Chapter 25 Equilibria Flashcards
complete
1
Q
Define conjugate acid.
A
- ex-base
- species that donates a proton
2
Q
Define conjugate base.
A
- ex-acid
- species that can accept a proton
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
4
Q
Define pH.
A
- a measure of the concentration of hydrogen ions (H+) in solution
5
Q
Define pH mathematically.
A
pH = -log10 [H+]
6
Q
What is the relationship of [H+] & pH?
A
- inversely propotional
- [H+] increases, pH decreases
7
Q
How to calculate the pH of a strong acid?
A
- Calculate [H+].
- pH = -log10 [H+]
- The stronger the acid, the higher the concentration of H+ ions in the solution, the lower the pH.
8
Q
Define Kw.
A
- ionic product of water
9
Q
Define Kw mathematically.
A
Kw = [H+][OH-]
10
Q
What is Kw of pure water at 25°C?
A
1.00 x 10^-14 mol^2 dm^-6
11
Q
Define pKw mathematically.
A
pKw = -log10 [Kw]
12
Q
What is pKw at 25°C?
A
pKw = 14
13
Q
What is the concentration of [H+] at 25°C?
A
1.0 x 10^-7 mol dm^-3
- [H+] = [OH-]
- pH = 7
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.
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-]
16
Q
Relationship of pH with [OH-].
A
- the higher the pH
- the higher the [OH-] ions in solution
17
Q
How to calculate the pH of a strong base?
A
- Kw = [H+][OH-] to calculate [H+].
- pH = -log10 [H+]
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
- Concentration of H+ ions due to the ionisation of water is negligible.
- 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
Relationship between pKa & strength of the acid.
- the less positive the pKa value
- the more acidic the acid is
26
What is a buffer solution?
A solution in which the pH does not change significantly when a small amount of acid or alkali is added.
27
What does a buffer solution consist of ?
- 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
The 7 strong acids?
- H2SO4
- HNO3
- HCl
- HBr
- HI
- HClO4
- HClO3
If any of these are present, no buffer.
29
State the equilibrium reactions in an ethanoic acid & sodium ethanoate buffer solution.
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
Why does an ethanoic acid & sodium ethanoate buffer solution contain high [CH3COOH] & [CH3COO-]?
- 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
Describe what happens when H+ ions are added to a CH3COOH & CH3COONa buffer solution.
- 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
Describe what happens when OH- ions are added to a CH3COOH & CH3COONa buffer solution.
- 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
Formula for calculating pH of a buffer solution.
pH = pKa + log10 [salt] / [acid]
34
Define solubility product, Ksp.
- 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
Ksp for:
C (s) <> aA^x+ (aq) + bB^y- (aq)
Ksp = [A^x+ (aq)]^a [B^y- (aq)]^b
36
Smaller Ksp indicates?
Lower solubility of the salt
37
Saturated solution
A solution that contains the **maximum** amount of dissolved salt
38
Common Ion Effect
The solubility of a sparingly soluble salt will decrease when the solution already contains one of its ions.
39
What is Ksp affected by?
- Temperature.
- When temperature increases, Ksp increases.
40
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
- Solution is saturated
- Solution is not saturated
- Precipitation occurs
41
Define partition coefficient, Kpc.
Ratio of the concentrations (of a solute between) 2 solvents (at equilibrium).
42
Describe & explain the uses of buffer solutions, including the role of HCO3- in controlling pH in blood. State the chemical equation.
- 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
Titration curve endpoints & equivalence points for:
- Strong acid & Strong base
- Strong acid & Weak Base
- Weak acid & Strong base
- 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
When a strong base is added to a weak acid, what happens at the point halfway to the equivalence point? (5)
- **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.
