3.1.12 Acids and Bases Flashcards
1
Q
What is a Brønsted–Lowry acid?
A
A substance that donates a proton (H⁺).
2
Q
What is a Brønsted–Lowry base?
A
A substance that accepts a proton (H⁺).
3
Q
What happens when an acid (HA) reacts with water?
(write an equation)
A
Acid donates a proton to water, forming H₃O⁺ and A⁻:
HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)
4
Q
What happens when a base (B) reacts with water?
(write an equation)
A
Base accepts a proton from water, forming BH⁺ and OH⁻:
B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)
5
Q
Can acids release protons without a base present?
A
No, proton transfer requires a base to accept the proton.
6
Q
Write an equation for the general proton transfer reaction between acid and base?
A
HA(aq) + B(aq) ⇌ BH⁺(aq) + A⁻(aq)
7
Q
What shifts acid-base equilibrium to the right?
A
Increasing [HA] or [B] - more products (BH⁺ and A⁻) formed.
8
Q
What shifts acid-base equilibrium to the left?
A
Increasing [BH⁺] or [A⁻] - more reactants produced.
9
Q
Finish equation
Acid + reactive metal →?
A
Salt + Hydrogen gas
10
Q
Finish equation
Acid + metal carbonate →?
A
Salt + water + carbon dioxide
11
Q
Finish equation
Acid + alkali (soluble base) → ?
A
Salt + water
12
Q
Finish equation
Acid + insoluble base (metal oxide) → ?
A
Salt + water
13
Q
When you react a weak acid with water, where does the equilibrium lie?
A
Far to the left
14
Q
When you react a strong acid with water where does the equilibrium lie?
A
Far to the right.
15
Q
What does it mean for water to be amphiprotic?
A
Can behave as both an acid and a base.
16
Q
What is the ionic product of water (Kw)?
A
Kw = [H⁺][OH⁻]
17
Q
Why is the concentration of water considered essentially constant in the Kc equation?
A
Water is present is vast excess compared to [H⁺] and [OH⁻]
18
Q
What is the value of Kw at 298 K in pure water?
A
Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶
19
Q
What happens to the concentrations of [H⁺] and [OH⁻] in pure water?
A
They are equal.
20
Q
What is the pH scale?
A
It is a logarithmic scale that measures the concentration of hydrogen ions in a solution.
21
Q
What is the formula for calculating pH?
A
pH=−log10 [H⁺]
22
Q
How do you calculate [H⁺] from pH?
A
[H⁺] = 10 (to the power of) −pH
23
Q
How does [H⁺] change with each unit decrease in pH?
A
It increases by a factor of 10.
24
Q
For strong acids, like HCl or HNO3, how does [H⁺] relate to the [acid]?
A
[H⁺] = [acid]
Strong acids fully ionise in solutions
25
What is the pH of a 0.005 mol dm⁻³ HCl solution?
pH = 2.30
26
Calculate the [H⁺] of a HCl solution with a pH of 1.60.
Give answer to 3 significant figures
[H⁺] = 2.51x10-2 moldm-3
27
In pure water at 25°C, what are the concentrations of [H⁺] & [OH⁻]?
Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶
Both are 1.0x10-7 moldm-3
28
What happens to Kw as temperature increases?
H₂O ⇌ H⁺ + OH⁻
Forward reaction is **endothermic**
Increasing temperature shifts equilibrium to **right**, as it favours endothermic forward reaction
Producing more H⁺ and OH⁻ ions, **increasing Kw**.
29
How does a change in [H⁺] or [OH⁻] affect Kw?
No change but equilibrium will shift to keep Kw consant.
30
For strong bases, like NaOH, how does [OH⁻] relate to [base]?
[OH⁻] = [base]
As the base fully ionises in water
31
What is the pH of a 0.025 mol dm⁻³ NaOH solution at 25°C?
[OH⁻] = 0.025 mol dm⁻³
Kw = [OH⁻] [H⁺]
Rearrange equation
[H⁺] = 1.0 x10-14 divided by 0.025
= 4.0x10-13 mol dm⁻³
do pH equation
pH = 12.40
32
What does the acid dissociation constant, Ka, represent?
Ka measures how much a weak acid dissociates in aqueous solution.
It is the equilibrium constant for the dissociation of a weak acid.
33
Write the Ka expression for a weak acid, HA.
Ka =
34
What does a larger Ka value indicate about an acid?
Stronger acid, disscociates more in the solution.
35
What are the units of Ka?
mol dm⁻³
36
What two assumptions are made when calculating pH of a weak acid using Ka?
[HA] at equilibrium ≈ [HA] initial
As ionisation of weak acid is so small that con of undissociated HA molecules at eqb is ≈ the same as inital concentration of acid.
[H⁺] at eqb ≈ [A⁻] at eqb
As ionisation of water is negligable
37
What is the simplified Ka expression based on assumptions?
same as normal Ka expression but it would be [H⁺] squared as numerator.
38
What is the pH of 0.0100 mol dm⁻³ ethanoic acid if
Ka = 1.75x10-5 mol dm⁻³ ?
[H⁺] = 4.18x10-4 mol dm⁻³
pH = 3.38
39
Given a propanoic acid solution's **pH is 2.89** and its Ka is **1.34x10-5 mol dm⁻³** at 298 K, calculate the acid's concentration. Give your answer to 3 significant figures.
[H⁺] = 1.29x10-3 mol dm⁻³
conc of acid = 0.124 mol dm⁻³
40
Given a hydrofluoric acid (HF) solution's **pH is 3.14** and its concentration is **0.100 mol dm⁻³**, calculate the Ka for hydrofluoric acid. Give your answer to 3 significant figures.
[H⁺] = 7.24 x 10-4 mol dm⁻³
Ka = 5.25 x 10-6 mol dm⁻³
41
How is pKa related to Ka (equation)?
pKa = -log(Ka)
42
How do you calculate Ka from pKa?
Ka = 10⁻ᵖᴷᵃ
43
What does a smaller pKa value indicate about the acid?
Stronger acid
44
45
Given the Ka of carbonic acid is 4.5 x10-7 mol dm⁻³, calculate its pKa. Give your answer to 2 significant figures.
pKa = 6.3
46
What determines the shape of a pH curve in a titration?
The strengths of the acid and base involved (strong/weak).
47
What determines the initial pH in a titration?
The strength of the acid.
Strong acids start lower in pH than weak acids.
48
What happens to pH in the early stages of titration?
pH changes slowly with small additions of base.
49
What is the equivalence point in a titration?
Vertical line on pH curve.
The point where moles of acid equal moles of base Complete neutralisation occurs.
50
What happens to pH at the equivalence point?
A very **rapid & large pH change** occurs
(except in weak acid/weak base titrations).
51
What determines the **final pH** of a titration?
The strength of the **base**.
Strong bases give higher final pH.
52
How do the pH curves differ between titration types + draw each curve?
* Strong acid/strong base - **Sharp** jump in pH
* Strong acid/weak base - More weak base is needed to reach equivalence point, **less steep** jump
* Weak acid/strong base - Less trong base needed to cause big pH change, **steep jump**, but higher starting pH
* Weak acid/weak base - **Gradual curve**, no sharp pH change
53
What does the choice of indicator depend on?
The **pH range** over which the indicator changes colour.
Whether the **indicator’s colour change range** falls within the steep, **vertical part** of the pH curve (the equivalence point).
54
What occurs at the half the equivalence point in a titration of a weak acid with a strong base and why?
pH = pKa
At half the equivalence point, exactly half of the acid (HA) has turned into its conjugate base (A⁻).
55
Why can’t indicators be used for weak acid/weak base titrations?
There’s **no sharp pH change** at the equivalence point, instead it is a gradual pH change.
A **pH meter** is needed instead.
56
What steps are needed to calculate concentration of an acid or base from titration?
1. Accurately measure the **neutralisation volume**.
2. **Repeat** the titration at least three times and calculate the average.
3. Ignore any results that don't match; all results should be **concordant**.
57
30.0 cm3 of 1.00 mol dm-3 HCl was required to neutralise 50.0 cm3 of NaOH solution:
HCl(aq) + NaOH(aq) ➔ NaCl(aq) + H2O(l)
Calculate the concentration of the sodium hydroxide solution. Give your answer to 3 significant figures.
mol of HCl = 1 x 0.03 = 0.03 mol
mol of NaOH neutralised = 0.03 mol
conc. of NaOH = 0.600 mol dm⁻³
58
36.25 cm3 of 0.200 mol dm⁻³ sodium hydroxide solution are added to 25.00 cm3 of 0.150 mol dm⁻³ hydrochloric acid.
Calculate the pH of the final solution at 25 ºC
Kw = 1.00 × 10–14 mol2 dm–6 at 25 ºC (5)
* mol of OH⁻ = 0.03625 x 0.2 = 7.25 x10-3 mol
mol of H⁺ = 0.025 x 0.15 = 3.75 × 10–3 mol
* Amount of XS OH⁻ = 7.25 × 10–3 – 3.75 × 10–3
= 3.50 × 10–3 mol
* [OH⁻] = 3.50 × 10–3 mol ÷ total vol (0.06125) = ÷ 0.0571
* [H⁺] = Kw ÷ 0.0571 = 1.75 × 10–13
* pH = 12.76
59
What is a buffer solution and what does it do?
A buffer is a solution that minimises alteration in pH when small quantities of acid or base are introduced
It doesn't completely stop pH change, but significantly reduces it.
60
What are the two main types of buffer solutions?
Acidic buffers and Basic buffers
61
What are acidic buffers made from?
A weak acid and one of its salts
(e.g. ethanoic acid and sodium ethanoate)
62
Write two equations to show how ethanoic acid and sodium ethanoate dissociate in an **acidic buffer**.
The weak acid only **slightly dissociates**:
CH₃COOH ⇌ H⁺ + CH₃COO⁻
The salt **fully dissociates** into its ions:
CH₃COONa → CH₃COO⁻ + Na⁺
63
How does an **acidic buffer** respond when a small amount of **acid** is added?
The H⁺ ion concentration increases
Most of the added H⁺ combines with CH₃COO⁻ ions to form CH₃COOH.
This shifts the equilibrium to the **left** (reverse direction), which reduces the H⁺ concentration to a value close to its original value.
64
How does an **acidic buffer** respond when a small amount of **base** is added?
The OH⁻ ions concentration increases
Most of the added OH⁻ ions react with the H⁺ ions (from the weak acid dissociation) to form water & removing H⁺ ions from the solution.
This shifts the equilibrium to the **right** (forward), causing more CH₃COOH to dissociate.
The [H⁺] increases until it is close to its original value, maintaining a stable pH.
65
What are basic buffers made from?
A weak base and one of its salt
(e.g. ammonia & ammonium chloride)
66
Write two equations to show how NH₃ and NH₄Cl dissociate in a basic buffer?
The salt fully dissociates:
NH₄Cl → NH₄⁺ + Cl⁻
Some of the ammonia molecules react with water:
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
67
How does a **basic buffer** respond when small quantities of **acid** is added?
The [H⁺] increases
Some of the H⁺ ions react with OH⁻ ions to form water
This causes the equilibrium (of the ammonia dissociation) to shift to the **right** to increase the [OH⁻]
& Some of the H⁺ ions react with NH₃ to form NH₄⁺
Both reactions help minimise pH.
68
How does a **basic buffer** respond when small quantities of a **base** is added?
The OH⁻ concentration increases
Most of the added OH⁻ will react with NH₄⁺ to form NH₃ and water.
This causes the equilibrium to shift to the **left** (reverse) to lower the [OH⁻], minimising the change in pH.
69
What are three practical applications of buffer?
* **Shampoos** - most contain a 5.5 pH buffer
* **Biological washing powders** - maintain optimal pH for enzymes.
* **Biological systems** - e.g. blood must remain at pH close to 7.4.
70
What assumptions are made when calculating the pH of a buffer solution using Ka?
* The salt of the conjugate base is fully dissociated - equilibrium [A-] is equal to initial concentration of salt
* HA is only slightly dissociated, so assume that equilibrium conc. is equal to its initial conc.
71
A buffer solution is made by adding 0.50 mol of ethanoic acid (CH3COOH) and 0.50 mol of sodium ethanoate (CH3COONa) to enough water to make 1.0 dm3 of solution. The Ka for ethanoic acid is 1.8 × 10-5 mol dm-3.
Calculate the pH of this buffer. Give your answer to 2 decimal places.
pH = 4.74
you dummy
do Ka expression
72
A buffer solution initially contains 0.50 mol of ethanoic acid (CH3COOH) and 0.50 mol of sodium ethanoate (CH3COONa) in 1.0 dm3 of solution. The Ka for ethanoic acid is 1.8x10-5 mol dm-3.
Calculate the new pH of this buffer after adding 10 cm3 of 2.0 mol dm-3 hydrochloric acid (HCl). Give your answer to 1 decimal place.
**The added H⁺ will react with CH₃COO⁻ to form CH₃COOH**
initial mol of CH₃COOH = 0.50
initial mol of CH₃COO⁻ = 0.50
mol of HCl = 0.01 x 2 = 0.02
mol of CH₃COOH after acid added = 0.52
mol of CH₃COO⁻ after base added = 0.48
[H⁺] = 1.949×10-5
pH = 4.7
73
Consider a buffer solution containing 0.40 mol dm-3 methanoic acid (HCOOH) and 0.60 mol dm-3 sodium methanoate (HCOONa) in 500 cm3 of solution. The Ka for methanoic acid is 1.6 x10-4 mol dm-3.
Calculate the new pH of this buffer after adding 25 cm3 of 0.30 mol dm-3 sodium hydroxide (NaOH). Give your answer to 1 decimal place.
**The added OH- react with HCOOH (acid) to form HCOO⁻**
initial mol of HCOOH = 0.2
initial mol of HCOO⁻ = 0.3
mol of OH- added = 0.0075
mol of HCOOH after base added = 0.1975
mol of HCOO⁻ after base added = 0.3075
[H⁺] = 1.001×10-4
pH = 4.0
