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A-level Chem > Acids, bases and buffers > Flashcards

Acids, bases and buffers Flashcards

1
Q

Definition of a Bronsted-Lowry acid.

A

Proton donor.

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2
Q

Definition of a Bronsted-Lowry base.

A

Proton acceptor.

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3
Q

Definition of a strong acid.

A

A proton donor that fully dissociates into its ions in solution.

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4
Q

Definition of a weak acid.

A

A proton donor that partially dissociates into its ions in a solution.

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5
Q

What do the terms monobasic, dibasic and tribasic refer to?

A

The total number of hydrogen ions in the acid that can be replaced per molecule in an acid-base reaction.

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6
Q

Definition of a conjugate acid-base pair.

A

A set of two species that transform into each other by loss or gain of a proton, i.e they consist of the same species apart from the H+.

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7
Q

What is the equation for find the pH?

A

pH = -log10[H+]

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8
Q

What is the equation to find [H+]?

A

10^-pH

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9
Q

Write a Kc expression for the dissociation of water.

A

*Refer to ABB notes.

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10
Q

What is the value of Kw at 25 degrees?

A

1 x 10^-14 mol^2dm^-6.

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11
Q

Calculate the pH of 0.11moldm^-3 of HCl(aq).

A

pH = 0.959

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12
Q

Calculate the pH of 2moldm^-3 HNO3(aq).

A

pH = -0.301

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13
Q

Calculate the pH of 0.5moldm^-3 of HCl(aq).

A

pH = 0.301

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14
Q

Cacluate the pH of 0.21moldm^-3 of H2SO4(aq).

A

pH = 0.377

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15
Q

Calculate the pH of 0.01moldm^-3 of H3PO4(aq).

A

pH = 1.52

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16
Q

Find [H+] in a solution of HCl(aq) of pH 2.05.

A

[H+] = 8.91 x 10^-3 moldm^-3

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17
Q

Find [H+] in a solution of CH3CH3COOH(aq) of pH 3.40.

A

[H+] = 3.98 x 10^-4 moldm^-3

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18
Q

Find [H+] in NaOH9aq) with a pH of 13.5.

A

[H+] = 3.16 x 10^-14 moldm^03

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19
Q

Find the pH of water at 298K given that Kw = 1 x10^-14 mol^2dm^-6 at 298K.

A

pH = 7.

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20
Q

Find the pH of water at 398K given that Kw = 2.5 x 10^-14 mol^2dm^-6.

A

pH= 6.80

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21
Q

Calculate the pH of 1moldm^-3 NaOH(aq).

A

pH = 14

22
Q

Calculate the pH of 0.03moldm^-3 of KOH(aq).

A

pH = 12.5

23
Q

Calculate the pH of 0.025moldm^-3 CaOH2(aq).

A

12.7

24
Q

What is the equation to calculate pKa?

A

pKa = -log10Ka

25
Q

What is the equation to calculate Ka?

A

Ka = 10^-pKa

26
Q

How does a larger Ka value affect the equilibrium?

A

The larger the numerical value of Ka, the more the equilibrium lies to the right, the greater the concentration of the dissociated ions H+ and A- (numerator), and the stronger the acid.

27
Q

How does a smaller Ka value affect equilibrium?

A

The smaller the numerical value of Ka, the more the eqilibrium is to the left, the smaller the concentration of the dissociated ions H+ and A- (nuerator, and the weaker the acid.

28
Q

Why is [H+] = [A-] only an approximation when calculating the pH of weak monobasic acids using Ka?

A

Water also dissociates forming a very small concentration of H+ but this is so small it can be considered to be negligible compared with the [H+] from the acid.

29
Q

When will the approximation: [HA]eqm = [HA]start (undissociated), be true when calculating the pH of a weak monobasic acid using Ka?

A

[HA]eqm = [HA]start - [H+]eqm. As the dissociation of weak acids is small, can assume [H+]eqm is very small so [HA]&raquo_space; [H+]eqm and the decrease in the concentration of HA from the dissociation is negligible.

30
Q

What is the simplification of the Ka expression when using the approximations to calculate the pH of a weak monobasic acid?

A

*Refer to ABB notes.

31
Q

Find the pH of a weak acid; HA, which has a concentration of 0.1moldm^-3 and a Ka of 1.7 x 10^-5 moldm^-3.

A

2.88

32
Q

Explain the limitation of the approximation: [H+] = [A-] when calculating the pH of a weak monobasic acid.

A

This assumes that the [H+] from water is negligible. However, at 25 degrees, the [H+] from the dissociation of water is 1 x 10^-7. For a very weak acid with a pH>6, the [H+] from the dissociation of water will be significant compared with the dissociation of the weak acid. For example: pH = 6.1, the [H+] = 10^-6.1= 7.94 x 10-5 moldm^-3.

33
Q

Explain the limitation of the approximation: [HA]eqm = [HA] start (undissociated) when calculating the pH of a monobasic weak acid.

A

This assumes [HA]&raquo_space; [H+]. This is not true when [H+] becomes significant. As the acid has dissociated more (stronger acid). Dissociation of water is significant. There is now a significant difference between [HA]eqm and [HA]start. So, the approximation is not valid for a stronger weak acid and for very dilute solutions.

34
Q

Definition of a buffer

A

A system that resists change in pH when small amounts of acid or base are added.

35
Q

What does a weak acid buffer solution contain?

A

A weak acid and the salt of a weak acid, for example: CH3COOH and CH3COO- 9ethanoic acid and ethanoate).

36
Q

Describe two methods of making a weak acid buffer solution.

A
  1. The weak acid and one of its salts. For example: CH3COOH and CH3COO-.
  2. Adding an alkali and an excess of the weak acid. For example: NaOH and CH3COOH.
37
Q

How does the buffer work on addition of acid? Use CH3COOH/CH3COO- as the conjugate base pair.

A

H+ from the acid reacts with CH3COO-.
CH3COO- + H+ …. CH3COOH
Equlibrium shifts to the left.
H+ ions being removed.

38
Q

How does the buffer work on addition of an alkali? Use CH3COOH/CH3COO- as the conjugate base pair.

A

Added OH- reacts with H+.
H+ + OH- ….H2O
Equilibrium shifts to the right.
CH3COOH …. CH3COO- + H+
H+ ions are being replaced.

39
Q

What is the Henderson-Hasselbach equation?

A

*Refer to ABB notes

40
Q

Calculate the pH of a buffer solution made from 0.60moldm^-3 propanoic acid and 0.80moldm^-3 moldm^-3 sodium propanoate.

A

pH = 3.99

41
Q

The pH of a buffer is desired to be 3.76. What concentration of methanoic acid must be mixed with 0.5moldm^-3 sodium methanoate to achieve this? Ka of methanoic acid is 1.78 x 10^-4.

A

[HCOOH] = 0.489 moldm^-3

42
Q

Write the reaction equation for water and carbon dioxide.

A

CO2 + H20 …/… H2CO3

43
Q

Write the dissociation equation for carbonic acid.

A

H2CO3 …/… H+ + HCO3-

44
Q

Describe the equilibrium of blood on addition of acid.

A

H+ from the acid reacts with HCO3-.
H+ + HCO3- … H2CO3
Equlibrium shifts to the left.
H+ ions are removed.

45
Q

Decribe the equlibrium of blood on addition of alkali.

A

OH- ions react with acid (the H+ of H2CO3).
H+ + OH- …. H2O
Equlibrium shifts to the right.
Realces the H+ ions.

46
Q

Definition of equivalnce point

A

The volume of one solution that exactly reacts with the volume of the other solution.

47
Q

What are the three key points for drawing pH curves?

A
  1. pH at the start.
  2. The volume added to reach the end point / neutralisation (the titre).
  3. pH at the end.
48
Q

Draw a titration curve for the reaction between a strong acid, e.g. HCl, and a strong base, e.g. NaOH.

A

*Refer to ABB notes.

49
Q

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

A

*Refer to ABB notes.

50
Q

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

A

*Refer to ABB notes.

A-level Chem (30 decks)

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