acids, bases and buffers Flashcards

1
Q

What is a Brønsted–Lowry base?

A

A proton acceptor

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

What is a Brønsted–Lowry acid?

A

A proton donor

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

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

A

A monobasic acid can release one proton per molecule.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

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

A

A dibasic acid can release two protons per molecule.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

What does it mean when an acid is described as tribasic

A

A tribasic acid can release three protons per molecule.

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

Give an example of a tribasic acid.

A

H3PO4 / Phosphoric acid

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

What is meant by a conjugate acid–base pair?

A

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

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

What is the conjugate base of ethanoic acid, CH3COOH?

A

Ethanoate ion / CH3COO–

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

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

A

a) H2O
b) NH4+

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

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

A

H2SO4 + CaCO3 -> H2O + CO2 + CaSO4

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

What is meant by a ‘strong acid’?

A

An acid that completely dissociates in aqueous solution

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

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

A
  • H2SO4 -> H+ + HSO4 - AND HSO4– -> H+ + SO4 2–
    OR
  • H2SO4 -> 2H+ + SO4
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

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

A

A neutralisation reaction

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

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

A

H+ + OH- -> H2O

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

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

A

Ka =
[H+][A−]
[HA]

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Give the expression for pKa.

A

pKa = −log10Ka

17
Q

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

A

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

18
Q

Give the equation for calculating pH from hydrogen ion
concentration.

A

pH = −log10[H+]

19
Q

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

] concentration of

1.995 × 10–3 mol dm–3
.

A

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

20
Q

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

Calculate the pH of the acid.

A

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

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

pH= 1.82

21
Q

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

A

[H+] = 10−pH

22
Q

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

A

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

23
Q

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

A

CH3COOH(aq) ⇌ H+(aq) + CH3COO– (aq)

24
Q

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.

A

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

25
Q

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

A

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

26
Q

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.

A

[H+] =Kw/[OH−]

1.0 × 10−14/0.75 = 1.33 × 10−14

pH = −log10(1.33 × 10−14) = 13.9

27
Q

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

A

[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

28
Q

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

A

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.

29
Q

What does a buffer do?

A

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

30
Q

Describe two ways a buffer can be formed.

A

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.

31
Q

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
.

A

[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

32
Q

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

A
  • 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.
33
Q

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

A
  • 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.
34
Q

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

A

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

35
Q

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

A
  • 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).
36
Q

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

A

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.

37
Q

What is a pH meter and how is it used?

A

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+].

38
Q

What is involved in calibrating a pH meter?

A

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.

39
Q

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

A

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.