kb 10-5.04/0.5 1.82 x 10-5 mol dm-3

3
Second dissociation constant of sulphuric acid

In dilute aqueous solution, sulphuric acid can be
regarded as totally ionized to H3O HSO4-. The
hydrogensulphate ion is itself a weak acid with a
dissociation constant of 1.2 x 10-2.
Calculate the concentrations of ions present in a
0.1 M H2SO4
H2SO4(aq) H2O(l) H3O(aq) HSO4-(aq)
K1 is large HSO4-(aq) H2O(l) H3O(aq)
SO42-(aq) K2 0.012 0.10 - x
(0.1 x) x
K2 x (0.1 x) /(0.1 - x) 0.012,
x2 0.112 x - 0.0012 0, x SO42- 0.00985
or 0.00990 M
HSO4- 0.10 - 0.00985 0.09015 M or 0.0902 M
H3O 0.10 x 0.1099 M or 0.110 M
OH- Kw/H3O 1 x 10-14/0.110 9.1 x 10-14
M
The pH of 0.10 M H2SO4 -log 0.11 0.9 .

4
Question on Partition Equilibrium

At room temperature, the partition coefficient of
an organic compound X between ether and water is
3.10.
1) Calculate the mass of X extracted into the
ether layer when 100 cm3 of water containing 10.0
g of X is shaken with 100 cm3 of ether.
7.56 g
2) Calculate the mass of X which would be removed
from the aqueous layer in each of TWO further
extractions using 100 cm3 of ether each time.
1.84 g 0.45 g
3) Compare the total mass of X extracted in these
3 successive extractions with that obtained from
a single extraction using 300 cm3 of ether.
Comment on the difference. 9.85
g(three) 9.03 g(one) 88II2a 6

5
Calculation on Partition Equilibrium

At room temperature, the partition coefficient of
an organic compound X between ether and water is
3.10.
Kpartition Xether/Xaq 3.10,
a g of X in 100 cm3 water goes to
the 100 cm3 ether layer 3.10
(a/100)/(10-a)/100 a/(10-a)
a 7.56 g (out of 10.0 g of X in 100 cm3 water),
fraction of solute extracted r 7.56/10.0
0.756
2.44 g of X remain in the 100 cm3 aqueous layer
In a further extraction using 100 cm3 of ether,
2.44 x 0.756 or 1.845 g of X will be extracted
into the 100 cm3 ether layer in a third
extraction using 100 cm3 of ether, (2.44-1.845) x
0.756 or 0.45 g of X will be extracted into the
100 cm3 ether layer.
Amount of X extracted in 3 extractions using 100
cm3 ether 7.56 1.845 0.45 9.855 g
b g of X in water was extracted using 300 cm3
ether 3.10 (b/300) /(10-b)/100 b/3
(10-b), b 9.03

6

General formula for solvent extraction

100 cm3 water
solute extracted into 100 cm3 ether original
a g solute After 1st extract a
r g solute
1st extract a (1 - r) g solute After 2nd
extract a (1 - r) r g solute 2nd extract
a (1 - r)2 g solute After 3rd extract a (1 - r)2
r g solute
nth extract a(1 - r)n g solute After
(n1)th extract a (1 - r)n r g solute where r
is the fraction of solute extracted
Mass of solute remained after n extractions with
100 cm3 ether a (1 - r)n g
Total mass of solute extracted after n
extractions with 100 cm3 ether a r 1 (1 -
r) (1 - r)2 (1 - r)n-1 g

7
Using Kc in a Partition Equilibrium

4 g of a solute X is dissolved in 25.0 cm3 of
aqueous solution and this solution is in
equilibrium at R.T.P. with an ethereal solution
of X containing 80 g of X in 100 cm3.
mass of X in 25 cm3 water
mass of X in 100 cm3 ether
original 4 g
80 g
Kpartition Xaq/Xether
(4/25)/(80/100) 0.2
mass of X in 100 cm3 water
mass of X in 50 cm3 ether
original 5 g
0 g
1st extract (5 - x)
x g
0.2 (5-x)/100/(x/50) (5-x)/2x x
3.57 g
mass of X in 100 cm3 water
mass of X in 25 cm3 ether
original 5 g
0 g
1st extract (5 - x)
x g
0.2 (5-x)/100/(x/25) (5-x)/4x x
2.777 g (r 2.777/5)
2nd extract (5-2.777) - y
y
0.2 (2.223 - y)/100/(y/25) y
1.235 g 2.777(1-r)
Total mass of solute X extracted 2.777 1.235
4.01 g

8
Hydrolysis of salt

HA(aq) H2O(l) H3O(aq) A-(aq) Ka
H3OA-/HA
A-(aq) H2O(l) HA(aq) OH-(aq)
Kh Kb HAOH-/A-
HAOH-H/A-H
Kw/A-H/HA Kw/Ka
The weaker the acid, the stronger the hydrolysis
because a weak acid has a strong conjugate base.
CN-(aq) H2O(l) HCN(aq) OH-(aq)
1.0 - x x
x
Kb Kw/Ka 10-14/(4 x 10-10) 2.5 x 10-5
x2/(1.0-x) 2.5 x 10-5, x 5 x 10-3 M, pH 14
log x 11.7

9
The Common Ion Effect
In 0.10 M ammonia solution, OH- 1.35 x 10-3 M
(0.10 -x) x
x
NH3(aq) H2O(l) NH4(aq) OH-(aq)
(0.10 -x) M (0.10 x) M
x M
Kb 1.8 x 10-5 (0.1x) x/ (0.1-x), Kb x if
x ltlt 0.10 pH 14 log 1.8 x 10-5 9.26
In the presence of a common ion, the ionization
of ammonia solution is suppressed, resulting in
a decreased OH- a lower pH
Neutralizing ammonia with dilute nitric acid
produces the common ion
If 25 cm3 of 0.1 M HNO3 is run into 50 cm3 of 0.1
M NH3(aq) in a conical flask, half of the ammonia
will be neutralized and by stiochiometry, NH3o
and NH4o become 1/30 M. The pH of the
resulting solution in the flask is 9.26
10
Buffer Solution (1)

Solution A 0.50 dm3 of 0.30 M methanoic acid,
HCO2H
Solution B 0.50 dm3 of 0.15 M sodium hydroxide,
NaOH
Stoichiometric cosideration
HCO2H(aq) NaOH(aq) HCO2Na(aq) H2O(l)
0.150 0.075
0 0
0.15-0.075 0.075-0.075 0.075
0.075 moles
HCO2H 0.075/1 0.075 M, HCO2Na 0.075 M
Equilibrium consideration
HCO2H(aq) HCO2-(aq)
H(aq) (0.075- x) M
(0.075 x) M x M
Ka 1.75 x 10-4 x if x ltlt 0075
pH - log (1.75 x 10-4) 3.76
Mixing solutions A and B produces equal
concentrations of HCO2H
and HCO2Na,
that is a buffer solution. The 1st stage of an
acid-base titration
corresponds to a buffer.

11
Buffer System (2)

In a buffer system, there is a mixture of a weak
acid its conjugate base (or a weak base its
conjugate) and the concentrations of the
constituents are in simple ratio. A buffer system
resists any change in pH upon the addition of a
strong acid or alkali.
As a strong acid/alkali is fully ionized, any
addition of strong alkali/acid effectively lowers
the H/OH-
Adding H to a buffer converts A- to HA and
adding OH- to a buffer converts HA to A-. In both
cases the A-/HA remains fairly constant. As
KaHA-/HA, a constant at fixed
temperature, H and pH is thus kept constant.
Calculate the pH of a one dm3 buffer mixture
containing 0.1 moles of HA and 0.05 moles of
NaA.
HA(aq) H2O(l) H3O(aq) A-(aq)
0.10-x x
0.05x
Ka x(0.05x)/(0.1-x) x/2 if x ltlt 0.05 H3O
2 Ka

12

Buffer Solution (3)

A strong alkali MOH(aq) was run from a burette
into a conical flask with 40.0 cm3 of an aqueous
solution of a weak acid, HA(aq)
Calculate the initial concentration of HA(aq)
if the initial pH is 2.70
HA(aq) H(aq) A-(aq) ,
Ka 1.8 x 10-5
(a - x) M 10-2.7 10-2.7
where x 10-2.7
Ka 1.8 x 10-5 10-5.4/a if 10-2.7 ltlt a,
a 5.6 x 10-1.4 0. .
MOH(aq) HA(aq) MA(aq) H2O(l)
By stoichiometry, 40.0 cm3 of 0.2212 M HA(aq)
required V cm3 of b M MOH(aq). The resulting
mixture, corresponding to the equivalence point
of the titration, was found to have a pH of 8.90
Concentration of the strong base MOH(aq),
b(40/V)(0.2212)8.848/V
A-(aq) H2O(l)
HA(aq) OH-(aq)
0.2212x 40/(40V)
10-5.10 10-5.10
Kh 10-14/1.8 x 10-5 5.6 x 10-10
10-10.20/0.2212 40/(40V)
40 V 5.556 x 10-10 x 0.2212 x 40 (1010.2)
4.916 x10-9(1010.2) 77.9
V 37.9 b 0.2335