Transcript Topic 18. Acids and bases

Topic 18- Acids and bases
18.1 Calculations involving acids and bases
18.2 Buffer solutions
18.3 Salt hydrolysis
18.4 Acid-base titrations
18.5 Indicators
The relation between oxonium ions and hydroxide ions
18.1 Calculations involving acids and
bases
• pH = -log[H3O+]
pOH = -log[OH-]]
• [H3O+] = 10-pH
[OH-] = 10-pOH
• pH + pOH = 14
[H+]=[H3O+]
Calculate the pH in following solutions:
• 1M HCl
pH = -log[1] = 0
• 0.001 M HNO3
pH = -log [0.001] = 3
• 0.5 M H2SO4  2H+
pH = -log [2*0.5]= 0
• 0.15 M NaOH
pOH =-log[0.15]= 0.82
pH = 14-0.82 = 13.18
Calculate the [H3O+] in following
solutions
• pH = 5,5
[H3O+] = 10-5.5 = 3.2*10-6
• pH =- 1
[H3O+] =10-(-1) = 101 = 10 M
Autoprotolysis of water
H2O + H2O
H3O+ + OH-
Kc= [H3O+] . [OH- ]
.
[H2O] [H2O]
The concentration of water
is not changing- it is constant
Kw= K . [H2O] . [H2O] = [H3O+] . [OH-]
Kw = the dissociation constant of water
Kw = dissociation constant of water
H2O + H2O
H3O+ + OHIn 25oC pure water:
[H3O+] = 10-7 mol/dm3
[OH-] = 10-7 mol /dm3
Kw = [H3O+]*[OH-] = 10-7*10-7 = 10-14 mol2/dm6
-log Kw = -log ([H3O+]*[OH-])= -log [H3O+]+ -log[OH]=-log 10-14
pKw = pH + pOH = 14
The effect of temperature on the water dissociation
constant
Temperature
(◦C)
0
15
25
50
100
Kw
pKw
0,11 . 10 -14
0,45 . 10 -14
14,96
14,35
. 10 .-14 vid-14
Kw= 1,01,0
10 25 C
14,00
13,26
12,29
H2O + H2O
5,5 . 10 -14
51 . 10 -14
H3O+ + OH- DH> 0
Weak acids
HA + H2O
K= [H3O+] . [A- ]
[HA] . [H2O]
H3O+ + AKa=K . [H2O] = [H3O+] . [A- ]
[HA]
Ka = acid dissociation constant
Ka => acid strength => higher value => stronger acid.
pKa = -log Ka
=> lower value => stronger acid
Some acid dissociation constants
Ka
104
pKa
-4
103
-3
Hydrochloric acid
Sulphuric acid
HCl
H2SO4
Nitric acid
Oxalic acid
Phosphoric acid
HNO3
H2C2O4
H3PO4
1
0.06
0.007
-1
1.25
2.15
Salicylic acid
Citric acid
C7H6O3
C6H8O7
0.001
7.10-4
2.97
3.13
Ascorbic acid
C6H4O2(OH)
1.10-4
4.17
4
Acetic (etanoic) acid HAc
Carbonic acid
CH3COOH 1.8.10-5
H2CO3
4.2.10-7
Ammonium ion
NH4+
4.75
6.37
5.5.10-10 9.26
Weak bases
BH + H2O
K= [BH2+] . [OH-]
.
[BH] [H2O]
BH2+ + OHKb=K . [H2O] = [BH2+] . [OH-]
[BH]
Kb = base dissociation constant
Kb => base strength => higher value => stronger base
pKb = -logKb
=> lower value => stronger base
Bases
Kb
pKb
Litium hydroxide
Sodium hydroxide
Ammonia
LiOH
NaOH
NH3
2.3
0.63
1.8.10-5
-0.36
0.2
4.74
Hydrogen
carbonate ion
Acetate ion
Nitrate ion
Chloride ion
HCO3-
2.4.10-8
7.62
CH3COONO3Cl-
5.5.10-10 9.25
10-15
15
10-18
18
Kw connects Ka and Kb for a
corresponding acid/ base pair, such as
CH3COOH/CH3COOKa * Kb = Kw = 10-14
pKa + pKb = pKw = 14
(at 25 ºC)
Calculate pKa of ethanoic acid, HAc
(CH3COOH)
We know that c= 0.01M We measure pH
HAc+H2O
Cstart
Ceq
0.1
0.1- 10-pH
H3O+ +Ac0
10-pH
0
10-pH
Ka = [H3O+]*[Ac-] / [HAc] = 10-pH* 10-pH /0.1- 10-pH=
pKa= -log Ka =
Calculate pH
in 0,1 M ethanoic acid, HAc
H3O+ +Ac-
HAc
Cstart
Ceq
0.1
0.1-X
0
X
0
X
pKa = 4.75
(see CDB)
Ka = [H3O+]*[Ac-] / [HAc] = 10-4.75 = X2/0.1-X ~ X2/0.1 if x is small
X2 = 0.1* 10-4.75
X = (0.1* 10-4.75 )½
pH = -log [X] = -log[(0.1* 10-4.75 )½] = 2.88
18.2 Buffer solutions
• In pure water pH= 7.
• Addition of small amounts of acid or base
gives big changes in pH
• That can be great problem, especially in biological systems. But there are
ways to make a solution that can be quite pH stable.
• A buffer resist changes in pH when a strong
acid or base is added
A Buffer: a mixture of a weak acid and
its corresponding base
The equilibrium:
HA(aq) +H2O
H3O+(aq) + A-(aq)
If strong acid is added (fully dissociated,
contains mainly H3O+) =>
reaction goes to the left (Le Chatelier’s principle) =>
little change in pH (-log [H3O+])
A Buffer: a mixture of a weak acid and
its corresponding base
The equilibrium: HA(aq) +H2O
H3O+(aq) + A-(aq)
• If a strong base is added => OH- reacts with
H3O+ to form water =>
• equilibrium goes to the right => Restore the
[H3O+] => little change in pH
How to prepare a buffer:
• Mix a weak acid and its corresponding (conjugate) base e.g.
CH3COOH and CH3COO-Na+.
• Mix a weak base and its corresponding (conjugate) e.g. NH3
and NH4Cl.
• Add strong base to an excess of weak acid.
• Add strong acid to an excess of weak base.
Add strong base to an excess of weak acid (HA)
The reaction of HA and water:
HA(aq) +H2O
H3O+(aq) + A-(aq)
The reaction of HA and strong base (NaOH):
HA + NaOH → H2O + Na+A-
When you add some NaOH to a HAsolution your mixture will consist of all the
above particles, but in particular the weak
acid and it’s corresponding base HA/A-
Add strong acid to an excess of weak base (BH)
The reaction of BH and water:
BH + H2O
BH2+ + OH-
The reaction of B and strong acid (HCl):
HCl + BH → BH2+Cl-
When you add some HCl to a BH-solution
your mixture will consist of all the above
particles, but in particular the weak base
and it’s corresponding acid BH/BH2+
The Hydrogen ion concentration and pH of a
buffer can be calculated from the expression for
the acid dissociation constant:
[H3O+] = Ka* [HA] /[A-]
take –log on both sides
-log [H3O+] = -log Ka+ -log [HA] /[A-]
identify
pH = pKa -log([HA] /[A-])
You have to be able to derive the equation
yourself!
Exercises
1. Which combination will form a buffer solution?
A. 100 cm3 of 0.10 mol dm–3 hydrochloric acid with 50 cm3 of 0.10 mol dm–3 sodium hydroxide.
B. 100 cm3 of 0.10 mol dm–3 ethanoic acid with 50 cm3 of 0.10 mol dm–3 sodium hydroxide.
C. 50 cm3 of 0.10 mol dm–3 hydrochloric acid with 100 cm3 of 0.10 mol dm–3 sodium hydroxide.
D. 50 cm3 of 0.10 mol dm–3 ethanoic acid with 100 cm3 of 0.10 mol dm–3 sodium hydroxide.
2. Buffer solutions resist small changes in pH. A phosphate buffer can be made by
dissolving NaH2PO4 and Na2HPO4 in water, in which NaH2PO4 produces the acidic ion and
Na2HPO4 produces the conjugate base ion.
(i)
Deduce the acid and conjugate base ions that make up the phosphate buffer and state the
ionic equation that represents the phosphate buffer.
(ii) Describe how the phosphate buffer minimizes the effect of the addition of a strong base, OH–
(aq), to the buffer. Illustrate your answer with an ionic equation.
(iii) Describe how the phosphate buffer minimizes the effect of the addition of a strong acid, H+(aq),
to the buffer. Illustrate your answer with an ionic equation.
Answers
1. Which combination will form a buffer solution?
B. 100 cm3 of 0.10 mol dm–3 ethanoic acid with 50 cm3 of 0.10 mol dm–3 sodium hydroxide.
You have both ethanoic acid and sodium ethanoate
2. (i) Acid: H2PO4–; (Conjugate) base: HPO42–; H2PO4–(aq) H+(aq) + HPO42–(aq);
(ii) strong base/OH– replaced by weak base (H2PO42–, and effect minimized) /
strong base reacts with acid of buffer / equilibrium in (i) shifts in forward
direction; OH–(aq) + H2PO4–(aq) → H2O(l) + HPO42–(aq);
(iii) strong acid/H+ replaced by weak acid (H2PO4–, and effect minimized) /
strong acid reacts with base of buffer / equilibrium in (i) shifts in
reverse direction;
H+(aq) + HPO42–(aq) → H2PO4–(aq);
18.3 Salt hydrolysis
• The positive and negative ion in a salt can be
neutral or act as an acid or a base
• Cations (positive ions) can act as acids and
anions (negative ions) can act as bases
The acetate (ethanoate) ion
HAc + H2O
H3O+ + Ac-
acid
Ac- + H2O
base
pKa(HAc)= 4.75
base
HAc + OH-
pKb(Ac-)= 9.25
acid
The ethanoate ion is salt of a WEAK acid (ethanoic acid)
and thus basic
More basic ions
•
•
•
•
CNHCO3CO32PO43-
Salts of weak acids
The chloride ion
HCl + H2O
H3O+ + Cl-
acid
Cl- + H2O
base
pKa(HCl)= - 4
base
HCl + OH-
pKb(Cl-)= 18
acid
The chloride ion is salt of a STRONG acid (hydrochloric acid)
and thus so week so it is neutral (>pKw)
More ions with no acid/base character
•
•
•
•
Na+
K+
Ca2+
NO3-
SO42ClO4Cl-
• Derives from strong acids and bases => no
acid-base activity
The ammonium ion
NH3 + H2O
NH4+ + OH-
base
acid
NH4+ + H2O
acid
H3O+ + NH3
pKb(NH3)= 4.74
pKa(NH4+)=
base
Ammonium ion is salt of a WEAK base (ammonia)
and thus acidic
Metallic ions with high charge
• Metallic ions with high charge, e.g. Al3+ and Fe3+,
form complexes with water:
• Al(H2O)63+ and Fe(H2O)63+. The electronegative effect
of the ion weakens the O-H bond in water molecules:
[Fe(H2O)6]3+(aq) + H2O
[Fe(OH)(H2O)5]3+(aq)+ H3O+(aq)
An acidic solution
18.4 Acid-base titrations
• Strong acid – Strong base
• Weak acid – Strong base
• Strong acid – Weak base
Strong acid – Strong base
HCl + NaOH
NaCl + H2O
Start with 10 ml 0.1 M HCl (pH=1). Titrate with 0.1M NaOH
• When 90% of the base been added: HCl ~0.01 => pH = 2
• When 99% of the base been added: HCl ~0.001 => pH = 3
• When 101% of the base been added: [OH-] = 0.001 pH =11
Weak acid – strong base
CH3COOH + NaOH
CH3COONa + H2O
• When strong base is added the pH gradually increase.
• At equivalence point all acid is consumed, pH increase rapidly.
• The salt of the weak acid is a weak base => pH > 7 at
equivalence point.
• At ½ equivalence point [HA] =[A-] => pH => pKa
Titration simulations at
http://chem-ilp.net/labTechniques/AcidBaseIdicatorSimulation.htm
18.5 Indicators
• A weak acid/base where the colours of the
protonated and ionized forms are different
HIn  H+ + InRed
Blue
The colour depends both on pH and the pKavalue. => Different indicators change their
colours at different pH
Structures of BTB at different pH
pH 7.3
Increasing [OH-]/pH
Indicators change colour around their
pKa-values
How to choose indicator?
• If you titrate CH3COOH with NaOH the pH will be above 7 at
the equivalence point => choose an indicator that change
colour above 7 e.g. phenolphthalein (pKa =9.6), range 8.3 –
10.0. Rapid pH changes in that area.
- If you titrate NH3 with HCl the pH will be under 7 at the
equivalence point => choose an indicator
that change colour under 7 e.g.
methyl orange(pKa = 3.7), range
3.1 – 4.4.
Rapid pH changes in that area.