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chemistry > O ijklm: Acid-base equilibria > Flashcards

O ijklm: Acid-base equilibria Flashcards

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1
Q

Bronsted-Lowry theory

Define:

  • Acids
  • Bases
A
  • Acids = proton donors
  • Bases = proton acceptors
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2
Q

Write the formula for the conjugate base of ethanoic acid.

A

CH3COO-

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3
Q

Write the formula for the conjugate base of:

  • Hydrochloric acid
  • Chloric(VII) acid
  • Sulfuric acid
  • Hydrogen sulfide
  • Water
A
  • HCl → Cl-
  • HClO4 →ClO4-
  • H2SO4 →HSO4-
  • H2S → SH-
  • H2O → OH-
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4
Q

Water may act as an acid or base. Write 2 equations, including hydrochloric acid and ammonia, to demonstrate this.

A

Acting as a base: accepting a proton from HCl

H2O(l) + HCl(aq) → H3O+(aq) + Cl-(aq)

Acting as an acid: donating a proton to ammonia

H2O(l) + NH3(aq) → OH-(aq) + NH4+(aq)

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5
Q
  • What equation is used to calculate pH?
  • How can [H+] be calculated from a pH value?
A

pH = -log10[H+]

[H+] = 10-pH

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6
Q

What is the [H+] in water?

A

pH = 7

-log[H+] = 7

[H+] = 10-7 mol dm-3

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7
Q

What can be said of [H+] in the case of a solution with pH < 0?

A

pH < 0

-log[H+] < 0

log[H+] > 0

[H+] > 100

[H+] > 1 mol dm3

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8
Q

Compare strong and weak acids. Use equations to demonstrate the comparison.

A

Strong acids dissociate protons completely in aqueous solution:

HA + H2O → H3O+ + A-

HA → H+ + A-

Weak acids do not dissociate protons completely in aqueous solution:

HA + H2O ⇌ H3O+ + A-

HA ⇌ H+ + A-

The further to the left the eq. position lies, the weaker the acid

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9
Q

Write equations to demonstrate why solutions of carbon dioxide are acidic.

A

CO2 (aq) + H2O ⇌ H+(aq) + HCO3-(aq)

HCO3-(aq) ⇌ H+(aq) + CO32-(aq)

Weak acidity; equilibrium lies to the left.

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10
Q
  • Write an equation for the equilibrium constant of a weak acid, used to find [H+].
  • Name this constant
A

HA ⇌ H+ + A-

Ka = [H+] [A-] / [HA]

Ka, acidity constant / acid dissociation constant

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11
Q

To find [H+] for a weak acid by calculating Ka, what 2 assumptions are made for simplicity?

A

1: [H+] = [A-]

Negate small number of protons added by ionisation of water: H2O ⇌​ H+ + OH-

2: [HAeq] = [HAinitial]

Negate fraction of HA which have dissociated H+ - small since weak acid

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12
Q

To find [H+] for a weak acid by calculating Ka, the following assumptions are made. When are they not good assumptions?

1: [H+] = [A-] Negate small number of protons added by ionisation of water: H2O ⇌​ H+ + OH-

2: [HAeq] = [HAinitial] Negate fraction of HA which have dissociated H+

A
  1. For very dilute solutions
  2. For stronger acids
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13
Q

Calculate the pH of a solution of methanoic acid made up by adding 0.25 mol to 100 cm3 of water. Ka at 298 K = 1.6 x 10-4 mol dm-3.

A

Ka = [H+]2 / [HA]

[H+]2 / (0.25/0.1) = 1.6 x 10-4

[H+] = √(1.6 x 10-4 x 2.5)

pH = -log(√(1.6 x 10-4 x 2.5)) = 1.7

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14
Q

How do values of Ka compare for strong and weak acids?

A

For stronger acids, [HA] is lower and [H+] is greater, so Ka is greater.

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15
Q

Since values of Ka for weak acids can be small, a log scale pKa is often used. How is pKa calculated?

A

pKa = -log10(Ka)

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16
Q

The weaker the acid, the __ the value of pKa

A

Greater

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17
Q

Define an:

  • Acidic solution
  • Alkaline solution
A
  • Acidic: [H+] > [OH-]
  • Alkaline: [OH-] > [H+]
18
Q

What is a strong base?

A

One which produces OH- ions completely in aqueous solution.

19
Q
  1. What is the ionisation product of water?
  2. How is it represented symbolically?
A
  1. [H+] [OH-], i.e. the product of the products of ionisation of water
  2. Kw
20
Q

Give the value and unit of the ionisation product of water, Kw, at 298 K.

A

Kw = 1 x 10-14 mol2dm-6

21
Q

Explain why the expression for Kw is [H+] [OH-].

A

Ka = [H+] [OH-] / [H2O]

Water is in excess so has no effect on equilibrium. Therefore, it is omitted:

Kw = [H+] [OH-]

[H2O] is ignored because it’s so large that it makes values small, and has a practically identical effect on all values

22
Q

Calculate the pH of 0.1 mol dm-3 NaOH(aq).

A

Strong base; dissociates completely. Neglect small amount of OH- formed from ionisation of water. So:

[OH-] = 0.1 mol dm-3

Kw = [H+] [OH-]

1 x 10-14 = 0.1 [H+]

pH = -log(1 x 10-14 / 0.1) = 13

23
Q

Calculate the pH of a 0.100 mol dm-3 barium hydroxide solution at 298 K.

A

Ba(OH)2 → Ba2+ + 2OH-

Strong base; dissociates completely. Neglect small amount of OH- formed from ionisation of water. So:

[OH-] = 0.2 mol dm-3

KW = [H+] [OH-]

1 x 10-14 = 0.2 [H+]

pH = -log(1 x 10-14 / 0.2) = 13.3

Don’t square [OH-] due to the 2OH-; this is not done for Kw as it is for Kc and Ksp

24
Q
A

a) Pink. H+ is removed by reaction with OH-, so equilibrium position shifts to right to restore the constant Ka, so [In-] increases.
b) Ka = [H+] [In-] / [HIn]
c) pKa = 9.3 so Ka = 10-9.3

At end point, [In-] = [HIn] so Ka = [H+] = 10-9.3

pH = -log(10-9.3) = 9.3

25
Q

Write the formula for a hydrogencarbonate ion.

A

HCO3-

26
Q

What is a buffer solution?

A

A solution which resists pH changes for small additions of acid or alkali.

27
Q

What do buffer solutions usually comprise of?

A
  • Weak acid and one of its salts (e.g. CH3COOH + CH3COONa)
  • Weak base and one of its salts (e.g. NH3 + NH4Cl)
28
Q

The action of a buffer solution depends on the weak acid equilibrium:

HA(aq) ⇌ H+(aq) + A-(aq)

What 2 assumptions are required to explain how buffers counteract small pH changes?

A
  • Almost no weak acid (HA) molecules dissociate
  • (And therefore) all A- ions come from salt; negate small amount from weak acid (HA)
29
Q

Explain how the pH of a buffer solution is reestablished when acid is added.

A

HA(aq) ⇌ H+(aq) + A-(aq)

  • Conjugate base suppresses ionisation of acid: H+ ions added react with A-, forming HA + water
  • H+ ions removed from solution
  • Equilibrium shifts to left
  • pH reestablished
30
Q

Explain how the pH of a buffer solution is reestablished when alkali is added.

A

HA(aq) ⇌ H+(aq) + A-(aq)

  • OH- ions added react with and remove H+ ions, forming H2O
  • Weak acid HA ionises further, regenerating H+
  • Equilibrium moves to right
  • pH reestablished
31
Q

Explain how the expression for Ka can be used to work out the pH of a buffer solution.

A

Ka = [H+] [A-] / [HA]

Assume that:

  • Almost no weak acid molecules dissociate
  • (And therefore) All A- ions come from salt

Ka = [H+] [salt] / [acid]

[H+] = Ka [acid] / [salt]

pH = -log(Ka [acid] / [salt])

32
Q

What factors does the pH of a buffer solution depend upon?

A
  • The value of Ka
  • The ratio of acid to salt

[H+] = Ka x [acid] / [salt]

33
Q

Explain why the pH of a buffer solution is not affected by dilution.

A

pH = -log[H+] = -log (Ka x [acid] / [salt])

  • Ka is a constant, so it is unaffected by dilution (overall change in concentration does not alter equilibrium position)
  • Dilution reduces concentrations of acid + salt by same proportion, so ratio of concentrations is unchanged
  • These factors are what affect [H+]; pH depends on H+ so is unchanged
34
Q

Calculate the pH of a buffer solution containing 0.1 mol dm-3 ethanoic acid and 0.2 mol dm-3 sodium ethanoate. Ka for ethanoic acid is 1.7 x 10-5 mol dm-3 at 298 K.

A

Ka = [H+] [A-] / [HA] = [H+] x [salt] / [acid]

[H+] = Ka x [acid] / [salt]

pH = -log(Ka x [acid] / [salt])

= -log(1.7 x 10-5 x 0.1 / 0.2) = 5.1

35
Q

A mass of a solid base is added to an acid of known concentration to make up a buffer solution. The mass of base, concentration of acid and volume of solution are used to work out the pH.

What assumption is made in doing this?

A

The total volume of the solution is not altered when the base is added.

36
Q

What is the relationship between Ka, Kb and Kw?

Kb is the weak base ionisation constant.

A

Kw = Kb x Ka

37
Q

Suggest why alkaline solutions dissolve CO2 more effectively than water.

A

CO2 is weakly acidic:

CO2(g) ⇌ CO2(aq)

CO2(aq) + H2O ⇌ H+(aq) + HCO3-(aq)

HCO3-(aq) ⇌ H+(aq) + CO32-(aq)

Overall: CO2(g) + H2O ⇌ CO32-(aq) + 2H+(aq)

  • OH- ions from alkali react with H+ ions
  • [H+] decreases
  • Equilibrium position moves right to restore Ka since it is a constant
  • More CO2 gas dissolves
38
Q

Sodium ethanoate is used in buffer solutions.

Calculate the mass of sodium ethanoate that would need to be added to 1.0 mol dm-3 ethanoic acid in order to make 250 cm3 of buffer solution of pH 4.80.

Ka of ethanoic acid is 1.74 x 10-5 mol dm-3.

A

[H+] required = 10-4.8

Ka = [H+] [ethanoate] / [acid]

1.74 x 10-5 = 10-4.8 [ethanoate] / 1

[ethanoate] = 1.74 x 10-5 / 10-4.8 = 1.098

Mol CH3COONa = conc x vol = 1.098 x 0.25 = 0.274 mol

Mass CH3COONa = mol x Mr = 0.274 x 82 = 23 g

39
Q
A

Acid: NaH2PO4 Salt: Na2HPO4

[H+] = 10-7 mol dm-3

Ka = [H+] [salt] / [acid] → [acid] = [H+] [salt] / Ka

Conc acid = (10-7 x 0.1) / (6.2 x 10-8) = 0.161 mol dm-3

Mol acid = conc x vol = 0.161 x 1 = 0.161

Mass acid = mol x Mr = 0.161 x 120 = 19 g

40
Q

A student titrates sodium hydroxide solution against hydrochloric acid and records a titre of 7.5 cm3.

They suggest that less than 7.5 cm3 of sodium hydroxide solution would be needed to neutralise H2PO4- of the same concentration, since it is a weak acid.

Discuss the student’s suggestion.

A
  • For a weak or strong acid of a given concentration, the total number of protons which will eventually be dissociated, and react with the base, is the same
  • It takes longer for a weak acid to reach the end-point, but the same volume of base is required
41
Q

Which statement is true of buffer solutions?

  1. A buffer does not change in pH when small amounts of acid or alkali are added.
  2. An acidic buffer is a mixture of a weak acid and a solution of its salt.
  3. All buffers are acidic.
  4. The pH of a buffer depends only on the ratio of its components’ concentrations.
A

2

  • 1 incorrect since pH of buffers does vary - just less than pH of things that aren’t buffers*
  • 3 incorrect since they can be basic*
  • 4 incorrect since also dependent on Ka*
42
Q
A

B

chemistry (50 decks)

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