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• 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:

NH_3(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 OH^–so the pH trends to remain almost the same

Acid Buffers: adding acid

• If H⁺ is added, the equilibrium will shift to the left as the H⁺ ions 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 K_w 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 K_a: 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 (K_c) is formed:

Kc = [H⁺_(aq)][A^–_(aq)]

[HA_(aq)]

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

K_a = [H⁺_(aq)][A^–_(aq)]

[HA_(aq)]

• The greater the value of K_a 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!

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 dm³ 

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

1^st stage) Equivalence point = pH 2.7

HOOC-COOH + OH^– à HOOC-COO- + H₂O

The first proton is lost to the base

2^nd 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

• 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 H⁺ compared to pH 4

Using K_w to determine pH of a base

• At a known temperature,
• At 298K, the K_w =[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(K_w) is used

K_w = [H⁺][OH^–]

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

e.g. The value of K_w 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

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 H⁺ than the acid itself
• g. 0.1 mol dm^-3 of H₂SO₄ is [H⁺] = 0.2 mol dm^-3

pH = log₁₀ [0.2] = 0.70

• 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 pK_a to be determined in weak acids

HA + OH^– à H₂O + A^–

• At the Half Neutralisation Point:

[HA] = [A^–]

K_a = [H⁺][A^–]

[HA] 

K_a = [H⁺]

-log₁₀K_a = -log₁₀[H⁺]

pK_a = pH

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) + NH_3(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

NH_3(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:

K_c = [H⁺_(aq)][OH^–_(aq)]

[H₂O_(l)]

• [H₂O_(l)] is a constant and is incorporated into a modified equilibrium constant:

K_w = K_c x [H₂O_(l)]

K_w = [H⁺_(aq)]_eqm[OH^–_(aq)] _eqm

• This new constant is called the ionic product of water (K_w)
• The units for K_w is always mol² dm^-6
• K_w always has the same value for an aqueous solution at a given temperature e.g. at 298K the K_w 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 K_w of pure water is K_w = [H⁺]²

The Dissociation of Weak acids

• If a weak acid of HA is dissociated:

HA_(aq) ⇋H⁺_(aq) + A^–_(aq)

• The equilibrium constant (K_c) is formed:

Kc = [H⁺_(aq)][A^–_(aq)]

[HA_(aq)]

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

K_a = [H⁺_(aq)][A^–_(aq)]

[HA_(aq)]

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

Units for K_a

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

e.g. K_a = mol dm^-3 x mol dm^-3 = mol dm^-3

mol dm^-3

• 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

How Buffers work

• Buffers are design to maintain the concentration of hydrogen ions and hydroxide ions
• 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
• e.g.

HA_(aq) ⇋ H⁺_(aq) + A^–_(aq)

• Therefore, [H⁺_(aq)] = [A^–_(aq)]
• As it is a weak acid, [H⁺_(aq)] = [A^–_(aq)] are both very small as they are most undissociated

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 OH^–so the pH trends to remain almost the same

Acid Buffers: adding acid

• If H⁺ is added, the equilibrium will shift to the left as the H⁺ ions combine with the A^– ions to produce undissociated HA.
• As the [A^–] is small the supply of A^– soon runs out and there is no subsequent A^– left to remove the added H⁺.
• Therefore, the buffer has stopped
• However, by the addition of extra A^– by adding a soluble salt of HA, which fully ionises
• This increased amount of A^– allows for more H⁺ to be used up
• An Acidic Buffer is made from a mixture of a weak acid and a soluble salt of that acid. It will thus maintain a pH of below 7 (acidic)
• The function of the weak acid allows there to be a source of HA which can remove any OH^–

HA_(aq) + OH^–_(aq) à H₂O_(aq) + A^–_(aq)

• The function of the salt allows there to be a source of A^– ions which can be removed by any added H⁺ ions:

H⁺_(aq) + A^–_(aq) à HA_(aq)

• Buffers do note ensure that there is no change in pH but rather that the addition of acid/alkali will only slightly change or the effects are reduced with the use of a buffer.
• If it possible to saturate a buffer by there being too much acid/alkali that all of the available HA or A^– is used up
• Another method of producing a buffer is the mixture of a weak acid with its salt with the addition of an alkali such as sodium hydroxide.
• By neutralising half of the acid the result is a buffer whose pH is equal to the pK_a of the acid

At Half Neutralisation: pH = pK_a

• These buffers are useful as they are equally efficient at resisting a change in pH whether it be an acid of an alkali

Basic Buffer

• Basic buffers are those that resist change but maintain a pH at above 7
• They are a mixture of weak base and a salt of that base
• A mixture of aqueous ammonia and ammonium chloride (NH₄⁺ Cl^–) act as a basic buffer
  • Aqueous ammonia removes added H⁺:

NH_3(aq) + H⁺_(aq) à NH₄⁺_(aq)

• The ammonium ion, NH₄⁺ removes added OH^–:

NH₄⁺_(aq)  + OH^–_(aq) à NH_3(aq) + H₂O_(l­)

Examples of Buffers

• Buffers are used in blood storage where they are maintained at approximately pH 7.4
• Blood must be buffered as a change in pH >0.5 can be fatal
• Blood is buffered through several mechanisms, however most importantly:

H⁺_(aq) + HCO₃^–_(aq) ⇋ CO_2(aq) + H₂O_(l­)­

• The addition of extra H⁺ ions moves this equilibrium to the right, therefore the removing the additional H⁺
• Adding exta OH^– ions removes the H⁺ by reacting to form water.
• The equilibrium moves to the left releasing more H⁺ions
• Buffers are also found in detergents and shampoos as if they become too acidic or to alkaline they can damage fabrics, skin or hair

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

e.g. A buffer consist of 0.1 mol dm^-3 ethanoic acid and 0.1 mol dm^-3 sodium ethanoate

(K_a for ethanoic acid = 1.7 x10^-5, pK_a = 4.77)

What is the pH of the buffer?

• Calculate the [H⁺_(aq)] of the equation

K_a = [H⁺_(aq)][A^–_(aq)]

[HA_(aq)]

• Sodium ethanoate is fully dissociated so [A^–_(aq)] = 0.100 mol dm^-3
• Ethanoic acid is almost undissociateed so [HA_(aq)] = 0.100 mol dm^-3

1.7 x 10^-5  = [H⁺_(aq)] x 0.100

0.100

1.7 x 10^-5  = [H⁺_(aq)] and pH -log₁₀ [H⁺_(aq)]

pH = 4.77

• When equal concentrations of acid and salts are used the pH of the buffer = pK_a of acid used (equal the half neutralisation point)
• Changing the concentration of HA to A^– will affect the pH of the buffer
• By using 0.2 mol dm^-3 ethanoic acid and 0.1 mol dm^-3 sodium ethanoate, the pH will be 4.50

pH change when an acid is added to a buffer

Example:

• In the above calculation, the pH was 4.77 after adding 1.00 mol dm^-3 of buffer solution of ethanoic acid at concentration 0.1 mol dm^-3 and sodium ethanoate at concentration 0.1 mol dm^-3, k_a= 1.7 x 10 ^-5.
• If 10cm³of 1 mol dm^-3 HCl were added to the buffer all of the H⁺ ions will react with the ethanoate ions [A^–] to form a molecule of ethanoic acid [HA]
  • Number of moles of ethanoic acid = 0.1
  • Number of moles of sodium ethanoate = 0.1

Moles of HCl = M x V /1000

=0.010

• After adding the acid:
  • Number of moles of ethanoic acid [HA]= 0.110 mol dm^-3(increased by 0.010)
  • Number of moles of sodium ethanoate [A^–] = 0.090 mol dm^-3 (decreased by 0.010)
  • The concentration of [HA] is now 0.110 mol dm^-3
  • The concentration of [A^–] is now 0.090 mol dm^-3
• The above calculations do not factor in the volume of HCl added

K_a = [H⁺_(aq)][A^–_(aq)]

[HA_(aq)]

1.7 x 10^-5 = [H⁺_(aq)] 0.090

0.110

[H⁺_(aq)] = 1.7 x10^-5 x 0.090 = 2.08 x 10 ^-5

0.110

pH = 4.68

• The pH change is small from 4.77 à68

pH change when a base is added to a buffer

• if 10cm³ of 0.1 mol dm^-3 NaOH is added to the original buffer it will react with the H⁺ and more HA will ionise, thereby decreasing the concentration of the acid [HA] by 0.010 and increasing the concentration of ethanoate ions by 0.010