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Equilibrium Constant: Kp

Equilibrium Constant: Kp

Mole Fraction

  • The mole fraction is a unit of concentration, defined to be equal to the number of moles of a component divided by the total number of moles of a solution
  • The mole fraction is determined for all reactants in a mixture of gases
  • g. to find the mole fraction of A (Xa):

Partial pressure

  • Partial pressure it the pressure exerted by the single gas if it were isolated
  • The total pressure is the combined total of all the partial pressures
  • To determine the partial pressure for one gas, multiply the mole fraction by the total pressure
  • g. to find the partial pressure of A (PA):
  • Units for partial pressure Pascals or atmospheres

PA = Mole Fraction of A  x Total Pressure

PA = Xa  x   P

Kp in homogeneous gaseous equilibria

    • Homogeneous Equilibria refers to all involved reactants are in the same phase
    • For Kp to function, everything must be a gas

  • Kp will always be a constant except for a change in temperature

Kp in heterogeneous equilibria

  • Heterogeneous equilibrium will involve gases in contact with solids
  • As Kp functions only with a gas, the solid is not included
Properties of Hydrocarbons

Properties of Hydrocarbons

Polarity
• Alkanes are not polar, as there is no electronegativity between C-H bonds
• Alcohols and water are polar because they contain –OH bonds, with a difference in electronegativity
• The only force present in hydrocarbons are Van Der Waals
Reactivity
• Alkenes are more reactive than alkanes due to their double bonds which break in the reaction releasing more energy
• Alkanes are not reactive but can be combusted as fuels for engines.
Solubility
• Alkanes are non-polar and therefore do not mix and dissolve into polar liquids such as water. They are unable to form hydrogen bonds between each molecule. Water also has strong hydrogen bonds between each H2O molecule and therefore not want to move apart to allow for the alkane in
Boiling Points
• The increase in the chain of the atoms in an alkane, increases the boiling points as a result of there being more Van der Waal forces between each molecule. (A CH4 molecule has a boiling point of -161.5 °C, where are C10H22

Polarity
• Alkanes are not polar, as there is no electronegativity between C-H bonds
• Alcohols and water are polar because they contain –OH bonds, with a difference in electronegativity
• The only force present in hydrocarbons are Van Der Waals

Reactivity
• Alkenes are more reactive than alkanes due to their double bonds which break in the reaction releasing more energy
• Alkanes are not reactive but can be combusted as fuels for engines.

Solubility
• Alkanes are non-polar and therefore do not mix and dissolve into polar liquids such as water. They are unable to form hydrogen bonds between each molecule. Water also has strong hydrogen bonds between each H2O molecule and therefore not want to move apart to allow for the alkane in

Boiling Points
• The increase in the chain of the atoms in an alkane, increases the boiling points as a result of there being more Van der Waal forces between each molecule.

Polarity
• Alkanes are not polar, as there is no electronegativity between C-H bonds
• Alcohols and water are polar because they contain –OH bonds, with a difference in electronegativity
• The only force present in hydrocarbons are Van Der Waals

Reactivity
• Alkenes are more reactive than alkanes due to their double bonds which break in the reaction releasing more energy
• Alkanes are not reactive but can be combusted as fuels for engines.

Solubility
• Alkanes are non-polar and therefore do not mix and dissolve into polar liquids such as water. They are unable to form hydrogen bonds between each molecule. Water also has strong hydrogen bonds between each H2O molecule and therefore not want to move apart to allow for the alkane in

Name of Alkene

Number of Carbons

Chemical Formula

Boiling Point in °C
(degrees centigrade)

State at “Room Temperature” (20°C)

Melting Point in °C
(degrees centigrade)

Ethene

2

C2H4

-104

gas

-169

Propene

3

C3H6

-47

gas

-185

Z-Butene

4

C4H8

0.9

gas

-138.9

E-Butene

4

C4H8

3.7

gas

-139.7

1-Pentene

5

C5H10

30

gas

-165

Z-2-Pentene

5

C5H10

36

gas

-135

E-2-Pentene

5

C5H10

37

gas

-180

1-hexene

6

C6H12

63

liquid

−139.8

1-Heptene

7

CH14

115

liquid

-119

3-octene

8

C8H16

122

liquid

-101.9

3-nonene

9

C9H18

147

liquid

-84.4

5-decene

10

C10H20­

170

liquid

-66.3

 

Name of Alkane

Number of Carbons

Chemical Formula

Boiling Point in °C
(degrees centigrade)

State at “Room Temperature” (20°C)

Melting Point in °C
(degrees centigrade)

Methane

1

C H4

-162

gas

-183

Ethane

2

C2H6

-89

gas

-172

Propane

3

C3H8

-42

gas

-188

Butane

4

C4H10

0

gas

-138

Pentane

5

C5H12

36

liquid

-130

Hexane

6

C6H14

69

liquid

-95

Heptane

7

C7H16

98

liquid

-91

Octane

8

C8H18

126

liquid

-57

Nonane

9

C9H20

151

liquid

-54

Decane

10

C10H22

174

liquid

-30

Fractional Distillation of Crude Oil

Fractional Distillation of Crude Oil

Fractional distillation allows us to separate different lengths hydrocarbons, mainly alkanes, from a mixture, as they are found in crude oil. Different hydrocarbons have different uses:

Fraction

Length of atom chains

Uses

Gas

1-4

Bottled gases for camping etc.

Petrol

5-12

Petrol for combustion engines

Naphtha 

7-14

Petrochemicals

Kerosene

11-15

Plane fuels

Diesel

15-19

Combustion engines

Mineral Oils

20-30

Lubricating oils

Fuel Oils

30-40

Ship fuel and Power stations

Wax

40-50

Candles and lubrication

Bitumen

50+

Road surfaces

Cracking in Industry

  • Crude oil is heated to 350°c and pumped into the bottom of the distillation column
  • The crude oil vaporises, the smaller hydrocarbon molecules with their lower boiling points, rise and condense higher up the column
  • Going down the column the different hydrocarbon chains grow larger, going from gases to more viscous substances such as bitumen. The largest hydrocarbons do not vaporise as their boiling points are higher than 350°c and therefore collect at the bottom forming a thick residue.
  • As the crude oil rises it reaches its fractions, and being far away from the heat source and falls below its boiling point and condenses on to it. This follows a temperature gradient
  • Some gases do not do condense (e.g. propane, ethane) and are just collected at the top.

Cracking in a lab

The difference between industry and on a smaller scale in a lab is minimal however necessary due to the limitation of height and temperature.

  • Put the hydrocarbon mixture in the bottom of the distillation column.
  • Using an electric mantle, as they will heat the mixture to a constant set temperature which a Bunsen burner cannot achieve easily, heat the mixture
  • The mixture that you want to obtain from the set temperature will vaporise and rise up the column.
  • At the top of the column a condenser is used to lower the temperature of the vaporised liquid down, so it can therefore condense back into a liquid. The condenser has a cold circulation of water running through it.
  • A collection beaker is used to collect the required fraction.
  • After no more of the fraction is being collected, raise the temperature and use a different beaker to collect the next fraction.

Organic Chemical Tests

Organic Chemical Tests

by My 0 Comments

 

Test Method Positive Test Results Extra
Primary, Secondary, and tertiary Alcohols 1) Add potassium dichromate solution · Primary and secondary alcohols can be oxidised by the Cr2O72-

· Dichromate ions (Cr2O2-) are orange. On reduction Cr 3+ ions are formed which are green.

· Primary + secondary remain green

· Tertiary changes to orange

· Primary alcohols are oxidised to aldehydes and the carboxylic acids

· Secondary alcohols are oxidised to ketones.

· Tertiary alcohols cannot be oxidised by the dichromate ions.

Primary and Secondary alcohols Using Tollens’ reagent

1) diamminesilver(I) ion, [Ag(NH3)2]+ (containing silver(i) ions) is added

2) drop of NaOH gives a precipitate of silver(i) oxide ions

3) Add dilute NH3 to redissolve precipitate

4) Few drops of aldehyde (primary) or ketone (secondary) is added

5) Solution warmed gently in hot water bath for several minutes

· Aldehyde (primary alcohol): Colourless solution producing silver mirror

· Ketone (secondary alcohol): No change in the colourless solution

· Large enough of aldehyde (from oxidation of primary alcohol) or ketone (from secondary alcohol) to be able to test them.
Carbocyclic Acids 1) Add Sodium hydrogen carbonate

2) Add limewater

· Carbon Dioxide gas produced

· Limewater turns cloudy

Halogenoalkanes 1) Add Aqueous NaOH

2) Warm the solution

3) Acidify with nitric acid

4) Add aqueous silver nitrate

· Precipitate of silver halides formed
Alkenes 1) Mix with bromine water · Solution changes from orange to colourless
Aldehyde and Ketones Using Tollens’ reagent

1) diamminesilver(I) ion, [Ag(NH3)2]+ (containing silver(i) ions) is added

2) drop of NaOH gives a precipitate of silver(i) oxide ions

3) Add dilute NH3 to redissolve precipitate

4) Few drops of aldehyde or ketone is added

5) Solution warmed gently in hot water bath for several minutes

Fehling’s solution or Benedict’s solution

· Both contain complexed copper(II) ions in an alkaline solution.

· Fehling’s solution contains copper(II) ions complexed with tartrate ions in sodium hydroxide solution

· Benedict’s solution contains copper(II) ions complexed with citrate ions in sodium carbonate solution.

1) A few drops of the aldehyde or ketone are added to the reagent

2) Mixture is warmed gently in a hot water bath for a few minutes.

· Aldehyde: Colourless solution producing silver mirror

· Ketone: No change in the colourless solution

 

 

 

 

 

 

 

 

 

· Aldehyde: Blue solution -> Dark red ppt (Cu2O ppt)

· Ketone: No change (blue solution)

Metal Halides 1) Add dilute nitric acid

2) Add a few drops of silver nitrate

· Fluoride = no precipitate

· Chloride = White precipitate (able to dissolve in dil. NH3)

· Bromide = Cream precipitate (able to dissolve in conc. NH3)

· Iodide = Yellow precipitate (insoluble)

Acid, Bases and Buffers

Acid, Bases and Buffers

  • Bronsted-Lowry acids are proton donors
  • Bronsted-Lowry bases are proton acceptors

Strong and Weak Acids

  • Strong acids dissociate (or ionise) almost completely in water as nearly all of the H+ ions are released

e.g. HCl(g) + H2O(l) à H+(aq) + Cl(aq)

  • Strong bases dissociate (or ionise) almost completely in water too

e.g. NaOH(s) + H2O(l) à Na+(aq) + OH(aq)

  • In both Strong Bases and Acids, the equilibrium lies over to the right
  • Weak acids dissociate minimally in water, as only a few of the H+ ions are released

CH3COOH(aq) ⇋ CH3COO(aq) + H+(aq)

  • Weak bases dissociate minimally in water:

NH3(aq) + H2O(l) ⇋ NH4++ OH(aq)

  • In both weak Bases and Acids, the equilibrium lies over to the left

Buffer Solution

  • Buffers are solutions which can resist changes in acidity or alkalinity
  • When a small volume of acid/alkali is added to them their pH remains at a constant
  • They are based on an equilibrium reaction which will move in the direction to remove either additional hydrogen ions or hydroxide ions

Acidic Buffers

  • Acidic buffers are made from weak acids
  • The dissociation of a weak acid is an equilibrium reaction

HA(aq) ⇋ H+(aq) + A(aq)

Acid Buffer: adding alkali

  • If a little alkali is add, the OH ions react with the HA to produce water molecules and A:

HA(aq) + OH(aq) ⇋ H2O(aq) + A(aq)

  • This removes the added OHso the pH trends to remain almost the same

Acid Buffers: adding acid

  • If His added, the equilibrium will shift to the left as the Hions combine with the A ions to produce undissociated HA.

Calculations of Buffers

  • Different buffers can be made which will maintain different pH’s. when weak acid dissociates:

HA(aq) ⇋ H+(aq) + A­(aq)

  • The expression of this being:

[H+(aq)][A(aq)]

[HA(aq)]

  • Using this expression, the pH of the buffer can be calculated

pH Scale

  • The acidity of a solution varies on the concentration of H+ (aq) and is measured along the pH scale

pH = -log10[H+(aq)]

Calculating Hydrogen Ion Concentration from pH

To find the hydrogen ion concentration from a given pH then the inverse of the pH formula is used: [H+] = 10-pH

Calculating Hydroxide Ion Concentration from pH

  • The Concentration of OH ions is proportional to the concentration of the base
  • However, to find the pH the [H+] must be known, therefore the ionic product of water(Kw) is used Kw = [H+][OH]
  • If the [OH] for a strong aqueous base and Kw at a certain temperature are both known then the [H+] can be determined and therefore the pH

Strong Monoprotic Acids

  • Monoprotic refers to each molecule of an acid which will release 1 proton when it is dissociated
  • Examples of monoprotic acids are Hydrochloric and Nitric acid

Strong Diprotic Acids

  • Diprotic acids are those which release 2 protons when they dissociate
  • Examples of diprotic acids are Sulfuric acid and carbonic acid

Indicators for Titrations

  • The equivalence point is where there is an equal proportion of hydrogen ion added to hydroxide ions
  • The End point is the volume of an alkali or acid which when added together the indicator changes colour
  1. Strong acids and Strong Base
    • Methyl Orange and Phenolphthalein changes colour within the equivalence point and therefore suitable
  1. Phenolphthalein Weak Acid and Strong Base
    • Phenolphthalein
  1. Strong Acid and Weak Base
    • Methyl Orange will change sharply at the equivalence point
  1. Weak Acid and Weak Bases
    • Neither indicator is suitable
    • No indicator would be viable for the equivalence point over the two pH units

Calculating pH of a weak acid using Ka: The Dissociation of Weak acids

  • Weak acids do not ionise fully in a solution and therefore the [H+] is not proportional to the acid concentration
  • If a weak acid of HA is dissociated:

HA(aq) ⇋H+(aq) + A(aq)

  • The equilibrium constant (Kc) is formed:

Kc = [H+(aq)][A(aq)]

[HA(aq)]

  • For a weak acid, this is usually given the Ka symbol known as the acid dissociation constant

Ka = [H+(aq)][A(aq)]

[HA(aq)]

  • The greater the value of Ka the further the equilibrium is to the right, as more acid which is dissociated and the stronger it is

Become a master of the Rubik’s Cube with this online tutorial where you can easily learn how to solve the cube with the simple method!

Acid-Base Titrations

Acid-Base Titrations

pH changes during acid-base titrations

  • An acid of known concentration is added from a burette to a measured volume of base solution
  • An indicator is used, upon changing colour it allows a visual representation that the mixture has been neutralised
  • Similarly, a pH meter can be used to determine the solutions pH. This must be calibrated in a known pH solution; buffer solution.

Titration Curves

  • Each graph shows a curve which does not change in a linear fashion as the base is added
  • Each graph has a horizontal section where a large volume of a base can be added without the pH changing much
  • Each graph has a vertical section (known as the equivalence point) where a single drop of a base can change the pH several times (however not in the weak acid and weak base)
  • The Equivalence point is the point at which sufficient base has been added to just neutralise the acid (or acid to neutralise the base)
  • In all of the above graphs the equivalence point is found at the 25cm3 mark, however note that the pH is not neutral at 7
  • If the acid is added to the base then the graph is flipped

Calculating concentrations of acid/base from the equivalence points

  • To determine the concentration, the volume of the solutions involved must be known as well as the concentration of the opposite solution (i.e. if concentration for an acid is unknown, then the concentration for the base must be known)
  • Monoprotic Acids

Number of moles (acid) = Concentration x volume (cm3)

1000

  • Divide by 1000 to convert from cm3 to dm

e.g. If 25cm3 of 0.5 mol dm3 HCl was required to neutralise 35cm3 of NaOH, calculate the concentration of the NaOH

Number of moles (HCl) = (0.5 mol dm3)(25cm3) = 0.5 x 25 = 0.0125 moles

1000 1000

  • Therefore for 0.125 moles of HCl is required to neutralise 0.125 moles of NaOH
    • As the balanaced equation shows a 1:1 ratio

1HCl + 1NaOH à 1NaCl + 1H2O

Number of moles (NaOH) = 0.0125 x 1000 = 0.036 moles

35 cm3

  • Diprotic Acids
  • As Diprotic Acids release 2 protons when in a solution they have 2 equivalence points
  • An example of this is ethanedioic acid which when it reacts with a base, e.g. Sodium Hydroxide, it is neutralised. However, this process occurs over two stages as the two protons are removed from the acid separately

1st stage) Equivalence point = pH 2.7

HOOC-COOH + OH à HOOC-COO- + H2O

The first proton is lost to the base

2nd stage) Equivalence point = pH 8.4

HOOC-COO + OH à OOC-COO + H2O

The second proton is lost to the base

  • To calculate the concentration of a diprotic acid the moles of the base must be half of the moles of the acid

e.g. 25cm3 of ethanedioic acid is neutralised by 20cm3 of 0.1 mol dm-3 NaOH solution. Calculate the concentration of the ethanedioic acid solution

Number of moles (NaOH) = 0.1 mol dm3 x 20 cm3 = 0.002 moles

1000

  • As the solution is diprotic, the number of moles is divide by 2, therefore there is (0.002 / 2 = 0.001) 0.001 moles of ethanedioic acid#

Number of moles (C2H2O4) = 0.001 moles x 1000 = 0.04 moles dm-3

25 cm3

pH Scale

pH Scale

  • The acidity of a solution varies on the concentration of H+ (aq) and is measured along the pH scale

pH = -log10[H+(aq)]

  • On a pH scale:
    • The smaller the pH, the greater the concentration of H+(aq)
    • A difference of one pH increments means there is a tenfold difference in H+

i.e. pH 3 has 10x the amount of Hcompared to pH 4

Using Kw to determine pH of a base

  • At a known temperature,
  • At 298K, the Kw =[H+(aq)] [OH(aq)] = 1.0 x 10-14 mol2 dm-6

[H+(aq)] = [OH(aq)] = 1.0 x 10-7 mol dm-3 

pH = -log10[H+(aq)] = -log10 [1.0 x 10-7] = 7.00

pH = 7.00

Calculating pH from Hydrogen ion concentration

  • If the hydrogen ion concertation is given the pH can be calculated by using the formula: pH = -log10[H+(aq)]

e.g. A solution of Nitric acid has a hydrogen ion concentration of 0.01 mol dm-3

pH = -log10[H+(aq)] = -log10[0.01] = 1

pH= 1

Calculating Hydrogen Ion Concentration from pH

  • To find the hydrogen ion concentration from a given pH then the inverse of the pH formula is used: [H+] 10-pH

e.g. A solution of hydrochloric acid has a pH of 1.52

[H+] = 10-pH = 10-1.52 = 0.03 mol dm-3

Calculating Hydroxide Ion Concentration from pH

  • The Concentration of OH ions is proportional to the concentration of the base, for example if there is a 0.04 mol dm-3 sodium hydroxide solution there will be 0.04 mol dm-3of [OH]
  • However, to find the pH the [H+] must be known, therefore the ionic product of water(Kw) is used

Kw = [H+][OH]

  • If the [OH] for a strong aqueous base and Kw at a certain temperature are both known then the [H+] can be determined and therefore the pH

e.g. The value of Kw at 298K is 1.0 x 10-14 mol2 dm-6. Determine the pH of 0.1 mol dm-3 NaOH at 298K

[OH] = 0.1 mol dm-3 à [H+] = Kw

[OH]

= 1.0 x 10-14 = 1.0 x 10-13 mol dm-3

0.1

pH = -log10 = 1.0 x 10-13 = 13.0

pH and Temperature

  • The pH of water can vary depending on its temperature

H2O(l) ⇋ H+(aq) + OH(aq)  ΔH = + 57.3kJ mol-1

  • As the reaction is endothermic, by increasing the temperature the reaction will favour the right side, thus producing more H+ and OH
  • As there are more H+ ions, the pH will increase to 6
  • However, this does not mean the solution is more acidic, as although there are more H+ ions, there is in equal proportions OH ions which cancel the H+ ions out

Temperature (K)

Kw (mol2 dm‑6)

pH

273

0.114 x 10-14

7.47

283

0.293 x 10-14

7.27

293

0.681 x 10-14

7.08

298

1.008 x 10-14

7.00

303

1.471 x 10-14

6.92

313

2.916 x 10-14

6.77

323

5.476 x 10-14

6.63

373

51.30 x 10-14

6.14

Strong Monoprotic Acids

  • Monoprotic refers to each molecule of an acid which will release 1 proton when it is dissociated
  • Examples of monoprotic acids are Hydrochloric and Nitric acid
  • Both HCl and HNO3 will produce 1 mole of hydrogen ions, therefore the concentration of H+ is equal to the acids concentration
  • g. 0.1 mol dm-3 of HCl is [H+] = 0.1 moldm-3

pH = log10 [0.1] = 1.0

Strong Diprotic Acids

  • Diprotic acids are those which release 2 protons when they dissociate
  • Examples of diprotic acids are Sulfuric acid and carbonic acid
  • Diprotic acids produce two moles of hydrogen ions for each mole of acid, therefore there is twice the concentration of Hthan the acid itself
  • g. 0.1 mol dm-3 of H2SO4 is [H+] = 0.2 mol dm-3

pH = log10 [0.2] = 0.70

Indicators for Titrations

Indicators for Titrations

  • The equivalence point is where there is an equal proportion of hydrogen ion added to hydroxide ions
  • The End point is the volume of an alkali or acid which when added together the indicator changes colour

  • Suitable Indicators for particular Titrations require:
    • Colour change must be sharp rather than gradual.
    • The end point of the titration given by the indicator must be the same as the equivalence point, otherwise the titration will yield the incorrect answer
    • The different colours must be distinct (e.g. yellow à Orange is minor change, however colourless to pink is a major change)
  • The indicator must change in the equivalence point (the vertical sections)
  1. Strong acids and Strong Base
    • Methyl Orange changes colour within the equivalence point and therefore is suitable
    • Phenolphthalein changes colour within the equivalence point is also suitable
    • Phenolphthalein is preferred as the colour change is more obvious
  2. Weak Acid and Strong Base
    • Methyl Orange is not suitable as it does not change within the equivalence point of the curve
    • Phenolphthalein changes sharply at 25cm3, well within the equivalence zone, and is therefore suitable
  1. Strong Acid and Weak Base
    • Methyl Orange will change sharply at the equivalence point
    • Phenolphthalein would not change within the equivalence point
  2. Weak Acid and Weak Bases
    • Neither indicator is suitable
    • No indicator would be viable for the equivalence point over the two pH units

The Half-Neutralisation Point

  • The half neutralisation point is the distance half way between zero and the equivalence point
  • At this horizontal stage, additional acid has little effect on the pH
  • Therefore, by adding acid/base up until this point with the confidence that there will be insignificant change lays the basis of buffers
  • The half neutralisation point also allows for the pKa to be determined in weak acids

HA + OH à H2O + A

  • At the Half Neutralisation Point:

[HA] = [A]

Ka = [H+][A]

[HA] 

Ka = [H+]

-log10Ka = -log10[H+]

pKa = pH

Defining an Acids

Defining an Acids

An acid is a substance that can donate a proton (H+ ion) and a base is a substance that can accept a proton

Proton Transfer (Reactions of Acids and Bases)

  • Proton transfer is the movement of a proton from an acid to a base
  • Bronsted-Lowry acids are proton donors
  • Bronsted-Lowry bases are proton acceptors
  • g.

Hydrogen Chloride + Ammonia à Ammonium Chloride

HCl(g) + NH3(g) à NH4Cl(s)

  • The HCl is acting as the acid as it is donating a proton to the NH3
  • The NH3is acting as the base as it is accepting a proton from the HCl

Strong and Weak Acids

  • Strong acids dissociate (or ionise) almost completely in water as nearly all of the H+ ions are released

e.g. HCl(g) + H2O(l) à H+(aq) + Cl(aq)

  • Strong bases dissociate (or ionise) almost completely in water too

e.g. NaOH(s) + H2O(l) à Na+(aq) + OH(aq)

  • In both Strong Bases and Acids, the equilibrium lies over to the right
  • Weak acids dissociate minimally in water, as only a few of the H+ ions are released

CH3COOH(aq) ⇋ CH3COO(aq) + H+(aq)

  • Weak bases dissociate minimally in water

NH3(aq) + H2O(l) ⇋ NH4++ OH(aq)

  • In both weak Bases and Acids, the equilibrium lies over to the left

n.b. Strong acids and concentrated acids are different, likewise weak and dilute acids are different. Strong and weak acids determine how much an acid becomes ionises whereas the concentration, whether that be concentrated or dilute, is the number of moles of an acid.

Water as an Acid and a Base

  • HCl is able to donate a proton to a water molecule, therefore the water acts as the base

HCl + H2O à H3O+ + Cl

  • Oxonium Ion are ions with oxygen cation with three bonds in this case it is the H3O+ (This can also be called hydroxonium or hydronium ion)
  • Water can also act as an acid

H2O + NH3 à OH + NH4+

  • Water is donating a proton to the ammonia

The proton in an Aqueous Solution

  • H+ ions have a diameter of 1015m compared to a hydrogen atom which is 1010m, the small size of the H+ ion results in the ion having unusual properties
  • H+ ions are never found isolated
  • In an Aqueous solution, it is always bonded to at least one water molecule to form a H3O+ ion
  • The H+ ion has no electrons of its own and therefore can only form bonds with other species which have a lone pair of electrons

The Ionisation of Water

  • Water is slightly ionised:

H2O(l) ⇋ H+(aq) + OH(aq)

or

H2O(l) + H2O(l) ⇋ H3O+(aq) + OH(aq)

  • The equilibrium is established by in water and all aqueous solution:

H2O(l) + H2O(l) ⇋ H+(aq) + OH (aq)

  • The equilibrium expression:

Kc = [H+(aq)][OH(aq)]

[H2O(l)]

  • [H2O(l)] is a constant and is incorporated into a modified equilibrium constant:

Kw = Kc x [H2O(l)]

Kw = [H+(aq)]eqm[OH(aq)] eqm

  • This new constant is called the ionic product of water (Kw)
  • The units for Kw is always mol2 dm-6
  • Kw always has the same value for an aqueous solution at a given temperature e.g. at 298K the Kw is always 1.00×10-14 mol2 dm-6
  • In pure water there is awlys one H+ ion for each OH Therefore [H+] = [OH ] so the Kw of pure water is Kw = [H+]2
Calculating pH of a weak acid using K

Calculating pH of a weak acid using K

  • Weak acids do not ionise fully in a solution and therefore the [H+] is not proportional to the acid concentration
  • Ethanoic acid is an example of a weak acid, in 1 mol dm-3 solution only about 4 in every thousand ethanoic acid molecules are dissociated into ions and therefore the degree of dissociation is 4/1000
  •  

    CH3COOH(aq) ⇋ H+(aq) + CH3COO(aq)

     

    Ethanoic acid

    Hydrogen Ions

    Ethanoate Ions

    Before dissociation

    1000

    0

    0

    After dissociation

    996

    4

    4

    The Dissociation of Weak acids

    • If a weak acid of HA is dissociated:

    HA(aq) ⇋H+(aq) + A(aq)

    • The equilibrium constant (Kc) is formed:

    Kc = [H+(aq)][A(aq)]

    [HA(aq)]

    • For a weak acid, this is usually given the Ka symbol known as the acid dissociation constant

    Ka = [H+(aq)][A(aq)]

    [HA(aq)]

    • The greater the value of Ka the further the equilibrium is to the right, as more acid which is dissociated and the stronger it is
    • Acid dissociation constants:

    Acid

    Ka (mol dm-3

    Hydrocyanic

    4.9 x 10-10

    Benzoic

    6.3 x10-5

    Ethanoic

    1.7 x10-5

    Chloroacetic

    1.3 x10-5

    Units for Ka

    • The units for Ka are determined by the amount of molecules in the reaction

    e.g. Ka = mol dm-3 x mol dm-3 = mol dm-3

    mol dm-3