* Arrhenius
acid: generates [H+] in solution
base: generates [OH-] in solution
normal Arrhenius equation: acid + base <---> salt + water
example: HCl + NaOH <---> NaCl + H2O
* Bronsted-Lowery:
acid: anything that donates a [H+] (proton donor)
base: anything that accepts a [H+] (proton acceptor)
normal Bronsted-Lowery equation: acid + base <---> acid + base
example: HNO2 + H2O <---> NO2- + H3O+
Each acid has a conjugate base and each base has a conjugate acid. These
conjugate pairs only differ by a proton. In this example: HNO2 is the acid, H2O
is the base, NO2- is the conj. base, and H3O+ is the conj. acid.
* Lewis:
acid: accepts an electron pair
base: donates an electron pair
The advantage of this theory is that many more reactions can be considered
acid-base reactions because they do not have to occur in solution.
Wednesday 11 June 2008
pKa values of selected compounds
Measurements are at 25ºC in water for those with a pKa at or above -1.76:
* - 25.00: Fluoroantimonic acid
* - 15.00: Magic acid
* - 10.00: Fluorosulfuric acid
* - 10.00: Perchloric acid
* - 10.00: Hydroiodic acid
* - 9.00: Hydrobromic acid
* - 8.00: Hydrochloric acid
* - 3.00, 1.99: Sulfuric acid
* - 2.00: Nitric acid
* - 1.76: Hydronium ion
* 3.15: Hydrofluoric acid
* 3.75: Formic acid
* 4.04: Ascorbic acid (Vitamin C)
* 4.19: Succinic acid
* 4.20: Benzoic acid
* 4.63: Aniline*
* 4.76: Acetic acid
* 4.76: Dihydrogencitrate ion (Citrate)
* 5.21: Pyridine*
* 6.37: Carbonic acid
* 6.40: Monohydrogencitrate ion Citrate
* 6.99: Ethylenediamine*
* 7.00: Hydrogen sulfide, Imidazole* (as an acid)
* 7.50: Hypochlorous acid
* 9.25: Ammonia*
* 9.33: Benzylamine*
* 9.81: Trimethylamine*
* 9.99: Phenol
* 10.08: Ethylenediamine*
* 10.33: Bicarbonate
* 10.66: Methylamine*
* 10.73: Dimethylamine*
* 10.81: Ethylamine*
* 11.01: Triethylamine*
* 11.09: Diethylamine*
* 11.65: Hydrogen peroxide
* 12.50: Guanidine*
* 12.67: Monohydrogenphosphate ion (Phosphate)
* 14.58: Imidazole (as a base)
* 15.76: Water
* - 19.00 (pKb) Sodium amide
* - 37.00 (pKb) Lithium diisopropylamide (LDA)
* 45.00: Propane
* 50.00: Ethane
* - 25.00: Fluoroantimonic acid
* - 15.00: Magic acid
* - 10.00: Fluorosulfuric acid
* - 10.00: Perchloric acid
* - 10.00: Hydroiodic acid
* - 9.00: Hydrobromic acid
* - 8.00: Hydrochloric acid
* - 3.00, 1.99: Sulfuric acid
* - 2.00: Nitric acid
* - 1.76: Hydronium ion
* 3.15: Hydrofluoric acid
* 3.75: Formic acid
* 4.04: Ascorbic acid (Vitamin C)
* 4.19: Succinic acid
* 4.20: Benzoic acid
* 4.63: Aniline*
* 4.76: Acetic acid
* 4.76: Dihydrogencitrate ion (Citrate)
* 5.21: Pyridine*
* 6.37: Carbonic acid
* 6.40: Monohydrogencitrate ion Citrate
* 6.99: Ethylenediamine*
* 7.00: Hydrogen sulfide, Imidazole* (as an acid)
* 7.50: Hypochlorous acid
* 9.25: Ammonia*
* 9.33: Benzylamine*
* 9.81: Trimethylamine*
* 9.99: Phenol
* 10.08: Ethylenediamine*
* 10.33: Bicarbonate
* 10.66: Methylamine*
* 10.73: Dimethylamine*
* 10.81: Ethylamine*
* 11.01: Triethylamine*
* 11.09: Diethylamine*
* 11.65: Hydrogen peroxide
* 12.50: Guanidine*
* 12.67: Monohydrogenphosphate ion (Phosphate)
* 14.58: Imidazole (as a base)
* 15.76: Water
* - 19.00 (pKb) Sodium amide
* - 37.00 (pKb) Lithium diisopropylamide (LDA)
* 45.00: Propane
* 50.00: Ethane
Polyprotic acid
A polyprotic acid is one that has more than one proton to dissociate. Typical examples are malonic acid, which has two ionizable protons, and phosphoric acid, which has three. The constant for dissociation of the first proton may be denoted as pKa1 and the constants for dissociation of successive protons as pKa2, etc.
It is generally true that successive pK values increase (Pauling's first rule).[1] For example, for a diprotic acid, H2A, the two equilibria are
H2A <--> HA- + H+
HA- <--> A- + H+
It can be seen that the second proton is removed from a negatively charged species. Since the proton carries a positive charge extra work is needed to remove it. Therefore pKa2 > pKa1. There are a few exceptions to this rule which occur when there is a major structural change such as in the sequence
It is generally true that successive pK values increase (Pauling's first rule).[1] For example, for a diprotic acid, H2A, the two equilibria are
H2A <--> HA- + H+
HA- <--> A- + H+
It can be seen that the second proton is removed from a negatively charged species. Since the proton carries a positive charge extra work is needed to remove it. Therefore pKa2 > pKa1. There are a few exceptions to this rule which occur when there is a major structural change such as in the sequence
Monoprotic acid
When an acid, HA, dissolves in water, some molecules of the acid 'dissociate' to form hydronium ions and the conjugate base, (A-), of the acid.
HA --> H+ + A-
It is understood that H + stands for the hydronium ion and that each species in this equilibrium may solvate to a greater or lesser extent. The acid dissociation constant is defined as
Ka = [H+][A-]/[HA]
H2O is omitted from these expressions because in dilute solution the concentration of water may be assumed to be constant ([H2O] = 1). Values of Ka vary over many orders of magnitude, so it is common to take the negative logarithm to base ten of the value - this is just for ease of writing and calculations really
pKa = -log(ka)
It is easier to compare the strengths of different acids by comparing pKa values as they vary over a much smaller range.
HA --> H+ + A-
It is understood that H + stands for the hydronium ion and that each species in this equilibrium may solvate to a greater or lesser extent. The acid dissociation constant is defined as
Ka = [H+][A-]/[HA]
H2O is omitted from these expressions because in dilute solution the concentration of water may be assumed to be constant ([H2O] = 1). Values of Ka vary over many orders of magnitude, so it is common to take the negative logarithm to base ten of the value - this is just for ease of writing and calculations really
pKa = -log(ka)
It is easier to compare the strengths of different acids by comparing pKa values as they vary over a much smaller range.
Acid Dissociation constant Ka part 2
Acid dissociation constant, denoted by Ka, is an equilibrium constant for the dissociation of a weak acid. According to the Brønsted-Lowry theory of acids and bases, an acid is a proton donor (HA, where H represents an acidic hydrogen atom), and a base is a proton acceptor. In aqueous solution, water can function as a base, as in the following general example.
HA + H2O --> A- + H3O+
Acid dissociation constants are also known as the acidity constant or the acid-ionization constant. The term is also used for pKa, which is equal to the negative decimal logarithm of Ka (cologarithm of Ka).
pKa = -log (Ka)
HA + H2O --> A- + H3O+
Acid dissociation constants are also known as the acidity constant or the acid-ionization constant. The term is also used for pKa, which is equal to the negative decimal logarithm of Ka (cologarithm of Ka).
pKa = -log (Ka)
Acid Dissociation Constant Ka
From: ::Rose::
Date: 11 Jun 2008, 19:14
Q2. A solution of Phenol in (C6H5OH) in water has a concentration of 38g dm-3. The acid dissociation constant, Ka, of Phenol is 1.3 x 10 (to the -10) mol dm-3.
Write an expression for the acid dissociation constant of Ka, of Phenol.
Dissociation of Phenol:
HA + H2O = A- + H3O+
Ka = [H3O+][A-]/[HA][H2O]
Key points:
1) For every A-, there is one H3O+. Therefore [A-] is always equal [H3O+]
2) [H2O] is constant i.e. [H2O] = 1, because water is hardly altered by the ionisation of HA
Therefore the expression for the Ka of phenol is;
Ka = [H3O]2/[HA]
whereby Ka = 1.3 x 10-10 M (given by question itself)
[HA] = 0.4 M (see below)
[H3O]= 7.2 x 10-6 M (see below)
____________________________________________________________
_
Extra calculations for own understanding:
Converting g/dm3 into mol/dm3:
[HA] = 36 g/dm3
Divide this by the molecular mass of phenol (94 g/ml)
[HA] = 36g/dm3
---------
94g/mol
[HA] = 0.4 mol/dm3 (notice that the grams cancel out)
Finding the concentration of H30+:
Use the expression we already found above, just feed in the numbers
Ka = [H3O+]2/[Ha]
1.3 x 10-10 M = [H3O+]2/[0.4 M]
[H30+] = {(1.3 x 10-10)x(0.4)}to the power of 1/2
[H30+] = 7.2 x 10-6 M
If you want a more fancy eqn, use PhOH in place of HA, so
PhOH + H2O = PhO- + H30+
>>W....O....W !!!! I love you! You've just made my night! Lol!
Date: 11 Jun 2008, 19:14
Q2. A solution of Phenol in (C6H5OH) in water has a concentration of 38g dm-3. The acid dissociation constant, Ka, of Phenol is 1.3 x 10 (to the -10) mol dm-3.
Write an expression for the acid dissociation constant of Ka, of Phenol.
Dissociation of Phenol:
HA + H2O = A- + H3O+
Ka = [H3O+][A-]/[HA][H2O]
Key points:
1) For every A-, there is one H3O+. Therefore [A-] is always equal [H3O+]
2) [H2O] is constant i.e. [H2O] = 1, because water is hardly altered by the ionisation of HA
Therefore the expression for the Ka of phenol is;
Ka = [H3O]2/[HA]
whereby Ka = 1.3 x 10-10 M (given by question itself)
[HA] = 0.4 M (see below)
[H3O]= 7.2 x 10-6 M (see below)
____________________________________________________________
_
Extra calculations for own understanding:
Converting g/dm3 into mol/dm3:
[HA] = 36 g/dm3
Divide this by the molecular mass of phenol (94 g/ml)
[HA] = 36g/dm3
---------
94g/mol
[HA] = 0.4 mol/dm3 (notice that the grams cancel out)
Finding the concentration of H30+:
Use the expression we already found above, just feed in the numbers
Ka = [H3O+]2/[Ha]
1.3 x 10-10 M = [H3O+]2/[0.4 M]
[H30+] = {(1.3 x 10-10)x(0.4)}to the power of 1/2
[H30+] = 7.2 x 10-6 M
If you want a more fancy eqn, use PhOH in place of HA, so
PhOH + H2O = PhO- + H30+
>>W....O....W !!!! I love you! You've just made my night! Lol!
Henderson-Hasselbalch equation
From: ::Rose::
Date: 11 Jun 2008, 18:26
Q1. A buffer Solution contains HCOOH and HCOONa. Calculate the pH of a buffer solution containing equal volumes of 2.5mol dm-3 HCOONa and 1.0mol dm-3 HCOOH (Ka=1.6 x 10(to the -4) mol dm-3.
[Salt] = 2.5 M
[Acid] = 1.0 M
[Acid ka] = 1.6 x 10-4 M
First convert Ka into pKa
pKa of acid = -log(1.6 x 10-4) = 3.8
Now use the Henderson-Hasselbalch equation:
pH = pKa + log [salt]/acid]
pH = 3.8 + log 2.5/1.0
pH = 4.2
(I have rounded everything to 1 d.p, square brackets [] refers to concentration)
>> Brilliant! I owe you big time! I mean that!
Date: 11 Jun 2008, 18:26
Q1. A buffer Solution contains HCOOH and HCOONa. Calculate the pH of a buffer solution containing equal volumes of 2.5mol dm-3 HCOONa and 1.0mol dm-3 HCOOH (Ka=1.6 x 10(to the -4) mol dm-3.
[Salt] = 2.5 M
[Acid] = 1.0 M
[Acid ka] = 1.6 x 10-4 M
First convert Ka into pKa
pKa of acid = -log(1.6 x 10-4) = 3.8
Now use the Henderson-Hasselbalch equation:
pH = pKa + log [salt]/acid]
pH = 3.8 + log 2.5/1.0
pH = 4.2
(I have rounded everything to 1 d.p, square brackets [] refers to concentration)
>> Brilliant! I owe you big time! I mean that!
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