5.1

Question 1

5.1.1 a) Explain what the rate of reaction is.

Answer 1

The rate of reaction is the change in concentration of reactants or products per unit time.

Question 2

5.1.1 a) Explain what the term ‘order’ refers to. What is the overall order of a reaction?

Answer 2

If more than one reactant is involved in a reaction, each reactant can affect the rate of reaction differently.
The order, with respect to a reactant, indicates how the concentration of the reactant affects the rate. It is the power to which the concentration is raised in the rate equation. The order of a reactant can be either 0, 1 or 2.
The overall order of a reaction is the sum of the individual order.

Question 3

5.1.1 b) i) How does the order indicate how the concentration of the reactant affects the rate.

Answer 3

If the order is 0, with respect to reactant A, then:
rate ∝ [A]0
The rate is unaffected by changing the concentration of A.
{Note that any number to the power of 0 is equal to 1}

If the order is 1, with respect to reactant B, then:
rate ∝ [B]1
The rate is directly proportional to the concentration of B.

If the order is 2, with respect to reactant C, then:
rate ∝ [C]2
The change in rate will be equal to the change in concentration squared.

Question 4

5.1.1 b) ii) What is the rate equation?

Answer 4

The rate equation for a reaction A + B → C + D is given by:
rate = k[A]m[B]n
where m is the order of reaction with respect to A, and n is the order of reaction with respect to B.
{If the order of a reactant is zero, then it will not be in the rate equation: [A]0 = 1}

Question 5

5.1.1 a) Explain what the rate constant is.

Answer 5

The rate constant, k, is the constant that links the rate of reaction with the concentrations of the reactants raised to the powers of their orders in the rate equation.
The larger the rate constant, the faster the rate of reaction.

Question 6

5.1.1 c) How are the units for the rate constant determined?

Answer 6

The units of k are determined by substituting units for rate and concentration into the rate equation, and cancelling when applicable. Alternatively, they can also be determined from the overall order of a reaction:
If the overall order = 0 (rate = k[A]0), then:
k = mol dm-3 s-1
If the overall order = 1 (rate = k[B]1), then:
k = s-1
If the overall order = 2 (rate = k[C]2), then:
k = dm3 mol -1 s-1
If the overall order = 3 (rate = k[C]2[B]1), then:
k = dm6 mol-2 s-1

Question 7

5.1.1 d) i)Explain how you can deduce the order with respect to a reactant from the shape of a concentration-time graph.

Answer 7

The order of a reactant can be deduced from the shape of the graph:
A zero order reaction produce a diagonal straight line graph, with the concentration decreasing at a constant rate.
A first order reaction produces a curved graph, where the half-lives are constant (so the concentration of the reactant halves at equal time intervals).

Question 8

5.1.1 d) How can reaction rates be calculated from concentration-time graphs?

Answer 8

The concentration of a reactant decreases as the time increases. The concentration is on the y-axis, with the time of the x-axis.
The rate at any point in the reaction is given by the gradient at any point on the graph. If the graph is a curve, you have to draw a tangent to the curve and find the gradient of that:
gradient = (change in y)/(change in x)

Question 9

5.1.1 a) Explain what the term ‘half-life’ refers to.

Answer 9

The half-life (t1/2) of a reaction is the time it takes for half of the reactant to be used up. The half-life of a first order reaction is independent of the concentration: the half-lives are constant (so the concentration of the reactant halves at equal time intervals) i.e. each half-life is the same.

Question 10

5.1.1 e) Explain how half-lives in a concentration-time graph can indicate a reaction order of 1.

Answer 10

The half-life of a first order reaction is independent of the concentration. The half-lives are therefore constant (so the concentration of the reactant halves at equal time intervals) i.e. each half-life will be the same.

Question 11

5.11 f) Explain how you can determine the rate constant, k, from the concentration-time graph of a first order reaction?

Answer 11

You can determine the rate constant, k, from the constant half-life, t1/2, using the following relationship:
k = (ln 2)/(t1/2)
The units for k are (no units/s) = s-1 (the units for k in a first order reaction are always s-1)

Question 12

5.1.1 g) i) Explain how you can deduce the order with respect to a reactant from the shape of a rate-concentration graph.

Answer 12

If the order is 0 with respect to a given reactant, the rate-concentration graph will appear as a horizontal straight line. Changes in concentration of this reactant have no effect on the rate.
If the order is 1 with respect to a given reactant, the rate-concentration graph will appear as a diagonal straight line (non-linear) through the origin (with the rate increasing as the time increases). Changes in concentration of this reactant are directly proportional to the rate.
If the order is 2 with respect to a given reactant, the rate-concentration graph will appear as a curve (with the rate increasing as the time increases). The change in rate will be equal to the change in concentration squared.

Question 13

5.1.1 g) ii) Explain how you can determine the rate constant, k, for a first order reaction from its rate-concentration graph.

Answer 13

For a first order reaction, the rate constant, k, is equal to the gradient of the rate-concentration graph of that reactant:
rate = k[A]1
k = rate/concentration
k = change in y/change in x
k = gradient
{the rate is on the y-axis, with concentration on the x-axis}

Question 14

5.1.1 h) Which techniques and procedures are used to investigate reaction rates?

Answer 14

You can measure pH changes by carrying out titrations, or using a pH meter.
For reactions that produce gases, you can measure a either change in volume or pressure, or the loss in mass of a reactant.
For reactions that produce visual changes, you can observe either the formation of a precipitate or a colour change.
[You can measure colour changes using a colorimeter; the intensity of colour is usually related to the concentration of a substance - the more intense the colour, the higher the concentration, and the more light absorbed]

Question 15

5.1.1 h) How can the initial rate of a reaction be determined from a concentration-time graph?

Answer 15

Concentration-time graphs can be plotted from continuous measurements taken during the course of a reaction (continuous monitoring).
The initial rate of a reaction can be determined from a concentration-time graph: calculate the gradient of the tangent at time = 0.

Question 16

5.1.1 h) Outline a procedure used to investigate reaction rates by the initial rates method.

Answer 16

Rate-concentration data can be obtained from initial rates investigations of separate experiments using different concentrations of one of the reactants i.e:
Carry out an experiment where the concentration of a reactant is the independent variable. Draw concentration-time graphs, and calculate the initial rate, for each concentration. Plot the initial rate against the concentration in a rate-concentration graph to find out the order of a reactant.

Question 17

5.1.1 h) How can rate-concentration data be obtained from clock reactions?

Answer 17

Clock reactions are an approximation of this method where the time measured is such that the reaction has not proceeded too far i.e:
In a clock reaction, you measure how the time taken for a set amount of product to form changes as you vary the concentration of one of the reactants. There is usually an easily observable endpoint, such as a colour change, to tell you when the desired amount of product has formed.

Question 18

5.1.1 Outline the iodine clock reaction.

Answer 18

The reaction between hydrogen peroxide, H2O2, and potassium iodide, KI (in acid solution) is as follows:
H2O2(aq) + 2KI(aq) + 2HCl(aq) → I2(aq) + 2H2O(l) + 2KCl(aq)
or
H2O2(aq) + 2KI(aq) + H2SO4(aq) → I2(aq) + 2H2O(l) + K2SO4l(aq)

The ionic equation is:
H2O2(aq) + 2I-(aq) + 2H+ → I2(aq) + 2H2O(l)

The initial rate of this reaction can be measured by adding sodium thiosulphate (Na2S2O3) and starch to the hydrogen peroxide and potassium iodide beforehand.
The hydrogen peroxide and iodide ions react as per, producing iodine and water. The thiosulphate ions then reacts with the iodine (instantaneously), reproducing the iodide ions. This is a two-step reaction:
H2O2(aq) + 2I-(aq) + 2H+ → I2(aq) + 2H2O(l)
2S2O3(aq) + I2(aq) → 2I-(aq) + S4O62-(aq)
In the presence of the reformed iodide ions, the starch indicator will turn from yellow to blue-black. This is the end of the clock reaction. Varying either the iodide or the hydrogen peroxide concentration will produce different times for the colour change.

Question 19

5.1.1 i) What is the rate-determining step?

Answer 19

The slowest step in a multi-step reaction will dictate the rate of the overall reaction: this is the rate-determining step.

Question 20

5. 1.1 i) For a multi-step reaction, how can you predict:
   i) a rate equation that is consistent with the rate determining step?
   ii) the possible steps in a reaction mechanism from the rate equation and the balanced equation of the overall reaction?

Answer 20

The rate-determining step doesn’t have to be the first step in a reaction.
The order with respect to a reactant shows the number of molecules of that reactant which are involved in the rate-determining step.
If a reactant doesn’t appear in the rate equation, and has an order of 0, it doesn’t affect the rate and isn’t involved in the rate-determining step [and vice versa].
If the order is 1 with respect to a reactant, then 1 molecule of that reactant is involved in the rate-determining step [and vise versa].
If the order is 2 with respect to a reactant, then 2 molecules of that reactant are involved in the rate-determining step [and vice versa].
An intermediate formed in one step of a multi-step reaction is used up in a subsequent step. It is not seen as either a reactant or a product of the overall equation.

Question 21

5.1.1 j) Explain the effect of temperature change on the rate of reaction, and therefore the rate constant.

Answer 21

As the temperature of a reaction increases, the particles will, on average, have greater kinetic energy, moving faster, and resulting in more frequent collision. A higher proportion of molecule will have energy that surpasses the activation energy - this means that more collisions are likely to be successful. An increase in temperature will increase the rate of reaction.
According to the rate equation, the rate of reaction depends solely on the rate constant, k, and the concentration of the reactants:
rate = k[A]m[B]n
If the rate increases with increasing temperature when the concentrations are the same, then the rate constant must increase with temperature i.e:
A higher temperature = a higher rate constant, k = a faster rate of reaction.

Question 22

5. 1.1 k) i) Explain what the following abbreviations indicate in the Arrhenius equation:
   a) k
   b) Ea
   c) T
   d) R
   e) A
   f) e

Answer 22

k = rate constant
Ea = activation energy (J mol-1)
T = temperature (K)
R = gas constant (8.31 J K-1 mol-1)
A = the pre-exponential factor
e = mathematical constant (2.71828 or ‘e’ on your calculator)
The Arrhenius equation tell us that as that:
- temperature, T, and the rate constant, k, are related exponentially
- as temperature increases, the rate constant increases
- as activation energy increases, the rate constant increases

Question 23

5.1.1 k) ii) How can you determine Ea and A graphically?

Answer 23

You can determine Ea and A graphically using:
ln k = -Ea/RT + ln A
This logarithmic version of the Arrhenius equation follows the general pattern y = mx + c:
ln k = y
1/T = x
-Ea = m
ln A = c
You can create an Arrhenius plot by plotting ln k on the y-axis against 1/T on the x-acis.
This will produce a gradient of -Ea/R and a y-intercept of ln A.
e.g. the gradient, -Ea/R = change in y/change in x = m
So Ea = (m) x -R = (m) x -8.31
[If the gradient is negative, include the negative sign, i.e. -m x -8.31]
e.g. ln A = y-intercept = c
To calculate A (using a calculator):
Do shift ln, which should produce e, then do ec (which is e to the power of c)
[If the y-intercept is negative, include the negative sign, i.e. e to the power of -c.]

Question 24

5.1.2 b) How can you calculate the concentrations present at equilibrium, given the initial moles of the reactants and the equilibrium moles of one reactant.

Answer 24

aA + bB ⇄ cC
1. If given the initial moles of A and B, and the equilibrium mole of A:

• Calculate how many moles of A has reacted: this is the difference between the initial mole and the equilibrium mole of A.
• Calculate the mole of B that reacted: multiply the mole of A to get the mole of B that reacted (this depends on the molar ratio between A and B)
• Calculate the equilibrium mole of B: take away the mole of B that reacted from the initial mole of B.
• Calculate the mole of C produced: multiply the mole of A to get the mole of C produced (this depends on the molar ratio between A and B)
• Calculate the equilibrium mole of C: add the mole of C produced to the initial mole of C, which is 0 (i.e. the equilibrium mole of C = the mole of C produced)

Once all the equilibrium moles have been calculates, using the total volume to calculate the concentration for each.

Question 25

5. 1.2 c) What is the expression for Kc for homogeneous and heterogeneous equilibria?
   d) How do you calculate the units for Kc?

Answer 25

When a reaction reaches dynamic equilibrium, Kc (an equilibrium constant) gives a measure of where the equilibrium lies, whether it’s more to the right or the left.
aA + bB ⇄ cC + dD
Kc = [C]c[D]d / [A]a[B]b
The expression for Kc differs depending on the physical states of the reactants and products:
- For a homogeneous equilibrium, all the reactants and products are included in the expression for Kc
- For a heterogeneous equilibrium, only gases and aqueous substances go into the expression for the equilibrium constant (any solids or liquids get left out).
[If you are given the equilibrium amounts in moles, alongside the total volume, you must first calculate the concentrations: conc. = moles/(total) volume]
The units for Kc differ depending on the reaction: replace each concentration with its unit in the Kc expression, and cancel where applicable.

Question 26

5.1.2 a) Explain the term ‘mole fraction’ and how it is calculated.

Answer 26

The mole fraction, XA, of a substance is the proportion of that substance, A, in a reaction mixture.
mole fraction, XA = no. of moles of substance A / total number of moles of all substances.

Question 27

5.1.2 a) Explain the term ‘partial pressure’ and how it is calculated.

Answer 27

The partial pressure is the amount of pressure being exerted by an individual species within a reaction vessel. The total pressure of a reaction mixture is the sum of all the partial pressures of the individual species.
partial pressure, PA = mole fraction, XA x total pressure

Question 28

5. 1.2 c) What is the expression for Kp for homogeneous and heterogeneous equilibria?
   d) How do you calculate the units for Kp?

Answer 28

When a reaction reaches dynamic equilibrium, Kp (an equilibrium constant) can be calculated from the partial pressure of each species in the reaction mixture.
aA + bB ⇄ cC + dD
Kc = [C]c[D]d / [A]a[B]b
The expression for Kp differs depending on the physical states of the reactants and products:
- For a homogeneous equilibrium, all the reactants and products are included in the expression for Kc
- For a heterogeneous equilibrium, only gases and aqueous substances go into the expression for the equilibrium constant (any solids or liquids get left out).
The units for Kp differ depending on the reaction: replace each partial pressure with its unit in the Kc expression, and cancel where applicable.
The unit for pressure can vary:
1 atm = 101kPa = 101000Pa
1 Pa = 1 N m-2

Question 29

5.1.2 f) i) Explain the effect on equilibrium constants of changing the temperature for exothermic and endothermic reactions.

Answer 29

A change in temperature is counteracted by a shift in equilibrium:
- An increase in temperature shifts the position of equilibrium in the endothermic direction.
- A decrease in temperature shifts the position of equilibrium in the exothermic direction.
The shift in the position of equilibrium is controlled by the equilibrium constants:
- If the position of equilibrium is shifted to the right, then Kc/Kp has increased.
- If the position of equilibrium is shifted to the left, then Kc/Kp has decreased.

Question 30

5.1.2 f) ii) Explain the effect on Kc of changing the concentration of either a reactant or a product.

Answer 30

The value of Kc is unaffected by changes in concentration.
Increasing the concentration of a reactant, or decreasing the concentration of a product, results in:
- an increase in the concentration of the products
- and a decrease in the concentration of the reactants
While the position of equilibrium shifts to the right, Kc remains constant.
Decreasing the concentration of a reactant, or increasing the concentration of a product, results in:
- an increase in the concentration of the reactants
- and a decrease in the concentration of the products
While the position of equilibrium shifts to the left, Kc remains constant.

Question 31

5.1.2 f) ii) Explain the effect on Kp of changing the pressure.

Answer 31

The value of Kp is unaffected by changes in pressure.
Increasing the pressure of the system causes the position of equilibrium to counteract this change by shifting to side with fewer gas molecules, thus decreasing the pressure.
Decreasing the pressure of the system causes the position of equilibrium to counteract this change by shifting to side with more gas molecules, thus increasing the pressure.

Question 32

5.1.2 f) i) Explain the effect that a catalyst has on equilibrium constants.

Answer 32

Catalysts increase the rate of a reaction: they speed up both the forward and reverse reaction by the same amount. While equilibrium is reached more quickly, the position of equilibrium doesn’t change, and hence the equilibrium constants don’t change.

Question 33

5.1.3 a) i) Explain what a Brønsted–Lowry acid is.

Answer 33

A Brønsted–Lowry acid is a species that donates a proton.
[Aqueous acids, HA, release hydrogen ions, H+ - these combine with water, H2O, to form hydronium ion ions, H3O+, leaving a negative ion, A-]
HA(aq) + H2O(l) → H3O+(aq) + A-(aq)
= HA(aq) → H+(aq) + A-(aq)
e.g.
HCl(aq) + H2O(l) → H3O+(aq) + Cl-(aq)
= HCl(aq) → H+(aq) + Cl(aq)

Question 34

5.1.3 a) i) Explain what a Brønsted–Lowry base is.

Answer 34

A Brønsted–Lowry base is a species that accepts a proton.
[Aqueous bases, B, accept hydrogen ions, H+, from water, H2O - this leaves BH+ and OH- in solution]
B(aq) + H2O(l) → BH+(aq) + OH-(aq)
e.g.
NH3(aq) + H2O(l) → NH4+(aq) + OH-(aq)
= NH3(aq) + H+(aq) → NH4+(aq)

Question 35

5.1.3 a) ii) Explain what a conjugate acid-base pair is, using HA + B ⇄ BH+ + A- as an example.

Answer 35

A conjugate acid-base pair involves two species that transform into each other by the gain or loss of a proton.
A conjugate acid is the species that has gained a proton, and can therefore donate the proton (to form its conjugate base).
A conjugate base is the species that has lost a proton, and can therefore accept the proton (to form its conjugate acid).
HA + B ⇄ BH+ + A-
Acid + Base ⇄ Acid + Base
HA and A- are a conjugate acid-base pair:
- HA: the conjugate acid, donates a proton in the forward reaction (to form A-)
- A-: the conjugate base, accepts a proton in the reverse reaction (to form HA)
B and BH+ are a conjugate acid-base pair:
- B: the conjugate base, accepts a proton in the forward reaction (to form BH+)
- BH+: the conjugate acid, donates a proton in the reverse reaction (to form B)

Question 36

5.1.3 a) ii) Explain how in an conjugate acid-base pair, water can act as a base when reacted with an acid.

Answer 36

HA(aq) + H2O(l) ⇄ H3O+(aq) + A-(aq)
Acid + Base ⇄ Acid + Base
HA and A- are a conjugate acid-base pair:
- HA: the conjugate acid, donates a proton in the forward reaction (to form A-)
- A-: the conjugate base, accepts a proton in the reverse reaction (to form HA)
H2O and H3O+ are a conjugate acid-base pair:
- H2O: the conjugate base, accepts a proton in the forward reaction (to form H3O+)
- H3O+: the conjugate acid, donates a proton in the reverse reaction (to form H2O)

Question 37

5.1.3 a) ii) Explain how in an conjugate acid-base pair, water can act as an acid when reacted with a base.

Answer 37

B(aq) + H2O(l) ⇄ BH+(aq) + OH-(aq)
Base + Acid ⇄ Acid + Base
B and BH+ are a conjugate acid-base pair:
- B: the conjugate base, accepts a proton in the forward reaction (to form BH+)
- BH+: the conjugate acid, donates a proton in the reverse reaction (to form B)
H2O and OH- are a conjugate acid-base pair:
- H2O: the conjugate acid, donates a proton in the forward reaction (to form OH-)
- OH-: the conjugate base, accepts a proton in the reverse reaction (to form H2O)

Question 38

5.1.3 a) iii) Explain what monobasic, dibasic, and tribasic acids are, using HCl, H2SO4 and H3PO4 as examples.

Answer 38

Different acids can release different numbers of protons.
1. HCl is a monobasic acid, because each molecule can release one proton:
HCl(aq) → H+(aq) + Cl(aq)
2. H2SO4 is a dibasic acid, because each molecule can release two protons:
H2SO4(aq) → H+(aq) + HSO4-(aq)
HSO4-(aq) → H+(aq) + (SO4)2-(aq)
3. H3PO4 is a tribasic acid, because each molecule can release 3 protons:
H3PO4(aq) → H+(aq) + H2PO4-(aq)
H2O4-(aq) → H+(aq) + (HPO4)2-(aq)
(HPO4)2-(aq) → H+(aq) + (PO4)3-(aq)

Question 39

5.1.3 b) Outline the reaction between aqueous acids and solid carbonates, giving both the full equation and the ionic equation for the reaction between hydrochloric acid and calcium carbonate.

Answer 39

Aqueous acids react with solid carbonates, forming a salt, carbon dioxide and water.
The full equation:
2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l)
The species present:
2H+(aq) + 2Cl-(aq) + CaCO3(s) → Ca2+(aq) + 2Cl-(aq) + CO2(g) + H2O(l)
The ionic equation:
2H+(aq) + CaCO3(s) → Ca2+(aq) + CO2(g) + H2O(l)

Question 40

5.1.3 b) Outline the reaction between aqueous acids and aqueous carbonates, giving both the full equation and the ionic equation for the reaction between hydrochloric acid and sodium carbonate.

Answer 40

Aqueous acids react with solid carbonates, forming a salt, carbon dioxide and water.
The full equation:
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + CO2(g) + H2O(l)
The species present:
2H+(aq) + 2Cl-(aq) + 2Na+ + (CO3)2- → 2Na+(aq) + 2Cl-(aq) + CO2(g) + H2O(l)
The ionic equation:
2H+(aq) + (CO3)2- → CO2(g) + H2O(l)

Question 41

5.1.3 b) Outline the reaction between aqueous acids and solid metal oxides, giving both the full equation and the ionic equation for the reaction between nitric acid and magnesium oxide.

Answer 41

Aqueous acids react with solid metal oxides, forming a salt and water.
The full equation:
2HNO3(aq) + MgO(s) → Mg(NO3)2(aq) + H2O(l)
The species present:
2H+(aq) + 2(NO3)-(aq) + MgO(s) → Mg2+(aq) + 2(NO3)-(aq) + H2O(l)
The ionic equation:
2H+(aq) + MgO(s) → Mg2+(aq) + H2O(l)

Question 42

5.1.3 b) Outline the reaction between aqueous acids and aqueous alkalies, giving both the full equation and the ionic equation for the reaction between sulphuric acid and potassium hydroxide.

Answer 42

Aqueous acids react with aqueous alkalies, forming a salt and water.
The full equation:
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)
The species present:
2H+(aq) + (SO4)2-(aq) + 2K+(aq) + 2OH-(aq) → 2K+(aq) + (SO4)2-(aq) + 2H2O(l)
The ionic equation:
2H+(aq) + 2OH-(aq) → 2H2O(l)

Question 43

5.1.3 b) Outline the reaction between aqueous acids and solid metals, giving both the full equation and the ionic equation for the reaction between hydrochloric acid and magnesium.

Answer 43

Aqueous acids react with solid metals, forming a salt and hydrogen gas. The metal loses electrons and is oxidised, while the H+ ions accepts electrons and is reduced to hydrogen gas.
The full equation:
2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)
The species present:
2H+(aq) + 2Cl-(aq) + Mg(s) → Mg2+(aq) + 2Cl-(aq) + H2(g)
The ionic equation:
2H+(aq) + Mg(s) → Mg2+(aq) + H2(g)

Question 44

5.1.3 c) i) Explain the strength of acids in terms of acid dissociation.

Answer 44

Strong acids fully dissociate in water:
HA(aq) + H2O(l) → H3O+(aq) + A-(aq)
HA(aq) → H+(aq) + A-(aq)
A strong monobasic acid would product one mole of hydrogen ions from one mole of acid. The H+ concentration is equal to the concentration of the acid, HA.

Weak acids partially dissociate in water:
HA(aq) + H2O(l) ⇄ H3O+(aq) + A-(aq)
HA(aq) ⇄ H+(aq) + A-(aq)
The position of equilibrium lies to the left, with only small concentrations of H+ and A- (in comparison to the larger concentration of the undissociated acid, HA).

Question 45

5.1.3 d) What are the two pH expressions?

Answer 45

pH = -log[H+(aq)]
[H+(aq)] = 10-pH (10 to the power of -pH)

If you know the hydrogen ion concentration in a solution of acid, you calculate the pH (and vice versa) - this applies to both strong acids and weak acids.
pH is a measure of how acidic or basic something is. It measures the concentration of hydrogen ions in solution.
The lower the pH value, the higher the [H+].
The higher the pH value, the lower the [H+].
A pH increase of 1 cause the [H+] to decrease by 10:
pH 1 = 0.1
pH 2 = 0.01
pH 3 = 0.003

Question 46

5.1.3 f) i) How do you calculate the pH of a strong acid, given its concentration?

Answer 46

Strong acids fully dissociate in water:
HA(aq) → H+(aq) + A-(aq)
The H+ concentration is equal to the concentration of the acid, HA
i.e. [HA(aq)] = [H+(aq]
You can therefore calculate the pH of a strong acid from its concentration.
pH = -log[H+(aq)]
i.e. pH = -log[HA(aq)]

Question 47

5.1.3 c) i) Explain, and given an expression for, the acid dissociation constant, Ka.

Answer 47

The extent of acid dissociation is measured by the acid dissociation constant, Ka (this is an equilibrium constant).
The expression for the acid dissociation constant is:
Ka = [H+(aq)] [A-(aq)] / [HA(aq)]
A large Ka value indicates a large extent of dissociation: so the acid is strong.
A small Ka value indicates a small extent of dissociation: so the acid is weak.

Question 48

5.1.3 g) i) How do you calculate the Ka value for a weak acid, using approximations?
How do you calculate the pH of a weak acid, given its concentration and Ka value?

Answer 48

Weak acids partially dissociate in water:
HA(aq) ⇄ H+(aq) + A-(aq)
The expression for the acid dissociation constant is:
Ka = [H+(aq)] [A-(aq)] / [HA(aq)]
The approximations for weak acid calculations are as follows:
1. The equilibrium of a weak acid lies heavily to the left. Because only an insignificant amount of HA dissociates ( [HA(aq)]&raquo_space; [H+(aq)] ), it is assumed that the concentration of the undissociated acid is equal to the concentration of the acid at equilibrium:
[HA(aq)]undissociated ∼ [HA(aq)]equilibrium
2. When HA dissociates, H+ and A- are formed in equal quantities. It is assumed that the dissociation of any water present is negligible, and the concentration of H+ is unaffected. Therefore, the concentrations of H+ and A- are considered equal:
[H+]equilibrium ∼ [A-]equilibrium
Using these approximations, the expression for Ka now becomes:
Ka = [H+(aq)]2/ [HA(aq)]
Given the concentration of an acid, and the Ka value, you can find the pH of a weak acid using the following equations:
[H+(aq)] = √Ka x [HA(aq]
pH = -log[H+(aq)]

Question 49

5.1.3 h) What are the limitations of using approximations to Ka related calculations for ‘stronger’ weak acids?

Answer 49

Because only an insignificant amount of HA dissociates in weak acids ( [HA(aq)]&raquo_space; [H+(aq)] ), it is assumed that the concentration of the undissociated acid is equal to the concentration of the acid at equilibrium:
[HA(aq)]undissociated ∼ [HA(aq)]equilibrium
However, stronger acids dissociate more fully in solution: the difference between the concentration of the undissociated acid and the concentration of the acid at equilibrium is significant. The above assumption is no longer valid.

Question 50

5.1.3 c) ii) Give two expressions for the relationship between Ka and pKa.

Answer 50

pKa = -logKa
Ka = 10-pKa (10 to the power of -pKa)

The stronger the acid, the larger the Ka value and the smaller the pKa value.
The weaker the acid, the smaller the Ka value and the larger the pKa value.

Question 51

5.1.3 e) Give the expression for the ionic product of water, Kw, by explaining the ionisation of water.

Answer 51

Water can act as an acid by donating a proton:
H2O(l) → H+(aq) + OH-(aq)
Water can act as a base by accepting a proton:
H2O(l) + H+(aq) → H3O+(aq)
Overall:
2H2O(l) ⇄ H3O+(aq) + OH-(aq)
OR: H2O ⇄ H+(aq) + OH-(aq)
Consider the acid-base equilibrium above - by treating water as a weak acid, you can apply it to the Ka expression:
Ka = [H+(aq)] [OH-(aq)] / [H2O(l)]
The equilibrium lies well to the left: only an extremely small amount of water is dissociated at any given time. This means the concentration of water is effectively constant, and together with Ka, it can form another equilibrium constant: Kw
Ka x [H2O(l)] = [H+(aq)] [OH-(aq)]
Kw = [H+(aq)] [OH-(aq)]
Kw is the ionic product of water. The units for Kw are always mol2 dm-6.
Kw always has the same value for pure water or an aqueous solution at a given temperature.

Question 52

5.1.3 e) Using the Kw expression, outline why pure water has a pH of 7 at 25°C.
{At 25°C, Kw = 1 x 10-14 mol2 dm-6}

Answer 52

At 25°C, Kw = 1 x 10-14 mol2 dm-6
When pure, neutral water ionises, the H+ and OH- ions are released in a 1:1 ratio (i.e. in equal quantities).
[H+(aq)] = [OH-(aq)]
This means that Kw = [H+(aq)] [OH-(aq)] can be written as:
kw = [H+(aq)]2
If Kw = 1 x 10-14 mol2 dm-6, then:
[H+(aq)]2 = 1 x 10-14
[H+(aq)] = √1 x 10-14 = 1 x 10-7 mol dm-3
So the pH is:
pH = -log[1 x 10-7] = 7

Question 53

5.1.3 e) Explain how a change in concentration affects Kw.

Answer 53

Kw, like other equilibrium constants (e.g. Kc), is unaffected by changes in concentration.
When either [H+(aq)] or [OH-(aq)] changes, the position of equilibrium shifts to counteract this change and restore the value of Kw. The relative concentrations of H+(aq) and OH-(aq) ions are determined by Kw.

1) If the [H+(aq)] has increased (e.g. on addition of an acid, which would release protons), then the [OH-(aq)] will decrease until the Kw value has been restored: Kw = [H+(aq)] [OH-(aq)]. While the pH has decreased in this now acidic solution, Kw remains the same.
2) If the [H+(aq)] has decreased (e.g. on addition of a base, which would react with the protons), then the [OH-(aq)] will increase until the Kw value has been restored: Kw = [H+(aq)] [OH-(aq)]. While the pH has increased in this now alkaline solution, Kw remains the same.

All aqueous solutions contain H+(aq) and OH-(aq) ions:
- In water and neutral solutions, [H+(aq)] = [OH-(aq)]
- In acidic solutions, [H+(aq)] > [OH-(aq)]
- In alkaline solutions, [H+(aq)] < [OH-(aq)]
Kw is only affected by changes in temperature. While the concentration and pH of an aqueous solution might change, Kw will always remain the same at a given temperature.

Given the Kw and pH at a certain temperature, you can calculate the concentration of H+ ions and OH- ions:
[H+(aq)] = 10-pH
Kw = [H+(aq)] [OH-(aq)]
[OH-(aq)] = Kw / [H+ (aq)]

Question 54

5.1.3 c) Explain what happens to the Kw value, the pH value, and the concentration of H+(aq) and OH-(aq), on the addition of:
i) carbonic acid to pure water
ii) calcium carbonate to pure water
{At 25°C, Kw = 1 x 10-14 mol2 dm-6}

Answer 54

At 25°C, Kw = 1 x 10-14 mol2 dm-6. When carbonic acid, H2CO3, is added to pure water, it dissociates and releases H+(aq) ions. Because the [H+(aq)] has increased, the [OH-(aq)] must decrease until:
Kw = [H+(aq)] [OH-(aq)] = 1 x 10-14 mol2 dm-6
The pH has therefore decreased - this is because the [H+(aq)] has increased and the [OH-(aq)] has decreased - and the solution has become acidic. The Kw value has not changed, however.

At 25°C, Kw = 1 x 10-14 mol2 dm-6. When the base calcium carbonate, CaCO3, is added to pure water, it dissociates and releases (CO3)2-(aq) ions. They react with the protons, so that the [H+(aq)] has decreased - the [OH-(aq)] must therefore increase until:
Kw = [H+(aq)] [OH-(aq)] = 1 x 10-14 mol2 dm-6
The pH has therefore increased - this is because the [H+(aq)] has decreased and the [OH-(aq)] has increased - and the solution has become alkaline. The Kw value has not changed.

Question 55

5.1.3 e) With the knowledge that the ionisation of water is an endothermic process, explain how temperature change affects the Kw value.

Answer 55

Kw is only affected by changes in temperature.
Because the forward reaction is endothermic:
- an increase in temperature will cause the position of equilibrium to shift to the right, producing more H3O+(aq)/H+(aq) and more OH-(aq): Kw increases.
- a decrease in temperature will cause the position of equilibrium to shift to the left, producing less H3O+(aq)/H+(aq) and more OH-(aq): Kw decreases.
Remember that pure water must always be neutral: when the temperature changes, the Kw value, and therefore the pH value, will change. However, at this pH value, the [H+(aq)] and the [OH-(aq) values are still equal: so this pH is now neutral, and anything higher is alkaline, while anything lower is acidic (for this particular temperature, that is).

Question 56

5.1.3 f) How do you calculate the pH of a strong base, given its concentration and Kw value?

Answer 56

A strong base fully dissociates in water. This means that the concentration of OH-(aq) ions is equal to the concentration of the base.
BOH(aq) → B+(aq) + OH-(aq)
[BOH(aq)] = [OH-(aq)]
Given both the concentration of the base (equal to [OH-(aq)]) and the value for Kw, calculate the concentration of H+(aq) ions:
Kw = [H+(aq)] [OH-(aq)]
[H+(aq)] = Kw / [OH-(aq)]
The pH can then be calculated using:
pH = -log10[H+(aq)]

Question 57

5.1.3 i) What is a buffer solution?

Answer 57

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

Question 58

5.1.3 j) Outline two ways in which a buffer solution is formed.

Answer 58

A buffer solution is made by setting up an equilibrium between a weak acid, HA, and its conjugate base, A-. This can happen in two ways:

1. It can be made from a weak acid, HA, and a salt of the weak acid, A-X+
   - The salt fully dissociates into its ions, generating the conjugate base, A-:
   A-X+(aq) → A-(aq) + X+(aq)
   - The weak acid partially dissociates:
   HA(aq) ⇄ H+(aq) + A-(aq)
2. It can also be made from an excess of weak acid, HA, and a strong alkali, XOH
   - All of the base reacts with the acid:
   HA(aq) + XOH(aq) → A-X+(aq) + H2O(l)
   HA(aq) + OH(aq) → A-(aq) + H2O(l)
   - The excess weak acid partially dissociates:
   HA(aq) ⇄ H+(aq) + A-(aq)

In both cases, the resulting equilibrium mixture contains high concentrations of the undissociated weak acid, HA(aq), high concentrations of the conjugate base, A-(aq), which shifts the position of equilibrium to the left, so that the concentration of H+(aq) is very small.
HA(aq) ⇄ H+(aq) + A-(aq)

Question 59

5.1.3 k) Explain the role of the conjugate acid-base pair (in an acid buffer solution) when a small amount of acid is added.

Answer 59

On addition of an acid to a buffer solution, [H+(aq)] is increased. The conjugate base, A-(aq), in the buffer solution reacts with the added H+(aq) ions. This shifts the position of equilibrium to the left, removing most of the added H+(aq) ions:
HA(aq) ⇄ H+(aq) + A-(aq)
← Equilibrium shifts to the left

Question 60

5.1.3 k) Explain the role of the conjugate acid-base pair (in an acid buffer solution) to control pH when a small amount of acid is added.

Answer 60

On addition of an alkali, [OH-(aq)] is increased. The small concentration of H+(aq) ions in the buffer solution react with the added OH-(aq) ions, forming water:
H+(aq) + OH-(aq) → H2O(l)
This decreases the concentration of H+(aq) ions in the buffer solution. The weak acid, HA(aq), in the buffer solution dissociates to restore the concentration of H+(aq) ions. This shifts the position of equilibrium to the right:
HA(aq) ⇄ H+(aq) + A-(aq)
→ Equilibrium shifts to the right

Question 61

5.1.3 l) Explain how the pH of a buffer solution can be calculated from the Ka value of a weak acid and the equilibrium concentration of the conjugate acid-base pair.

Answer 61

The following equilibrium is set up in a buffer solution:
HA(aq) ⇄ H+(aq) + A-(aq)
The expression for the acid dissociation constant is:
Ka = [H+(aq)] [A-(aq)] / [HA(aq)]
The following approximations are made:
1. Because only an insignificant amount of HA dissociates ( [HA(aq)]&raquo_space; [H+(aq)] ), it is assumed that the concentration of the undissociated acid is equal to the concentration of the acid at equilibrium:
[HA(aq)]undissociated ∼ [HA(aq)]equilibrium
2. The salt of the weak acid fully dissociates. It is therefore assumed that the equilibrium concentration of the conjugate base is equal to the initial conentration of the salt of the weak acid:
[A-X+(aq)] = [A-(aq)]
Using these approximations, you can rearrange the expression for Ka and calulate [H+(aq)
Ka = [H+(aq)] [A-(aq)] / [HA(aq)]
[H+(aq)] = Ka x [HA(aq)] / [A-(aq)]
You can find then pH of the buffer solution:
pH = -log[H+(aq)]

Question 62

5.1.3 m) Explain how the pH of blood is controlled by a buffer solution.

Answer 62

The carbonic acid-hydrogencarbonate buffer system controls the pH of blood. It is present in blood plasma, and maintains a pH between 7.35 and 7.45.
The carbonic acid, H2CO3(aq), acts as the weak acid.
The hydrogencarbonate ion, HCO3-, acts as the conjugate base.
H2CO3(aq) ⇄ H+(aq) + HCO3-(aq)
1. On addition of an acid, [H+(aq)] is increased. The conjugate base, HCO3-(aq), reacts with the added H+(aq) ions. This shifts the position of equilibrium to the left, removing most of the added H+(aq) ions.
2. On addition of an alkali, [OH-(aq)] is increased. The small concentration of H+(aq) ions in the buffer solution react with the added OH-(aq) ions, forming water:
H+(aq) + OH-(aq) → H2O(l)
The weak acid, H2CO3 (aq), in the buffer solution dissociates to restore the concentration of H+(aq) ions. This shifts the position of equilibrium to the left.

The body produces far more acidic materials than alkaline, which the conjugate base, HCO3-, converts to H2CO3 (using the above equilibrium). This could lead to a build up of H2CO3, however the levels of carbonic acid are controlled by respiration. H2CO3 can be converted to CO2(aq), which can then be converted to CO2(g). By breathing out CO2(g), the following equilibrium shifts to the right and H2CO3 is reduced:
H2CO3(aq) ⇄ H2O(l) + CO2(aq)

Question 63

5.1.3 n) Outline the general shape of a pH titration curve (assuming a base is being added to an acid).

Answer 63

Consider a pH change when a base being added to an acid. Each pH titration curve has 3 distinct shapes:
1. A slight increase in pH occurs as the base is added. The acid is in such excess at this point that small amounts of base have little impact on the pH, so it only rises slightly.
2. A sharp rise in pH occurs: the acid is no longer is excess, so any base added has a larger impact on the pH. While the equivalence point is technically in the centre of the vertical section (when the acid is completely neutralised and [H+(aq)] = [OH-(aq)]), the rise is so sharp that the whole vertical section is often taken as an equivalence point (this why the end point of a suitable indicator can fall in the pH range of the vertical section in general, and not just match the exact equivalence point).
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 extra base has little impact on the pH.
The size of each distinct area differs depending on the strengths of the acid-base combination.

Question 64

5.1.3 n) iii) Explain how an indicator works.

Answer 64

An indicator is a weak acid, often represented as HIn(aq). A indicator is one colour in its acid form, HIn(aq), and a different colour in its conjugate base, In-.
HA(aq) → H+(aq) + A-(aq)
HIn(aq) → H+(aq) + In-(aq)
When there equal amounts of the weak acid and the conjugate base present, [HIn(aq)] = [In(aq)], the indicator is at its end point. For a titration, an indicator is chosen so that the pH value of its end point is as close as possible to the pH value of the titrations equivalence point. However, the rise is so sharp that the whole vertical section is often taken as an equivalence point (this why the end point of a suitable indicator can fall in the pH range of the vertical section in general, and not the exact equivalence point).
e.g. consider methyl orange:
HIn(aq) → H+(aq) + In-(aq)
HIn is red, and In- is yellow - so the end-point is orange, at a pH of 3 approx. As long as the pH of 3 falls in the pH range of the vertical section, the indicator can be used.

Question 65

5.1.3 n) i) Outline the shape of a strong acid-strong base titration.

Answer 65

For a strong acid-strong base titration:

• The curve starts low, because a strong acid has low pH. The slight increase in pH tends to be from 1 - 3.
• The vertical section covers a large pH change, from 3 - 11.
• The last slight increase is high up on the graph, because a strong base has a high pH. The slight increase in pH tends to be from 11-13.

Question 66

5.1.3 n) i) Outline the shape of a strong acid-weak base titration curve.

Answer 66

For a strong acid-weak base titration:

• The curve starts low, because a strong acid has low pH. The slight increase in pH tends to be from 1 - 3.
• The vertical section covers a pH change from 3 - 8.
• The last slight increase is relatively low (just above the middle) when compared to a strong base, because a weak base has a low pH. The slight increase in pH tends to be from 8 -10.

Question 67

5.1.3 n) i) Outline the shape of a weak acid-strong base titration curve.

Answer 67

For a weak acid-strong base titration:

• The curve starts relatively high (just below the middle) when compared to a strong acid, because a weak acid has higher pH. The slight increase in pH tends to be from 3 - 6.
• The vertical section covers a pH change from 6 - 11.
• The last slight increase is high up on the graph, because a strong base has a high pH. The slight increase in pH tends to be from 11-13.

Question 68

5.1.3 n) i) Outline the shape of a weak acid-weak base titration curve.

Answer 68

For a weak acid-weak base titration:

• The curve starts relatively high (just below the middle) when compared to a strong acid, because a weak acid has higher pH. The slight increase in pH tends to be from 3 - 6.
• There is no real vertical section, and no suitable indicator for this titration.
• The last slight increase is relatively low (just above the middle) when compared to a strong base, because a weak base has a low pH. The slight increase in pH tends to be from 8 -10.

Question 69

5.1.3 o) Outline the techniques and procedures used when measuring pH with a pH meter.

Answer 69

pH meters have a probe that you put into your solution and a digital display that shows the reading. Before you use a pH meter, it must be calibrated properly:
Place the probe of the pH meter in the following standard solutions, each with a known pH value, and adjust the reading to ensure that it displays the correct pH:
- 4
- 7 (i.e distilled/deionised water)
- 10