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Chemistry > Chapter 25 Equilibria > Flashcards

Chapter 25 Equilibria Flashcards

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1
Q

Define conjugate acid.

A
  • ex-base
  • species that donates a proton
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2
Q

Define conjugate base.

A
  • ex-acid
  • species that can accept a proton
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3
Q

What is a conjugate acid-base pair?

A
  • a pair of reactants & products
  • that are linked to each other by the transfer of a proton
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4
Q

Define pH.

A
  • a measure of the concentration of hydrogen ions (H+) in solution
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5
Q

Define pH mathematically.

A

pH = -log10 [H+]

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6
Q

What is the relationship of [H+] & pH?

A
  • inversely propotional
  • [H+] increases, pH decreases
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7
Q

How to calculate the pH of a strong acid?

A
  1. Calculate [H+].
  2. pH = -log10 [H+]
    - The stronger the acid, the higher the concentration of H+ ions in the solution, the lower the pH.
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8
Q

Define Kw.

A
  • ionic product of water
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9
Q

Define Kw mathematically.

A

Kw = [H+][OH-]

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10
Q

What is Kw of pure water at 25°C?

A

1.00 x 10^-14 mol^2 dm^-6

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11
Q

Define pKw mathematically.

A

pKw = -log10 [Kw]

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12
Q

What is pKw at 25°C?

A

pKw = 14

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13
Q

What is the concentration of [H+] at 25°C?

A

1.0 x 10^-7 mol dm^-3
- [H+] = [OH-]
- pH = 7

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14
Q

State the equilibrium in pure water. Explain how the equilibrium determines properties of water.

A
  • H2O <> H+ + OH-
  • This is why water is able to act as either an acid or base.
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15
Q

Explain the effect of temperature on pH of water.

A
  • Formation of H+ & OH- ions is an endothermic process.
  • When temperature increases, equilibrium shifts right.
  • [H+] & [OH-] increases.
  • [H+] increases, pH decreases. But water does not taste acidic because [H+] =[OH-]
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16
Q

Relationship of pH with [OH-].

A
  • the higher the pH
  • the higher the [OH-] ions in solution
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17
Q

How to calculate the pH of a strong base?

A
  1. Kw = [H+][OH-] to calculate [H+].
  2. pH = -log10 [H+]
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18
Q

Define Ka.

A
  • Acidic dissociation constant.
  • Equilibrium constant for the dissociation of a weak acid at 298K
19
Q

State the equilibrium expression for a weak acid, HA.

A

HA (aq) ⇌ H+ (aq) + A- (aq)

20
Q

Define Ka mathematically.

A

Ka = [H+][A-] / [HA]
- Ka = [H+]^2 / [HA]
- [H+] = square root of (Ka x HA)

21
Q

What are the assumptions made for the Ka expression?

A
  1. Concentration of H+ ions due to the ionisation of water is negligible.
  2. The dissociation of the acid is so small that the concentration of HA is approximately the same as the concentration of A-.
22
Q

What does a high value of Ka mean?

A
  • equilibrium position lies to the right
  • acid is almost completely ionised
  • acid is strongly acidic
23
Q

What does a low value of Ka mean?

A
  • equilibrium position lies to the left
  • acid is only slightly ionised
  • acid is weakly acidic
24
Q

Define pKa mathematically.

A

pKa = -log10 [Ka]

25
Q

Relationship between pKa & strength of the acid.

A
  • the less positive the pKa value
  • the more acidic the acid is
26
Q

What is a buffer solution?

A

A solution in which the pH does not change significantly when a small amount of acid or alkali is added.

27
Q

What does a buffer solution consist of ?

A
  • A buffer can consist of weak acid - conjugate base or weak base - conjugate acid.
  • a weak acid & one of its salt
  • E.g. ethanoic acid & sodium ethanoate (acidic buffer)
  • E.g. ammonia & ammonium chloride
28
Q

The 7 strong acids?

A
  • H2SO4
  • HNO3
  • HCl
  • HBr
  • HI
  • HClO4
  • HClO3
    If any of these are present, no buffer.
29
Q

State the equilibrium reactions in an ethanoic acid & sodium ethanoate buffer solution.

A

CH3COOH <> H+ + CH3COO-
CH3COONa <> Na+ + CH3COO-

  1. Ethanoic acid is a weak acid & partially ionises in solution to form a relatively low concentration of ethanoate ions.
  2. Sodium ethanoate is a salt which fully ionises in solution.
30
Q

Why does an ethanoic acid & sodium ethanoate buffer solution contain high [CH3COOH] & [CH3COO-]?

A
  • There are reserve supplies of the acid (CH3COOH) and its conjugate base (CH3COO-).
  • The buffer solution contains relatively high concentrations of CH3COOH (due to ionisation of ethanoic acid) and CH3COO- (due to ionisation of sodium ethanoate)
31
Q

Describe what happens when H+ ions are added to a CH3COOH & CH3COONa buffer solution.

A
  • Equilibrium position shifts to the left as H+ ions react with CH3COO- ions to form more CH3COOH until equilibrium is re-established.
  • A large reserve supply of CH3COO- the concentration of CH3COO- in solution doesn’t change much (because it reacts with the added H+ ions).
  • A large reserve supply of CH3COOH the concentration of CH3COOH in solution doesn’t change much (CH3COOH is formed from the reaction of CH3COO- with H+ ions).
  • pH remains reasonably constant.
32
Q

Describe what happens when OH- ions are added to a CH3COOH & CH3COONa buffer solution.

A
  • The OH- ions react with H+ ions to form water.
  • OH- (aq) + H+ (aq) → H2O (l)
  • Equilibrium position shifts to the right and more CH3COOH molecules ionise to form more H+ and CH3COO- until equilibrium is re-established.
  • A large reserve supply of CH3COOH so the concentration of CH3COOH in solution doesn’t change much when CH3COOH dissociates to form more H+ ions.
  • A large reserve supply of CH3COO- so the concentration of CH3COO- in solution doesn’t change much.
  • pH remains reasonably constant.
33
Q

Formula for calculating pH of a buffer solution.

A

pH = pKa + log10 [salt] / [acid]

34
Q

Define solubility product, Ksp.

A
  • The product of the concentrations of each ion in a saturated solution of a sparingly soluble salt
  • At 298 K
  • Raised to the power of their relative concentrations
35
Q

Ksp for:
C (s) <> aA^x+ (aq) + bB^y- (aq)

A

Ksp = [A^x+ (aq)]^a [B^y- (aq)]^b

36
Q

Smaller Ksp indicates?

A

Lower solubility of the salt

37
Q

Saturated solution

A

A solution that contains the maximum amount of dissolved salt

38
Q

Common Ion Effect

A

The solubility of a sparingly soluble salt will decrease when the solution already contains one of its ions.

39
Q

What is Ksp affected by?

A
  • Temperature.
  • When temperature increases, Ksp increases.
40
Q

The ionic product, K = [A^y+]^x [B^x-] ^y is same as the expression of Ksp. However, the concentrations used to calculate K are the initial values before precipitation occurs or before equilibrium is reached.

State what happens to the solution if:
- K = Ksp
- K < Ksp
- K > Ksp

A
  • Solution is saturated
  • Solution is not saturated
  • Precipitation occurs
41
Q

Define partition coefficient, Kpc.

A

Ratio of the concentrations (of a solute between) 2 solvents (at equilibrium).

42
Q

Describe & explain the uses of buffer solutions, including the role of HCO3- in controlling pH in blood. State the chemical equation.

A
  • pH of blood is maintained within a range of 7.35 to 7.45 by using the buffer in blood plasma.
  • CO2 + H2O <> H+ + HCO3-
  • If [H+] ion increases, position of equilibrium shifts to the left. H+ combines with HCO3- ions to form CO2 & H2O until equilibrium is restored.
  • If the [H+] ion decreases, the position of this equilibrium shifts to the right. CO2 & H2O combine to form H+ & HCO3- ions until equilibrium is restored.
43
Q

Titration curve endpoints & equivalence points for:
- Strong acid & Strong base
- Strong acid & Weak Base
- Weak acid & Strong base

A
  • Endpoint from pH 3 - 11. Equivalence point at pH 7.
  • Endpoint from pH 3 - 7. Equivalence point at pH 5.
  • Endpoint from pH 7 - 11. Equivalence point at pH 9.
44
Q

When a strong base is added to a weak acid, what happens at the point halfway to the equivalence point? (5)

A
  • Half of the acid has been neutralised.
  • For a weak acid, Ka = [H+] [A-] / [HA]
  • At half neutralisation point, [HA] = [A-], therefore Ka = [H+].
  • pKa = pH
  • At half neutralisation point, the mixture is a buffer solution because both the weak acid & conjugate base are present in equal & significant quantities.

Chemistry (13 decks)

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