Acids and Bases
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- Created by: JosephPHS
- Created on: 30-01-23 09:21
Bronsted-Lowry Acid
Proton Donor
HA (aq) + H₂O (l) --> H₃O⁺ (aq) + A⁻ (aq)
A = Acid
HA (aq) + H₂O (l) --> H₃O⁺ (aq) + A⁻ (aq)
A = Acid
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Bronsted-Lowry Base
Proton Acceptor
B (aq) + + H₂O (l) --> BH⁺ (aq) + OH⁻ (aq)
B = Base
B (aq) + + H₂O (l) --> BH⁺ (aq) + OH⁻ (aq)
B = Base
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Effect of Strength of Acids
Strong acids fully, such as HCl, dissociate in water whereas weak acids, such as ethanoic acid, only partially dissociate in water
HCL --> H⁺ + Cl⁻
CH₃COOH ⇌ CH₃COO⁻ (aq) + H⁺ (aq)
HCL --> H⁺ + Cl⁻
CH₃COOH ⇌ CH₃COO⁻ (aq) + H⁺ (aq)
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Monoprotic Acids vs Diprotic acids
Monoprotic acids have 1 H that dissociates such as HCl, Diprotic acids have 2 H that dissociate such H₂SO₄
Note acids such as CH₂COOH are monoprotic as only the H on the carboxyl group will dissociate
Note acids such as CH₂COOH are monoprotic as only the H on the carboxyl group will dissociate
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Effect of Base Strength
Strong bases, such as NaOH, fully ionise in water whereas weak bases, such as ammonia, only partially ionise in water
NaOH --> Na⁺ + OH⁻
NH₃ + H₂O ⇌ OH⁻ + NH₄⁺
NaOH --> Na⁺ + OH⁻
NH₃ + H₂O ⇌ OH⁻ + NH₄⁺
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General Reaction between Acids and Bases
HA (aq) + B (aq) ⇌ BH⁺ (aq) + A⁻ (aq)
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Ionic Product of Water (Reaction)
2H₂O ⇌ OH⁻ + H₃O⁺
OR
H₂O ⇌ OH⁻ + H⁺
OR
H₂O ⇌ OH⁻ + H⁺
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Equilibrium constant of water (Kc)
[H⁺][OH⁻]
Kc = --------------
[H₂O]
Kc = --------------
[H₂O]
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Ionic Product of water (Equation)
Kw = Kc X [H₂O]
Using the Kc equation this means:
Kw = [H⁺][OH⁻]
[x] = conc. x
Unit = mol² dm⁻⁶
Using the Kc equation this means:
Kw = [H⁺][OH⁻]
[x] = conc. x
Unit = mol² dm⁻⁶
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Kw in Pure Water
In pure water [H⁺]=[OH⁻]
So
Kw = [H⁺]² OR [OH⁻]²
So
Kw = [H⁺]² OR [OH⁻]²
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Calculating pH of strong acids
pH = -log₁₀ [H⁺]
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Calculating [H⁺] of strong acids
[H⁺] = 10⁻ᵖᴴ
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Calculating pH of strong bases
Use Kw equation to calculate [H⁺]
pH = -log₁₀ [H⁺]
pH = -log₁₀ [H⁺]
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Calculating pH of strong bases (e.g.)
Find the pH of a 0.1 mol dm⁻³ base given that Kw at 298 K = 1.0 X 10⁻¹⁴ mol² dm⁻⁶
Kw = [H⁺][OH⁻]
1.0 X 10⁻¹⁴ = [H⁺] X 0.1
Rearrange to give [H⁺] = 1.0 X 10⁻¹⁴/0.1
= 1.0 X 10⁻¹³
pH = -log₁₀(1.0 X 10⁻¹³) = 13.00
Kw = [H⁺][OH⁻]
1.0 X 10⁻¹⁴ = [H⁺] X 0.1
Rearrange to give [H⁺] = 1.0 X 10⁻¹⁴/0.1
= 1.0 X 10⁻¹³
pH = -log₁₀(1.0 X 10⁻¹³) = 13.00
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Acid Dissociation Constant (Ka)
[H⁺][A⁻]
Ka = ---------------
[HA](start)
Ka = ---------------
[HA](start)
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Using Ka to calculate pH of weak acids
In weak acids [H⁺]≈[A⁻]
So
[H⁺]² OR [A⁻]²
Ka = ---------------
[HA](start)
Rearrange to give [H⁺] = √(Ka X [HA(start)])
Then use -log₁₀ [H⁺] to calculate pH
So
[H⁺]² OR [A⁻]²
Ka = ---------------
[HA](start)
Rearrange to give [H⁺] = √(Ka X [HA(start)])
Then use -log₁₀ [H⁺] to calculate pH
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pKa
pKa is used when the value of Ka is very low
pKa = -log₁₀(Ka)
Ka = 10⁻ᵖᴷᵃ
pKa = -log₁₀(Ka)
Ka = 10⁻ᵖᴷᵃ
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pH curves
pH curves plot pH against volume of acid or base added
Strength of the base and acid used determine the shape that the graph takes
Acid strength determines how far towards 0 the leg in acidic pH travels
Base strength determines how far towards 14 the leg
Strength of the base and acid used determine the shape that the graph takes
Acid strength determines how far towards 0 the leg in acidic pH travels
Base strength determines how far towards 14 the leg
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Indicators and pH curves
The indicator used must have a colour change in the pH of the endpoint of the curve
Methyl orange - 3.1 - 4.4, Red --> Yellow (acidic)
Phenolphthalein - 8.3 - 10, Colourless --> Pink (basic)
If no indicator works a pH probe can be used to measure pH and
Methyl orange - 3.1 - 4.4, Red --> Yellow (acidic)
Phenolphthalein - 8.3 - 10, Colourless --> Pink (basic)
If no indicator works a pH probe can be used to measure pH and
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In the Titration:
HNO₃ + KOH --> KNO₃ + H₂O
HNO₃ = 40 cm³ at 0.75 mol dm⁻³
KOH = 60 cm³
Calculate conc. KOH
HNO₃ + KOH --> KNO₃ + H₂O
HNO₃ = 40 cm³ at 0.75 mol dm⁻³
KOH = 60 cm³
Calculate conc. KOH
HNO₃ = 0.04 dm³
mol HNO₃ = 0.04 X 0.75 = 0.03
mol ratio = 1:1
mol KOH = 0.03
KOH = 0.06 dm³
conc. KOH = 0.03/0.06 = 0.5 mol dm⁻³
mol HNO₃ = 0.04 X 0.75 = 0.03
mol ratio = 1:1
mol KOH = 0.03
KOH = 0.06 dm³
conc. KOH = 0.03/0.06 = 0.5 mol dm⁻³
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Creation of buffers
Acid + Conjugate Base/Base and Conjugate acid
OR
Acid and equivalent salt/Base and equivalent salt
e.g. CH₃COOH ⇌ CH₃COO⁻ + H⁺ (acid)
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
OR
Acid and equivalent salt/Base and equivalent salt
e.g. CH₃COOH ⇌ CH₃COO⁻ + H⁺ (acid)
NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
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Use of Buffers
Resist changes in pH when small amount/dilute acids or bases are added
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Acidic Buffer
pH<7
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Basic Buffers
pH>7
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Calculating pH of a buffer solution
Rearrange Ka equation to calculate [H⁺]
Calculate pH with -log₁₀[H⁺]
Calculate pH with -log₁₀[H⁺]
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Examples of buffers
Shampoos, washing powders, within the body
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Other cards in this set
Card 2
Front
Proton Acceptor
B (aq) + + H₂O (l) --> BH⁺ (aq) + OH⁻ (aq)
B = Base
B (aq) + + H₂O (l) --> BH⁺ (aq) + OH⁻ (aq)
B = Base
Back
Bronsted-Lowry Base
Card 3
Front
Strong acids fully, such as HCl, dissociate in water whereas weak acids, such as ethanoic acid, only partially dissociate in water
HCL --> H⁺ + Cl⁻
CH₃COOH ⇌ CH₃COO⁻ (aq) + H⁺ (aq)
HCL --> H⁺ + Cl⁻
CH₃COOH ⇌ CH₃COO⁻ (aq) + H⁺ (aq)
Back
Card 4
Front
Monoprotic acids have 1 H that dissociates such as HCl, Diprotic acids have 2 H that dissociate such H₂SO₄
Note acids such as CH₂COOH are monoprotic as only the H on the carboxyl group will dissociate
Note acids such as CH₂COOH are monoprotic as only the H on the carboxyl group will dissociate
Back
Card 5
Front
Strong bases, such as NaOH, fully ionise in water whereas weak bases, such as ammonia, only partially ionise in water
NaOH --> Na⁺ + OH⁻
NH₃ + H₂O ⇌ OH⁻ + NH₄⁺
NaOH --> Na⁺ + OH⁻
NH₃ + H₂O ⇌ OH⁻ + NH₄⁺
Back
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