5.1, Kp, Ka, Kw ect.

Question 1

What is the definition of an equilibrium law?

Answer 1

The equilibrium law states that for the equilibrium aA + bB ⇔ C + dD

K_c = [C]^c[D]^d/[A]^a[B]^b

Question 2

What is the definition of a homogeneous equilibrium?

Answer 2

A homogeneous equilibrium is an equilibrium in which all the species making up the reactants and products are in the same physical state.

Question 3

What is the definition of a heterogeneous equilibrium?

Answer 3

is an equilibrium in which species making up the reactants and products are in different physical states.

Question 4

How are the three meanings of K presented differently?

Answer 4

The equilibrium constant is represented by a capital K, followed by a subscript letter to indicate which equilibrium constant you are referring to a lower case k represents the rate constant and K is on its own is the symbol for potassium.

Question 5

What would Kc, and it’s units be for the reaction 2SO_{2 (g)} + O_{2 (g)} ⇔ 2SO_3(g) ?

Answer 5

1. Kc = [SO_2(g)]²/ [SO_2(g)]² [O_2(g)]
2. Units (moldm^-3)²/ (moldm^-3)³ = dm³mol^-1

Question 6

What needs to be known to determine concentrations at equilibrium?

Answer 6

To determine a value of Kc, the concentration of reactants and products need to be known.

Chemists determine the concentrations in a number of ways including a titration and using a colorimeter.

Question 7

What are the drawbacks of using titrations to determine concentrations at equilibrium?

Answer 7

• Titrating a reactant or product against a known concentration will show how much of the reactant or product os present.
• However the reactant or product has to be removed from the reaction mixture, and this will alter the position of equilibrium.
• Other substances may also be present, affecting the results (catalysts).

Question 8

What can monitor the concentrations of product molecules throughout reactions to determine the value of Kc?

Answer 8

A colorimeter can be used to monitor the concentrations of a reactant or product throughout.

This works because coloured substances alter the amount of light that can pass through.

Question 9

Calculating unknown equilibrium concentrations;

What can you tell from a balanced equation?

Answer 9

• the reacting quantities needed to prepare a required quantity of a product
• the quantities of products formed by reacting other known quantities of reactants.

Question 10

What are the steps involved in calculating Kc from unknown equilibrium concentrations?

Answer 10

1. Find the concentrations of all three components in the equilibrium mixture
2. Work out the equilibrium amounts of the reacting molecules.
3. You can use the balanced equation and stiochiochemistry to find out the moles of a reactant used a product formed.
4. Finally, convert the amounts to concentration.

Follow these steps on the example on page 25.

Question 11

What are the steps of calculating K_c from equilibrium concentrations?

Answer 11

1. Calculate the moles at equilibrium by giving the moles at equilibrium by the volume of substance or container.
2. Write the expression for K_c and work out the units. Then calculate K_c under these conditions.

Example on pg 24

Question 12

If you are given equilibrium amounts in moles and you know the total volume, what must you calculate first?

Answer 12

The concentrations.

concentrations = no. moles / volume in dm³

Question 13

What is affected by an equilibrium being heterozygous, instead of homozygous?

Answer 13

When a heterogeneous equilibrium is present, molar concentrations for solids and pure liquids do not change because their volume remains constant.

This means that the expression for K_c changes:

• the concentrations of solid substances are not included in the expression
• the concentrations of pure liquids are not included in te expression

Question 14

What is the expression for K_c for the reaction P_{4 (s)}+ 5O_{2 (g)} ⇔ P₄O_{10 (s)}?

Answer 14

K_c = [O₂]⁵

Question 15

What is a mole fraction?

Answer 15

The mole fraction of a substance is a measure of how much of a given substance is present in a reaction mixture.

Question 16

What is the partial pressure?

Answer 16

The partial pressure of a substance is the pressure an individual gaseous substance would exert if it occupied a whole reaction vessel on its own.

Question 17

What symbol is given to the mole fraction?

Answer 17

It is given the symbol X.

Question 18

How is the mole fraction of a substance calculated?

Answer 18

mole fraction, X_A = number of moles of substance A / total number of moles of all substances

Question 19

What is the pressure of any reaction mixture involving gases within a sealed vessel the sum of?

Answer 19

The pressure of any reaction mixture involving gases within a sealed vessel is effectively the sum of being exerted by each of the gaseous substances involved.

The amount of pressure being exerted by an individual species within a reaction vessel is called the partial pressure.

Question 20

What symbol is the partial pressure given?

Answer 20

partial pressure is given the symbol, P, followed by a subscript denoting the species to which is refers.

Question 21

How do you calculate partial pressure from the total pressure?

Answer 21

If we know the total pressure of a reaction along with the mole fraction of a given substance, we can calculate the partial pressure of that substance.

For substance A: partial pressure, P_A = mole fraction x total pressure.

Question 22

What equilibrium constant is used for reactions involving gases?

Answer 22

Equilibrium expressions can be written using partial pressures instead of concentrations.

Question 23

What would the equilibrium constant, K_p, be for the reaction aA + bB ⇔ cC + dD?

Answer 23

kp = (pC)c (PD)d / (PA)a (PB)b

Question 25

How does the expression for k_p change in a heterogenous reaction?

Answer 25

• Solids will not be included in the expression
• pure liquids will not be included in the expression

Question 26

What does the magnitude of the eqilibrium constants k_p and k_c indicate?

Answer 26

The extent of a chemical reaction.

A equilibrium constant, k, with a value of 1 would indicate of 1 would indicate that the position of equilibrium is halfway between reactants and products.

Question 27

What side of the reaction does equilibrium favour when K is less than 1?

Answer 27

• the reaction is reactant favoured
• the reactants on the left-hand side predominate at equiibrium.

Question 28

What side of the reaction does equilibrium favour when k_p is less than one?

Answer 28

When k_p is less than 1:

• the reaction is reactant-favoured
• the reactants on the left hand side predominate at equilibrium

Question 29

How do changes in temperature affect the enthalpy value?

Answer 29

• An increase in temperature shifts the position of equilibrium in the endothermic direction (ΔH + ve)
• A decrease in temperature shifts the position of equilibrium in the exothermic direction (ΔH - ve)

Question 30

How do the ΔH values for the forward and reverse reactions in equilibrium compare?

Answer 30

They have the same magnitude but have opposite signs.

Question 31

What can cause a shift in the position of equilibrium?

Answer 31

Shifts in the position of equilibrium are actually controlled by the rate constant, which changes its value only with changes in temperature. When any other changes occur to conditions, K will remain constant. The way that K changes is linked to ΔH​.

Question 32

How would equilibrium be affected by the reaction being exothermic?

Answer 32

• the equilibrium yield of the products, on the right-hand side, increases.
• the equilibrium yield of the reactants, on the left-hand side, decreases.

Question 33

How would equilibrium be affected by the reaction being endothermic?

Answer 33

• the equilibrium yield of product, on the right-hand side, decreases
• the equilibrium yield of the reactants, on the left-hand side, increases

Question 34

How do changes in concentration and pressure affect K?

Answer 34

Although the eqilibrium constant, k, is altered by changes in temperature, the value of K i s unaffected by changes in concentration and pressure.

Question 35

For the reaction N₂O₄ (aq) ⇔ 2NO_(g) Kc = 12.8 mol dm^-3, If the concentration of [N₂O_4(g)] is doubled from 2.00 mole dm^-3 to 0.400 mol dm^-3 how would Kc be affected?

Answer 35

The equilibrium position must shift to restore this ratio to the Kc value of 12.8 mol dm^-3.

The system must:

• increase NO_{2 (g)}
• decrease N₂O_4(g)

This causes a shift in the equilibrium poisition from left to right.

Will also be the same for Kp as it stays constant when concentrations, or the equivalent partcial pressures, are changed.

Question 37

What happens to the concentrations of a gaseous reactants and products if the pressure is doubled?

Answer 37

If the pressure is doubled, the concentrations of both the reactants and products (g) will double because there will be the same number of partciales in half the space.

Question 38

If the pressure of a reaction at equilibrium is doubled how will the equilibrium shift to restore the value of Kc for the reaction: N2O4 (aq) ⇔ 2NO(g) ?

Answer 38

• Decrease NO_2(g) on the top
• Increase N₂O_4(g) on the bottom

This causes a shift in the equilibrium position from right to left.

the same will be true for the rate constant K_P.

Question 39

How does the presence of a catayst affect K?

Answer 39

It doesnt.

Catalysts affest the rate of a chemical recation, but not the position of equilibrium.

Cataylsts speed up both the forward and reverse reactions in the equilibrium by the same factor. Equilibrium is reached more quickly, but the equilibrium position, and hence the value of the equilibrium constant, is unchanged by the action of a catayst.

Question 40

What is a Brønsted-Lowry acid?

Answer 40

What is a Brønsted-Lowry acid is a proton, H^{<span>+</span>}, donor.

Question 41

What is a Brønsted-Lowry base?

Answer 41

A Brønsted-Lowry base is any substance that can accept a proton.

Question 42

Why are acids and bases refered to acid Brønsted-Lowry acids or bases?

Answer 42

Because they behave in the way described by the Brønsted-Lowry model.

Question 43

Chemists have studied the behavior of acids and bases over many years, what discovery was made in the 1770’s?

Answer 43

In the 1770’s many chemist were investigating air and the gasses within it. Joseph Priestly and karl Scheele both reported findings that suggested the existence of the gas we now know as oxygen. Another chemist, Antoine-Laurent de Lavoisier, found, among othe rthings, that this gas was important in the rusting of metal and in 1778 he proposed that it was the source of acidity. Lavoisier as wrong, but it was an early step towards the understanding of acids.

Question 44

Chemists have studied the behavior of acids and bases over many years, what discovery was made in the 1815’s?

Answer 44

Humphrey Davy showed that some acidic substances, such as HCl, did not actually contain xoygen.

Question 45

Chemists have studied the behaviour of acids and bases over many years, what discovery was made in the 1830’s?

Answer 45

In 1832 Justus Liebig defined an acid as a substance containing hydrogen that could be replaced by a metal - very close to today’s definition.

Question 46

Chemists have studied the behaviour of acids and bases over many years, what discovery was made in the 1880’s?

Answer 46

In the late 1880’s Svante Arrhenius proposed that acids dissociated in water to form hydrogen ions, H⁺ and that bases dissociated in water to form hydroxide ions, OH-. This is true for many acids and bases, and they can be referred to as Arrhenius acids and bases. However, the model breaks down when acids and bases are not solutions in water or when bases are not soluble hydroxides.

Question 47

When was the Brønstead-Lowry definition proposed, by who?

Answer 47

In 1929 the Brønsted-Lowry definition was proposed by the Danish chemist Johannes Brønstead and the British chemist Thomas Lowry.

Question 48

What theory has been suggested as a step on from the Brønstead acid theory?

Answer 48

Even the Brønstead-Lowry acid can break down, such as for some solvents. The American chemist Gilbert Lewis proposed a more general theory that describes an acid (known as a Lewis acid) as an electron-pair donor. (the model isn’t needed for the exam)

Question 49

Brønstead Lowry acids;

What happens when an HCl is added to water?

Answer 49

HCl_(g) → H⁺ + Cl^-_(aq)

The HCl has donated an H⁺ ion because it is a Brønstead-lowry acid.

Question 50

How does ammonia dissociate?

Answer 50

NH_3(aq) + H⁺_(aq) → NH₄⁺_(aq)

The ammonia has accepted a proton (H⁺) - it is a Brønsted-Lowry base.

Question 51

Write the ionic equation for HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(l)?

Answer 51

H⁺_(aq) + OH^-_(aq) → H₂O_(l)

Question 52

Why can different acids release different numbers of protons depending on their formulae? Give examples.

Answer 52

• HCl is a monobasic acid because each molecule can release one proton;
  1. HCl _(aq) → H⁺_(aq) + Cl^-
• H₂SO₄ is a dibasic acid because each molecule can release two protons, this is done in two stages.

1. H₂SO_4(aq) → H⁺_(aq) + HSO₄^- _(aq)
2. HSO₄^-_(aq) → H⁺_(aq) + SO₄^2-_(aq)

• H₃PO₄ is a tribasic acid because each molecule can release three protons, and is done in three stages.

1. H₃PO_{4 (aq)} → H⁺_(aq) + H₂PO₄^-_(aq)
2. H₂PO₄^- _(aq) → H⁺_(aq) + HPO₄^2-_(aq)
3. HPO₄^2-_(aq) → H⁺_(aq) + PO₄^3-_(aq)

Question 53

What are conjugate acid-base pairs?

Answer 53

An acid-base pair is a set of two species that transform into each other by gain or loss of a proton.

Question 54

Give an example of a conjugate acid-base pair, for the dissociation of nitrous acid, HNO₂.

Answer 54

Question 55

Conjugate acid-base pairs;

What does acid base equilibria involve?

Answer 55

Acid base equilibria involves two acid-base pairs

Question 56

State the conjugate acid-base pairs for the equilibrium of the dissociation of notrous acid, HNO₂ in water:

HNO_2(aq) + H₂O_(l) ⇔ H₃O⁺_(aq) + NO₂^-_(aq)

Answer 56

HNO₂ and NO₂^- differ by H⁺ and make up one acid-base pair: acid 1 and base 1.

H₃O⁺ and H₂O differ by H⁺ and make up a second acid-base pair: acid 2 and base 2.

Question 57

What is the definition for an alkali?

Answer 57

An alkali is a base that dissloves in water forming OH^-_(aq) ions.

Question 58

What is the definition for neutralisation?

Answer 58

Neutralisation is a chemical reaction in which an acid and a base react togther to produce a salt and a water.

Question 59

What is the definition for a strong acid?

Answer 59

A strong acid is an acid that completely dissociates in solution.

Question 60

What is the definiton for a weak acid?

Answer 60

A weak acid is an acid that partially dissociates in solution.

Question 61

What is the definition for the acid dissociation constant?

Answer 61

K_a, of an acid HA is defined as K_a = [H⁺_(aq)][A^-_(aq)] / [HA_(aq)]

Question 62

What is the definition for pKa ?

Answer 62

pKa = - log₁₀ K_a

Question 63

What is the definition for K_a?

Answer 63

K_a = 10^-pka

Question 64

What reaction occurs when aqueous acids take part in typical acid-base reactions with carbonates, bases and alkalis.

Answer 64

In all of these reactions, a neutralisation reaction occurs and water is fomred as one of the products.

Question 65

What is the equation and ionic equation for hydrochloric acid, HCl, reacting with solid carbonates?

Answer 65

Symbol

2HCl_(aq) + CaCO_3(s) → CaCl_2(aq) + CO_2(g) + H₂O_(l)

Ionic

2H⁺ _(aq) + CaCO_3(s) → Ca²⁺ _(aq) + CO_2(g) + H₂O_(l)

Question 66

When acids release protons in water, what can they be accepted by?

Answer 66

When acids release protons in water, they are ususally accepted by the water to form a hydromium ion, H₃O⁺ (sometimes called the oxonium ion).

Question 67

Give an example of hydrchloric acid dissociating and forming an oxonium ion.

Answer 67

HCl_(aq) + H₂O_(l) → H₃O⁺_(aq) + Cl^-_(aq)

But as H⁺ and H₃O⁺ can be used interchangaebly when describing aqueous acid solutions, and H⁺ is usually used in ionic equaitions for acid-base reactons.

This is why the above equation would be simplified to: HCl_(aq) → H⁺_(aq) +Cl^-_(aq)

Question 68

What are the equation and ionic equation for hydrochloric acid, HCl, reacting with aqueous carbonates?

Answer 68

Symbol equation

2HCl_(aq) + Na₂CO_3(aq) → 2NaCl + CO_2(g) + H₂O_(l)

Ionic equation

2H⁺_(aq) + CO_3(s) → CO_2(g) + H₂O_(l)

If the carbonate is in solution, the final ionic equation simplifies beacause the carbonate is dissociated.

Question 69

What are spectator ions?

Answer 69

In ionic equations species which remain unchanged are being cancelled out, these are specator ions.

Question 70

Why is it important to include state symbols in ionic equations?

Answer 70

it is especially important when an acid reacts with a solid, as solids will not be dissociated into ions.

Question 71

What do aqueous acids form when they react with bases?

Give an symbol and ionic equation.

Answer 71

Aqueous acids react with alkalis, forming salt and water.

Full equation: 2HNO_3(aq) + MgO_(s) → Mg(NO₃)_2(aq) + H₂O_(l)

Ionic equations: 2H⁺_(aq) + MgO_(s) → Mg²⁺_(aq) + H₂O_(l)

Question 72

What do aqueous acids form when they react with alkalis?

Give a symbol equation.

Answer 72

Aqueous acids react with alkalis, forming a salt and a water.

Full equation: H₂SO_4(aq) + 2KOH_(aq) → K₂SO_{<span>4 (aq) </span>}+ 2H₂O (l)

Question 73

What do aqueous acids form when they react with a metal?

Answer 73

The reaction of an acid with a metal is a redox reaction, so does not fit in with the acid-base model. however, they are common acid reaction.

The general equation is;

acid + metal → salt + hydogen

Question 74

Give the full, all ions and ionic equation for the recation of hydrochloric acid and magnesium to show the process of cancelling out unchanged species to create a ionic equation.

Answer 74

Question 75

What may be an acception to acids reacting with metals to form a salt and hydrogen?

Answer 75

Some acids such as sulphuric an nitric acids, are powerful oxidising agents and other reactions may also take place, especially when acids are concentrated. It is best to sterr clear of clear of H₂SO₄ and HNO₃ when giving examples of acid reactions with metals.

Question 76

What are the common strong acids?

Answer 76

These acids 100% dissociate in an aqueous solution.

Question 77

Show the dissociation of ethanoic acid in water, and describe how a weak acid dissociates.

Answer 77

CH₃COOH^-_(aq) ⇔ H⁺_(aq) + CH₃COO^-_(aq)

• The equilibrium position lies well over to the left
• There are only small concentrations of dissociated ions, H⁺_(aq) undissociated CH₃COO^-_(aq) compared with the concentration of undissociated CH₃COOH_(aq)

We can actually say that CH₃COO^- is a very good base - it is very good at accepting the dissociated H⁺ back.

Question 78

What is the actual extent of acid dissociation meausred by?

Answer 78

An equilibrium constant calle dthe acid dissociation constant, Ka.

Question 79

How do the terms ‘strong’ and ‘weak’ describe the extent of dissociation?

Answer 79

• a large K_a value indiactes a large extent of dissociation - the acid is strong.
• a small Ka value indicates a small extent of dissociation - the acid is weak.

Question 80

How is pKa related to ka?

Answer 80

The range of values of Ka are vast, for this reason a logarithmic scale is used to decribe them. pKa is a more manageable number than Ka, and the two quamtities are related as follows;

• pk_a = - log₁₀ K_a
• K_a = 10^-pka
• A low value of K_a matches a high value of pK_a.
• A high value of K_a matches a matches a low value of pK_a.
• The smaller the pK_a value, the stornger the acid.

Question 81

What is the defintion for pH?

Answer 81

-log[H⁺_(aq)]

Question 82

What is the definition for [H⁺_(aq)]?

Answer 82

10^-pH

Question 83

Who came up with the pH scale?

Answer 83

Søren Sørensen, a Danish chemist.

Question 84

Describe the work of Søren Sørensen, on creating the pH scale.

Answer 84

Søren Sørensen, a Danish chemist, carried out research that involved measuring hyrogen ion concentrations. These numbers could be cumbersome to measure and record.

Aqueous solutions have concentrations of H⁺_(aq) ions in the range:

• from about 10 mol dm^-3 (10¹ mol dm^-3)
• to about 0.0000000000000001 mol dm^-3 (10^-15 mol dm^-3).

Sørensen came up with an easier way to measure and denote these concentrations - a logarithmic scale called the pH scale.

Question 85

What are are [H⁺] values and corrosponding pH values that make up the pH scale.

Answer 85

Question 86

[H⁺_(aq)] What do the [] stand for?

Answer 86

[] are shorthand for ‘the concentration of’

Question 87

Describe the relationship between pH and [H⁺_(aq)].

Answer 87

• A low pH value means a large [H⁺_(aq)].
• A high pH value means a small [H⁺_(aq)].
• A pH change of 1 changes [H⁺_(aq)] by 10 times.
• An acid with a pH of 2 contains 1000 times the [H⁺_(aq)] of an acid with a pH of 5.

Question 88

How do you calculate the pH of strong acids?

Answer 88

A strong monobasic acid, HA, has virtually complete dissociation in water:

HA_(aq) → H⁺_(aq) + A^-_(aq)

This means the [H⁺_(aq)] of a strong acid is approximately equal to the concentration of the acid, [HA_(aq)]:

[H⁺_(aq)] = [HA_(aq)]

The pH can then be calculated using pH = -log[H⁺_(aq)]

Question 89

Show the working out required to answer the question;

A sample of HCl has a concentration of 1.22 x 10^-3 mol dm^-3, what is it, pH?

Answer 89

HCl is a strong acid, and therfore completely dissociates:

HCl → H⁺ + Cl^-

Therefore [H⁺] = [HCl] = 1.22 x 10-3 mol dm-3

pH = -log[H⁺]

= -log(1.22 x 10^-3)

=2.91

Question 90

How do you work out the pH of a weak acid?

Answer 90

• Firts we find the H⁺,

K_a = [H⁺] [A^-] / [HA]

• If we assume that the H⁺ and A^- ions only come from the dissociation of [HA, then this approximates to [H⁺] / [A^-], therefore;

[H⁺]² = K_a x [HA]

[H⁺] = _{^{(the square root of)}} K_a x [HA]

pH = - log [H⁺]

Question 91

When calculating the pH of a weak acid why do you have to calculate K_a?

Answer 91

In aqueous solution, a weak monobasic acid, HA, partially dissociates, setting up the equilibrium: HA⇔ H⁺ + A^-

Because you can no longer assume [H⁺] is equal to the concentration of the acid, we need to work out the value for K_a for the reaction - to find the extent of dissociation that has occurred, as this will dictate the value of [H⁺].

Question 92

When calculating the pH of weak acids, how do you work out the value of K_a?

Answer 92

K_a = [H⁺][A^-] / [HA] is simplified to K_a = [H⁺]² / [HA]

Question 93

Why can the equaltion for K_a be simplified to K_a = [H⁺]² / [HA]?

Answer 93

• For weak acids, [H⁺] is much less than [HA] becuase the extent of dissociation is so small.
• [H⁺] and [A^-] can be considered equal as when HA molecules dissociate H⁺ and A^- ions are formed equally (any water present is expected to have disaccoiated into a negligable amount that will not affect the concentration of H⁺)
• This means the expression [H⁺] [A^-] in the k_a calculation becomes [H⁺]².
• As some HA molecule have dissociated;

[HA] _{undissociated} - [H⁺]

• Finally we make an approximation to simplifiy the pH calculation. We assume that such a small amount of the acid has dissociated that:

[HA]_{undissociated} - [H⁺] = [HA] _{undissociated}

• The K_a expression now becomes:
• K_a = [H⁺]² / [HA]

Question 94

When calcualting the pH of a weak acid how do we rearrange the equation to find [H⁺]?

Answer 94

The equation K_a = [H⁺]² / [HA] is rearranged to give: [H⁺] = -/ K_a x [HA].

Thw pH can then be calculated using pH = - log[H⁺]

Question 95

Recap; what is the calculation for H+ for strong and weak acids?

Answer 95

For strong acids, HA, [H⁺] = [HA]

For weak acids, HA, [H⁺] = -/K_a x [HA]

Question 96

When can you not use the approximation - used in the calculating the pH of a weak acid - that so little of the original acid has dissociated that the concentration of the acid at qulibrium is effectively the same as the concentration of the orginal amount of acid?

Answer 96

It is actually the case that there will be some acid which does dissociate, even if it is a small amount. If this amount is less than 5% of the total of [HA] _{undissociated} then it is deemed safe to use for the above approximation. This will b the case for very weak acids, with very low K_a values.

If, however, more than 5% of [HA] _{undissociated} dissociates, then the above approximation cannot be use. This would occur with stronger examples of weak acids, with higher K_a values.

In this instance K_a is solved by a quadratic equation, which we dont need to know.

Question 97

How do you determine k_a for a weak acid?

Answer 97

To determine K_a for a weak acid, we needs to measure the pH of a solution of the weak acid using a pH meter. We also need the concentration of the weak acid.

1. First we find the [H^+] from the pH: [H⁺] = 10^-pH
2. Now we can calculate K_a: K_a = [H⁺][A^-]
3. which approximates to ka = [H⁺]² / [HA]

Question 98

What is the definition for the ionic product of water?

Answer 98

K_w, is defined as K_w = [H⁺][OH^-]. At 25^oC, K_w = 1.00 x 10^-14mol²dm^-6

Question 99

Describe teh ionisation of water.

Answer 99

Water can act as an acid by donating a proton

H₂O → H⁺ + OH^-

Water can act as a base by accepting a proton

H₂O + H⁺ → H₃O⁺

This back and fourth transfer of protns happens continuosly in water. In fact, water exisits a equilibrium: H₂O ⇔ H⁺ + OH^-

Question 100

Why is [H₂O] considered a constant, just like K_c.

Answer 100

• The equilbrium of the dissociation of water lies well to the left and only an extremenly small amount of water is dissociated at any given time.
• Because the amount of disociation, or ionisation, of water is so small, [H₂O] is considered constant, just like K_c. We rearrange the equation as follows to combine the two constants.
• K_c = [H⁺][OH^-] / [H₂O] ⇒ Kc x [H₂O] = [H⁺][OH^-]
• This can be further simplified by reffering to the expression K_c x [H₂O] as the new constant, K_w:
• K_w = K_c x [H₂O] = Kc = [H⁺][OH^-] and is simplified to K_w = [H⁺][OH]

Question 101

What are the units for K_w?

Answer 101

The units for K_w are always mol^-2dm^-6 because [H⁺] mol dm^-3 x [OH^-3] mol dm^-3 = mol²dm^{<span>-6</span>}

Question 102

Why does K_w = 1.00 x 10^-14 at 25^oC? The pH of water is 7, and the [H⁺] = 10^-7 mol dm^-3.

Answer 102

• At 25 ^oC, the measured pH of water is 7 and [H⁺] = 10^-7 mol dm^-3.When pure water ionises, te smae number of OH^- and H⁺ ions are released:

[H⁺] = [OH^-]

• This means that the concentration of OH^- and H⁺ ions are released:

[H⁺] = [OH^-]

• This means that the concentration of OH^- is also equal to 10^-7 mol dm^-3
• so the calculation for K_w becomes :

K_w = [H⁺][OH^-] = 1.0 x 10^-7 x 1.0 x 10^-7 = 10^-14mol²dm^-6

• Therefore, at 25^oC, K_w = 1.00 x 10^-14mol²dm^-6

Question 103

Why would the ionisation of water shift?

Answer 103

The balance between [H⁺] and [OH^-] can be affected by the addition of extra H⁺ or OH^- ions that cause the equilibrium to shift. (the equilibrium will shift to restore change)

Question 104

What will cause K_w to change?

Answer 104

K_w changes only if the temperature is altered.

Question 105

When are the concentrations of OH^- and H⁺ balanced in water?

Answer 105

At 25^oC, the pH value of 7 is the neutral point at which H⁺ and OH^- concentrations are the same equal to 10^-7 mol dm^-3 . This applies to neutral solutions.

Question 106

At 25^oC what must K_w always equal for water?

Answer 106

K_w = [H⁺] x [OH^-] = 1.00 x 10^-14 mol²dm^​-6

Question 107

Describe in terms of the dissociation of water, why rainwater becomes more acidic with the addition of carbon doixide.

Answer 107

At 25^oC rainwater is acididc, CO₂ dissolves forming a weak acid called carbonic acid, H2CO3, which dissociates to release H+ions into the water.

Because [H⁺] has increased [OH^-] must fall until:

[H⁺][OH^-] = 1.00 x 10^-14mol²dm^-6.

The overall effect is the pH decreases. Typical rainwater has a pH value of about 5.6.

Question 108

How does the pH of mineral water differ from normal rainwater?

Answer 108

Mineral water contains dissolved ions such as carbonate, CO₃^2-. These ions react with acids and remove H⁺ ions and lower the [H+].

Because [H⁺] has decreased, [OH-] must increase untill:

[H+][OH-] = 1.00 x 10^-14 mol^-2dm^-6.

pH increases (7.0 - 8.0)

Question 109

How can you find the concentrations of H⁺ and OH^- iions in solution using the link between [H+] and [OH-]?

Answer 109

• using [H+][OH-] = 1.00 x 10^-14mol^-2dm^-6.
• The indicies of [H+] and [OH-] will always add to -14.

Question 110

What are the most common strong bases?

Answer 110

• Strong bases tend to be hydroxides of the metals in groups 1 and 2 in the periodic table.
• NaOH, KOH, and Ca(OH)₂

Question 111

How do strong bases dissociate? Use NaOH as an example.

Answer 111

• In aqueous solution strong bases dissociates completely:
• NaOH_(aq) → Na⁺_(aq) + OH^-_(aq)
• Bases that dissociate in water to release hydroxide ions are called alkalis.
• Strong bases are alkalis - they are 100% dissociated in aqueous solution.

Question 112

How does a weak base dissociate?

Answer 112

Ammonia, NH₃, is a weak base. In aqueous solution, an equilibrium is set up. The equilibrium position lies well to the left hand side:

NH₃ + H₂O ⇔ NH₄⁺ + OH^-

Question 113

To work out the pH of a strong base we need to know [H⁺] and this depends on:

• the concentration of the base
• the ionic product of water, K_w = 1.00 x 10^-14 mol²dm^-6 (because this links [H⁺] with [OH^-]).

Question 114

Show the calculations for finding the pH of a strong monobasic alkali such as NaOH.

Answer 114

1. The [OH-] of a strong base is equal to the concentration of the base.
2. so [OH-] = [NaOH]
3. we can find [H+] from K_w and [OH-] K_w = [H+] [OH-]
4. [H+] = K_w / [OH-]
5. The pH can then be calculated using pH = -log₁₀[H⁺].

(worked example on pg 41)

Question 115

Recap: what can we use K_w to find?

Answer 115

[H+] = K_w / [OH^-]

We can use K_w to find either the concentraion of H+ or OH-, as long as we know the value for K_w and the other concentration.

Question 116

What is the definition of a buffer solution.

Answer 116

A buffer solution is a mixture that minimises pH changes on addition of small amounts of acid or base.

The word ‘minimses’ is essential to this definition.

Question 117

How can a buffer solution be made?

Question 118

Describe the carbonic acid buffer system.

Answer 118

• The weak acid, CH₃COOH, dissociates partially;

CH₃COOH ⇔ H⁺ + CH₃COO^-

• The salt dissociates completely, generating the conjugate base, CH₃COO^-:

CH₃COO^-Na⁺ → CH₃COO^- + Na⁺

The equilibrium mixture formed contains a high concentration of the undissociated weak acid, CH₃COOH, and its conjugate base, CH₃COO^-. The high concentration of the conjugate base pushes the equilibrium to the left, so the concentration of H⁺ ions is very small.

The resulting buffer solution contains large reservoirs of the weak acid and its conugate base.

Question 119

What is an alternative method of producing a buffer than from a weak acid and a salt?

Answer 119

Buffers can be made from a weak acid and a strong alkali. In this situation, a solution containing a mixture of the salt and the excess of weak acid is formed. For example, a weak acid such as methanoic acid, HCOOH, can be partially neutralised by an aqueous alkali, such as NaOH.

Question 120

How does a buffer solution react to the addition of an acid?

Answer 120

• [H⁺] is increased
• The conjugate base, A^- reacts with H⁺ ions
• the equilibrium shifts to the left, removing most of the added H⁺ ions.​

Question 121

How does a buffer solution react to the addition of an alkali to the buffer mixture

Answer 121

• [OH^-] is increased
• The small concentration of H⁺ ions reacts with the OH^- ions.
• HA dissociates, shifting the equilibrium to the right to restore most of the H⁺ ions that have reacted.

Question 122

When calculating pH of buffer solutions what does the pH of a buffer solution depend on?

Answer 122

The pH of a buffer solution depends on:

• the acid dissociation constant, K_a, of a buffer system.
• the concentration ratio of the weak acid and its conjugate base.

Question 123

What equation do we use for calculating the pH of buffer solutions?

Answer 123

K_a = [H⁺] [A^-] / [HA] therefore [H⁺] = K_a x [HA] / [A^-]

We need to check on the concentrations of HA and A^-. Only a very small proportion of HA dissociates so we can assume that [HA]_equilibrium = [HA]_{undissociated}

The salt of the weak acid is ionic and dissociated completely in aqueous solution, for example:

CH₃COO^-Na⁺ → CH₃COO^- + Na⁺

so effectively [CH₃COO^-] = [CH₃COO^-Na⁺​]

(example on page 43)

Question 124

What buffer system controls the pH of blood?

Answer 124

The carbonic acid- hydrogen carbonate buffer system.

Question 125

What is the importance of the carbonic acid - hydrogen carbonate buffer system?

Answer 125

Healthy human blood plasma needs to have a pH between 7.35 and 7.45. If the pH falls below 7.35, a condition called acidosis occurs. If the pH rises above 7.45, then the condition is called alkalosis.

The pH of blood is controlled by the mixture of buffers. The carbonic acid-hydrogen carbonate ion buffer, which is present in blood plasma, is the most important buffer system in the blood:

Question 126

How does the carbonic acid - hydrogen carbonate buffer system work?

Answer 126

• carbonic acid, H₂CO₃, acts as the weak acid
• the hydrogencarbonate ion, HCO₃^- , acts as the conjugate base.

Question 127

How does the carbonic acid - hydrogencarbonate buffer system react to an increase in [H⁺] ions?

Answer 127

Any increase in [H⁺] ions in the blood is removed by the conjugate base, HCO₃^-.The equilibrium shifts to the left, removing most of the H⁺ ions: H₂CO₃ ⇔ H⁺ + HCO₃^-

Question 128

How does the carbonic acid - hydrogencarbonate buffer system react to an increase in [OH^-] ions?

Answer 128

• Any increase in [OH^-] ions in the blood is removed by the weak acid, H₂CO₃. The small concentration of H⁺ ions reacts with the OH^- ions: H⁺ + OH^- → H₂O
• H₂CO₃ dissociates, shifting the equilibrium to the right to restore most of the H⁺ ions:

Question 129

What is the acid dissociation constant, K_a, for the carbonic acid - hydrogencarbonate ion equilibrium?

Answer 129

4.3 x 10^-7 mol dm^-3 ​

Question 130

In the carbonic acid - hydrogencarbonate ion buffer system what happens to the H₂CO₃ when acid is added?

Answer 130

The carbonic acid is converted into aqueous carbon dioxide through the action of an enzyme. In the lungs, the dissolved carbon dioxide is converted into carbon dioxide gas, which is then exhaled.

Question 131

What is the definition of the equivalence point?

Answer 131

The equivalence point is the point in a titration at which the volume of one solution has reacted exactly with the volume of the second solution. This matches the stoichiometry of the reaction taking place.

Question 132

What is the definition of the end point?

Answer 132

The end point is the point in a titration at which there are equal concentrations of the weak acid and its conjugate base forms of the indicator.the colour at the endpoint is midway between the colours of the acid and conjugate base forms.

Question 133

Describe the first section of this titration curve.

Answer 133

Section 1 - a slight increase in pH occurs as the base is added. The increase is only slight because the acid is in such excess. It is important to note that this section of the curve is not horizontal - pH is rising slightly.

Question 134

Describe the middle section of the titration curve.

Answer 134

Section 2 - a sharp rise in pH occurs. The acid is no longer in excess so any base added has a large impact on the pH.

The equivalence point is the centre of the verticle section of the titration curve.

Question 135

Describe the last section of the titration curve.

Answer 135

Section 3 - a slight increase in pH occurs as further base is added. The increase is only slight because the base is in excess now and an extra base has little impact on the pH. It is important to note however that this section is not horizontal - pH is rising slightly.

Question 136

How can the data that is used to plot a titration curve be recorded?

Answer 136

pH meters or data loggers can be used to measure the pH of the reaction mixture as a solution from the burette is added over time. The varying pH values can then be plotted as a graph - known as an acid-base titration curve. The curve may be plotted by hand or by a computer attached to a pH data logger.

Question 137

What symbols are often used to represent an indicator in an acid-base titration?

Answer 137

An acid-base titration is a weak acid, often represented as HIn. An indicator has one colour in its acid form (HIn) and a different colour in its conjugate base form (In^-​).

Question 138

What are the advantages of using a pH meter?

Answer 138

An indicator such as universal indicator can only give a general indication of the pH of a solution. A pH meter can give an accurate reading for pH, usually to two decimal places.

Question 139

What does a pH meter contain?

Answer 139

pH meters contain a protected electrode (the pH probe), which is placed into the solution in question, and a small computer display, which gives the pH reading.

Question 140

When using a pH probe, it is important to calibrate it first do that the results are accurate. How do you calibrate a pH probe?

Answer 140

Usually, this is done in a two or three step calibration;

• The probe is removed from its storage solution and rinsed with deionized water.
• The probe is blotted dry and then placed into a solution of a known pH, often starting with a solution of pH4. The pH reading is allowed top settle before checking that a pH of 4 has been registered.
• This process is repeated with other solutions of known pH, often using prepared solutions of pH 7 and pH 10.

Question 141

The ranges of the indicators are not actually on the equivalence points on most titration curves, so how can they show the reaction has reached this point?

Answer 141

Becuase the equivalence points occurs halfway through an incredibly sharp rise (or fall) in pH, the moment at which the indicator passes through its end point is effectively the same as the moment that the equivalence point is effectively the same moment that the equivalence point takes place.

Question 142

What do you notice from the graph? The indicators used are phenolphthalein pink and methyl orange.

Answer 142

• The verticle section of the graph covers a large change in pH, starting around pH 3 and ending around pH 11, with an equivalence point at pH7.
• Both indicators, methyl orange and phenolphthalein, have end points that fall within its pH range.
• Either indicator would be suitable to use in a titration between a strong acid and a strong base.

Question 143

Draw a strong acid - strong base titration curve.

Answer 143

Question 144

Draw a strong acid- weak base titration curve.

Answer 144

Question 145

What do you notice from the graph? The indicators used are phenolphthalein pink and methyl orange.

Answer 145

• The verticle section of the graph covers a smaller change in pH, starting around pH 3 and ending around pH 7.5, with the equivalence point occurring at a more acidic value, i.e. at pH lower than 7.
• Methyl orange has an end point that falls within its pH range.
• Phenolphthalein does not have an end point that falls within this pH range.
• Only methyl orange would be a suitable indicator for a titration between a strong acid and a weak base.

Question 146

Draw a titration curve for a weak acid - strong base titration.

Answer 146

Question 147

What do you notice from the graph? The indicators used are phenolphthalein pink and methyl orange.

Answer 147

• The vertical section id the graph covers a smaller change in pH, and occurs further up towards the higher pH values, starting around pH 6.5 and ending up around 11.5.
• The equivalence point occurs at a more basic pH, i.e. a pH value above 7.
• Methyl orange has an end point that falls outside this pH range.
• Phenolphthalein has an end point that falls within this pH range.
• Only phenolphthalein would be a suitable indicator for a titration between a strong acid and a weak base.

Question 148

Draw a titration curve for the reaction between a weak acid and a weak base.

Answer 148

Question 149

What do you notice from the graph? The indicators used are phenolphthalein pink and methyl orange.

Answer 149

• There is no real vertical section.
• Neither indicator has an end point near the equivalence point.
• Neither indicator is suitable.
• An indicator would change colour gradually over a few cm³ of base added.
• No indicators are really suitable for a weak acid - weak base titration.