acids, bases and buffers

Question 1

What is a Brønsted–Lowry base?

Answer 1

A proton acceptor

Question 2

What is a Brønsted–Lowry acid?

Answer 2

A proton donor

Question 3

What does it mean when an acid is described as Monobasic?

Answer 3

A monobasic acid can release one proton per molecule.

Question 4

What does it mean when an acid is described as Dibasic ?

Answer 4

A dibasic acid can release two protons per molecule.

Question 5

What does it mean when an acid is described as tribasic

Answer 5

A tribasic acid can release three protons per molecule.

Question 6

Give an example of a tribasic acid.

Answer 6

H3PO4 / Phosphoric acid

Question 7

What is meant by a conjugate acid–base pair?

Answer 7

Two species that can be converted to one another by transfer of a proton

Question 8

What is the conjugate base of ethanoic acid, CH3COOH?

Answer 8

Ethanoate ion / CH3COO–

Question 9

What is the conjugate acid of
a) OH–?
b) NH3?

Answer 9

a) H2O
b) NH4+

Question 10

Give the equation to show the reaction of sulfuric acid with calcium carbonate.

Answer 10

H2SO4 + CaCO3 -> H2O + CO2 + CaSO4

Question 11

What is meant by a ‘strong acid’?

Answer 11

An acid that completely dissociates in aqueous solution

Question 12

Sulfuric acid is a strong acid. Give an equation to show the
dissociation of sulfuric acid in aqueous solution.

Answer 12

• H2SO4 -> H+ + HSO4 - AND HSO4– -> H+ + SO4 2–
  OR
• H2SO4 -> 2H+ + SO4

Question 13

An acid reacts with a base/alkali/carbonate in what type of
reaction?

Answer 13

A neutralisation reaction

Question 14

Write an ionic equation to show the reaction of hydrochloric
acid with sodium hydroxide.

Answer 14

H+ + OH- -> H2O

Question 15

Write the general expression for the acid dissociation
constant, Ka.

Answer 15

Ka =
[H+][A−]
[HA]

Question 16

Give the expression for pKa.

Answer 16

pKa = −log10Ka

Question 17

Calculate the pKa for an acid with an acid dissociation
constant of 6.70 × 10–4 mol dm–3

Answer 17

pKa = -log10 Ka = -log10 (6.70×10–4) = 3.17

Question 18

Give the equation for calculating pH from hydrogen ion
concentration.

Answer 18

pH = −log10[H+]

Question 19

Calculate the pH for an acid that has an [H+

] concentration of

1.995 × 10–3 mol dm–3
.

Answer 19

pH = -log10[H+] = -log10[1.995×10–3] = 2.70

Question 20

A solution of H2SO4 has a concentration of 7.60 × 10–3 mol dm–3
.

Calculate the pH of the acid.

Answer 20

H2SO4 is a strong dibasic acid, therefore [H+] = 2 × [H2SO4]

pH = -log10[H+] = -log10[2 × 7.6×10–3] = 1.818

pH= 1.82

Question 21

How can concentration of [H+] for an acid be calculated
from pH?

Answer 21

[H+] = 10−pH

Question 22

Calculate the [H+] concentration for an acid with a pH of
3.82.

Answer 22

[H+] = 10–pH = 10–3.82 = 1.51 × 10–4mol dm–3

Question 23

Ethanoic acid is a weak acid. Write an equation to show the
partial dissociation of ethanoic acid.

Answer 23

CH3COOH(aq) ⇌ H+(aq) + CH3COO– (aq)

Question 24

A sample of 0.100 M ethanoic acid is found to have
Ka = 1.50 × 10–6 mol dm–3 at 25.0 °C. Determine its pH.

Answer 24

Ka =[H+][CH3COO−]
[CH3COOH]
=
[H+]2
[CH3COOH]

[H+]2 = Ka × [CH3COOH]
[H+] = √Ka × [CH3COOH] = √1.5 × 10−6 × 0.1 = 3.87 × 10−4
pH = -log[H+] = -log(3.87x10–4) = 3.41

Question 25

Give an equation and value for the ionic product of water at 25.0 °C.

Answer 25

Kw = KC × [H2O] = [H+] [OH-] = 1.00× 10-14 mol2 dm-6

Question 26

NaOH is a strong base. Using the relationship between Kw, [OH–(aq)] and [H+(aq)], calculate the pH of a solution of 0.750 mol dm–3 NaOH.

Answer 26

[H+] =Kw/[OH−] = 1.0 × 10−14/0.75 = 1.33 × 10−14 pH = −log10(1.33 × 10−14) = 13.9

Question 27

Calculate the concentration of OH– at 25.0 °C in a solution with a pH of 13.39.

Answer 27

[H+] = 10−pH = 10−13.39 = 4.07 × 10−14 [OH−] =Kw/[H+] = 1.00 × 10−14/4.07 × 10−14 = 0.2457 = 0.246 mol dm−3

Question 28

Why is the use of the assumption [HA]initial = [HA] equation considered a limitation for some weak acids?

Answer 28

Some weak acids are stronger than others. The assumption that [HA]initial = [HA]equation relies on the limited dissociation of HA into H+ and A– ions in weak acids. The stronger the weak acid, the less accurate this assumption is, because more of HA will be dissociated.

Question 29

What does a buffer do?

Answer 29

A buffer is a solution that minimises a change in pH on addition of a small amount of H+ions or OH– ions.

Question 30

Describe two ways a buffer can be formed.

Answer 30

A buffer can be made from a solution of a weak acid and the salt of the weak acid, e.g. ethanoic acid and sodium ethanoate OR from an excess of a solution of a weak acid with a strong alkali, e.g. ethanoic acid and NaOH.

Question 31

A buffer solution at 25.0 ̊C contains 0.500 mol dm–3 of a weak acid and 0.750 mol dm–3 of the salt of the weak acid. Determine the concentration of H+ ions and hence calculate the pH of the buffer solution given that the Ka of the weak acid is 1.15 × 10–6 mol dm–3 .

Answer 31

[H+] = Ka ×[HA]/[A−] = 1.15 × 10−6 ×0.500/0.750 = 7.67 × 10−7 −log10[H+] = −log10(7.67 × 10−7) = 6.115 pH = 6.12

Question 32

Describe the action of a buffer in terms of conjugate acid and base pairs on addition of H+

Answer 32

* Adding H+ ions increases the concentration of H+ ions in the solution. * Conjugate base reacts with the added H+ ions. * The position of equilibrium moves to reduce the number of H+ ions as part of the reaction.

Question 33

With the help of an equation, explain how buffers act to maintain the pH of blood.

Answer 33

* Carbonic acid (H2CO3)–Hydrogencarbonate ion (HCO3–) buffer solution controls pH of blood. * If pH decreases ( [H+] increases): * HCO3–ions react with added H+ ions and position of equilibrium shifts left to maintain a stable pH. * If pH increases ( [OH–] increases): * H+ ions react with OH– ions. Position of equilibrium shifts right to regenerate H+ and maintain a stable pH.

Question 34

What is meant by the ‘equivalence point’ for a titration?

Answer 34

The point at which a volume of a solution reacts exactly with the volume of another solution

Question 35

The point at which a volume of a solution reacts exactly with the volume of HA ⇌ H+ + A– another solution

Answer 35

* HA and A– have different colours (or one is coloured and one isn’t). * Adding H+ shifts the equilibrium left, forming more HA of one colour (colour at low pH). * Removing H+ shifts the equilibrium right, forming more A– of a different colour (colour at high pH).

Question 36

Why can’t all titration reactions be followed with an indicator?

Answer 36

Weak acid–weak base titrations do not have a vertical section in the graph, and do not give a sharp change in pH at the equivalence point. This means that there will not be a sharp change in colour at the equivalence point, and it cannot be followed with an indicator.

Question 37

What is a pH meter and how is it used?

Answer 37

A pH meter measures the pH of a solution by submerging an electrode in the solution being measured. A voltage is produced dependent on [H+].

Question 38

What is involved in calibrating a pH meter?

Answer 38

The pH meter must first be calibrated by inserting it into buffers of known pH and plotting a calibration curve. This means that any difference between thepH meter reading and the true value can be determined. Any readings made by the pH meter can then be adjusted by adding or subtracting this difference.

Question 39

What are some of the advantages of using a pH meter instead of indicators?

Answer 39

A pH meter typically gives a reading to two decimal places and is, therefore, much more accurate than an indicator or litmus paper. A pH meter not only allows accurate pH measurement, but also enables continuous pH measurement throughout an experiment.