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A LEVEL - Chemistry (Inorganic) > Topic 12 - Acid-Base Equilibria > Flashcards

Topic 12 - Acid-Base Equilibria Flashcards

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1
Q

Who came up with the modern definition of an acid?

A

Brønsted-Lowry

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2
Q

What is an acid?

A
  • Proton donors

* Release H+ ions when they’re mixed with water

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3
Q

Are H+ ions found in solution when an acid is dissolved in water?

A
  • No, any H+ ions are combined with water to give H3O+ ions.

* HA(aq) + H2O(l) -> H3O+(aq) + A-(aq)

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4
Q

What is the equation for water reacting with an acid?

A

HA(aq) + H₂O(l) -> H₃O⁺(aq) + A⁻(aq)

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5
Q

What is the name for H3O+?

A

Hydroxonium ion

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6
Q

What is a base?

A

• Proton acceptor

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7
Q

What is the equation for a base reacting with water?

A

B(aq) + H2O(l) -> BH+(aq) + OH-(aq)

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8
Q

What is a strong acid/base?

A

One that dissociates almost completely in water.

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9
Q

Give an example of a strong acid and base.

A
  • Acid - HCl

* Base - NaOH

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10
Q

What is a weak acid/base?

A

One that dissociates only slightly in water.

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11
Q

Give an example of a weak acid/base.

A
  • Acid - CH3COOH

* Base - NH3

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12
Q

Give the equation for the dissociation of HCl in water.

A

HCl -> H⁺ + Cl⁻

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13
Q

Give the equation for the dissociation of NaOH in water.

A

NaOH -> Na⁺ + OH⁻

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14
Q

Give the equation for the dissociation of CH₃COOH in water.

A

CH₃COOH {-} CH₃COO⁻ + H⁺

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15
Q

Give the equation for the dissociation of NH₃ in water.

A

NH₃ + H₂O {-} NH₄⁺ + OH⁻

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16
Q

What is it important to remember about weak acid dissociation in water?

A

It is a reversible reaction, so an equilibrium is set up.

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17
Q

Is acid and base reactions, how is an acid and base symbolised?

A
  • Acid = HA

* Base = B

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18
Q

In order for an acid to lose its proton, what is required?

A

A base to accept the proton.

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19
Q

Give the general equation for the transfer of a proton between and acid and base.

A

HA(aq) + B(aq) {-} BH⁺ + A-

NOTE: This is reversible.

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20
Q

What is a conjugate pair?

A

Two species that a linked by the transfer of a proton.

e.g. HA and A⁻

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21
Q

In a reaction, how do you look for a conjugate pair?

A
  • Species on OPPOSITE sides of the reaction equation
  • Species differ by a proton
  • One is an acid, one is a base
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22
Q

Where are conjugate pairs in a reaction equation?

A

On opposite sides of the reaction equation.

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23
Q

What is a conjugate base?

A

A species that has lost a proton.

NOTE: It is quoted with reference to something - like, A⁻ is the conjugate base of the acid HA.

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24
Q

What is a conjugate acid?

A

A species that has gained a proton.

NOTE: It is quoted with reference to something - like, HA is the conjugate acid of A⁻.

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25
Q

When an acid is added to water, what does the water do?

A

It acts as the base.

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26
Q

Give the general equation for an acid reacting with water and state the conjugate pairs.

A

HA(aq) + H₂O(l) {-} H₃O⁺(aq) + A⁻(aq)

  • Pair 1 = HA and A⁻
  • Pair 2 = H₂O and H₃O⁺
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27
Q

Give the general equation for an acid reacting with water and state whether each of the species is an acid or a base.

A

HA(aq) + H₂O(l) {-} H₃O⁺(aq) + A⁻(aq)

  • HA = Acid
  • H₂O = Base
  • H₃O⁺ = Acid
  • A⁻ = Base
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28
Q

Is H₃O⁺ an acid or base?

A

Acid

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29
Q

What is another name for conjugate pairs and why?

A

Acid-base conjugate pairs, because one is always an acid, while the other is a base.

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30
Q

State the relationship between HA and A⁻.

A
  • HA is the conjugate acid of A⁻.

* A⁻ is the conjugate base of HA.

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31
Q

Give two equations demonstrating the conjugate pairs in this reaction.

HA(aq) + H₂O(l) -> H₃O⁺(aq) + A⁻(aq)

A

HA {-} H⁺ + A⁻

H⁺ + H₂O {-} H₃O⁺

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32
Q

HCl(aq) + H₂O(l)

A

HCl(aq) + H₂O(l) {-} H₃O⁺(aq) + Cl⁻(aq)

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33
Q

State the conjugate pairs.

HCl(aq) + H₂O(l) H₃O⁺(aq) + Cl⁻(aq)

A
  • HCl and Cl⁻

* H₂O and H₃O⁺

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34
Q

When are conjugate pairs formed?

A

When an acid or a base dissolves.

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35
Q

State the conjugate pairs.

B + H₂O OH⁻ + BH⁺

A
  • B and BH⁺

* H₂O and OH⁻

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36
Q

When a base dissolves in water, what does the water do?

A

It acts as an acid.

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37
Q

Give the general equation for a base reacting with water and state whether each of the species is an acid or a base.

A

B + H₂O {-} OH⁻ + BH⁺

  • B = Base
  • H₂O = Acid
  • OH⁻ = Base
  • BH⁺ = Acid
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38
Q

Acid + Base ->

A

Acid + Base -> Salt + Water

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39
Q

What is a neutral solution?

A

One where [H⁺] = [OH⁻], since all the H⁺ ions react with OH⁻ ions to give H₂O.

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40
Q

What is the standard enthalpy change of neutralisation?

A

The enthalpy change when solutions of an acid and base react together, under standard conditions, to produce 1 mole of water.

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41
Q

Is neutralisation endothermic or exothermic?

A

Exothermic

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42
Q

Do all weak acid and weak base neutralisation reactions have similar enthalpies of neutralisation? Why?

A
  • No
  • Because the acid/base dissolves only partially, meaning that the weak acid/base keeps dissociating throughout the reaction to maintain the equilibrium as the H⁺ or OH⁻ are used up.
  • This dissociation requires enthalpy, so it must be accounted for.
  • So the overall enthalpy of neutralisation includes both the reaction between H⁺ and OH⁻ ions AND the enthalpy of dissociation.
  • Since the enthalpy of dissociation varies depending on the weak acid/base, the standard enthalpy of neutralisation varies too.
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43
Q

What enthalpies must be factored into the standard enthalpy of neutralisation for weak acids/bases?

A

• Enthalpy of reaction between H⁺ and OH⁻ ions
AND
• Enthalpy of dissociation of the weak acid/base

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44
Q

Do all strong acid and weak base neutralisation reactions have similar enthalpies of neutralisation? Why?

A
  • Yes
  • Because the acid/base dissolves completely, meaning there is no dissociation enthalpy.
  • So the overall enthalpy of neutralisation includes only the reaction between H⁺ and OH⁻ ions, which is almost constant regardless of the strong acid or base.
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45
Q

What enthalpies must be factored into the standard enthalpy of neutralisation for strong acids/bases?

A

• Enthalpy of reaction between H⁺ and OH⁻ ions

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46
Q

What is the pH scale?

A

A measure of the hydrogen ion concentration.

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47
Q

What type of scale is the pH scale?

A

Logarithmic

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48
Q

What does the pH scale go from and to?

A

0 to 14

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49
Q

What is the neutral pH?

A

7

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50
Q

What is the equation for pH?

A

pH = -log₁₀[H⁺]

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51
Q

What do square brackets indicate?

A

Concentration of…

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52
Q

What is a monoprotic acid?

A

One that releases only 1 proton into solution when it dissociates.

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53
Q

What is a polyprotic acid?

A

One that releases more than 1 proton into solution when it dissociates.

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54
Q

What is a diprotic acid?

A

One that releases 2 protons into solution when it dissociates.

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55
Q

How is the pH of a strong acid calculated when given the concentration?

A
  • Since it fully dissociates, [H] = Acid concentration

* pH = -log₁₀[H⁺] = -log₁₀(Acid concentration)

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56
Q

What is the formula for the pH of a strong acid?

A

pH = -log₁₀(Acid concentration)

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57
Q

Calculate the pH of 0.050 mol/dm³ nitric acid.

A
  • [H⁺] = 0.050

* pH = -log₁₀(0.050) = 1.30

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58
Q

For a strong acid, how do you work out [H⁺] when given the pH?

A

[H⁺] = 10^-pH

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59
Q

An acid solution has a pH of 2.45. What is the hydrogen ion concentration, or [H⁺], of the acid?

A

[H⁺] = 10^-2.45 = 3.5 x 10^-3 mol/dm³

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60
Q

Will you be asked to calculate [H⁺] and pH for polyprotic strong acids?

A

No

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61
Q

Why is difficult to work out the concentration of a weak acid?

A

It does not dissociate completely, so the [H⁺] is not the same as acid concentration. So pH is tricker to find and we must use Ka.

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62
Q

What is Ka?

A
  • Acid dissociation constant

* It is a type of equilibrium constant that shows how much a weak acid will dissociate.

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63
Q

Do strong acids have a Ka?

A

No, we only use Ka for weak acids, since they set up an equilibrium.

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64
Q

Give the general formula for the dissociation equilibrium of an acid.

A

HA(aq) {-} H⁺(aq) + A⁻(aq)

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65
Q

Give the equation for Ka.

A

Ka = [H⁺][A⁻] / [HA]

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66
Q

Ka is an equilibrium constant. Why is H₂O missing from the bottom of the expression?

A

It’s concentration does not change very much. (CHECK THIS)

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67
Q

How can the expression for Ka be simplified and why?

A
  • Ka = [H⁺]² / [HA]

* Because dissociation of the acid produces the same amount of H⁺ and A⁻, so [H⁺] = [A⁻].

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68
Q

What are the units for Ka?

A

Mol/dm³

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69
Q

Under what assumption does the equation for Ka work?

A
  • [HA] at the start is approximately the same as [HA] at equilibrium.
  • This is assumed because an equilibrium constant works with the concentrations of the products and reactants at equilibrium. However, the concentration is usually quoted as the concentration before the reaction, so that value can only be used is [HA] doesn’t change much before equilibrium is reached.
  • Therefore, the Ka expression doesn’t work for strong acids.
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70
Q

What does the Ka value for a weak acid depend on?

A

The temperature.

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71
Q

Does the Ka value for an acid depend on concentration?

A

No

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72
Q

Describe how to calculate the pH of a weak acid when given the concentration and Ka.

A

1) Use Ka = [H⁺]² / [HA].
2) Sub in the correct values.
3) Solve for [H⁺].
4) pH = -log₁₀[H⁺]

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73
Q

Calculate the hydrogen ion concentration and the pH of a 0.02 mol/dm³ solution of propanoic acid (CH₃CH₂COOH). Ka for propanoic acid in this temperature is 1.30 x 10^-5 mol/dm³.

A
  • Ka = [H⁺]² / [CH₃CH₂COOH]
  • [H⁺]² = Ka x [CH₃CH₂COOH] = 1.30 x 10^-5 x 0.02 = 2.60 x 10^-7
  • [H⁺] = 5.10 x 10^-4 mol/dm³
  • pH = -log₁₀(5.10 x 10^-4) = 3.29
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74
Q

The pH of an ethanoic acid (CH₃COOH) solution was 3.02 at 298K. Calculate the molar concentration of this solution. Ka of ethanoic acid is 1.75 x 10^-5 mol/dm³ at 298K.

A
  • [H⁺] = 10^-pH = 10^-3.02 = 9.55 x 10^-4 mol/dm³
  • Ka = [H⁺]² / [CH₃CH₂COOH]
  • [CH₃CH₂COOH] = [H⁺]² / Ka = (9.55 x 10^-4)² / (1.75 x 10^-5) = 0.0521 mol/dm³
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75
Q

Does water act like an acid or base?

A

It can act as either.

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76
Q

How does water act as an acid?

A

By donating a proton.

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77
Q

How does water act as a base?

A

By accepting a proton.

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78
Q

What ions are found in water?

A
  • Hydroxonium ions (H₃O⁺)

* Hydroxide ions (OH⁻)

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79
Q

Can both hydroxonium and hydroxide ions be found in water at the same time?

A

Yes

80
Q

Give the equilibrium that exists in pure water.

A

• H₂O(l) + H₂O(l) {-}H₃O⁺(aq) + OH⁻(aq)
OR MORE SIMPLY:
• H₂O(l) {-} H⁺(aq) + OH⁻(aq)

(This is essentially the dissociation of water.)

81
Q

Give the Kc equation for the dissociation of water.

A

Kc = [H⁺][OH⁻] / [H₂O]

82
Q

What is Kw?

A
  • Ionic product of water

* It is essentially a dissociation constant for water, except adjusted

83
Q

How is Kw found?

A
  • H₂O(l) {-} H⁺(aq) + OH⁻(aq)
  • Kc = [H⁺][OH⁻] / [H₂O]
  • The water only dissociates a tiny amount, so the equilibrium is very much to the left. There’s so much water relative to the ions, that water is said to have a constant value.
  • Multiplying through by this constant gives:
  • Kw = [H⁺][OH⁻]
84
Q

What is the ionic product of water?

A

Kw

85
Q

What is the equation for Kw?

A

Kw = [H⁺][OH⁻]

86
Q

Is the dissociation of water affected by whether it is pure water or a solution?

A

No, Kw is the same either way at a given temperature.

87
Q

What are the units of Kw?

A

mol²/dm^-6

88
Q

In pure water, what is the ratio of H⁺ to OH⁻ ions?

A

1 : 1

89
Q

For pure water, how can the equation for Kw be simplified?

A
  • [H⁺] = [OH⁻]
  • So Kw = [H⁺]²

(NOTE: This is only for pure water.)

90
Q

How can pH of pure water at any temperature be found when given the Kw value?

A
  • Kw = [H⁺]²
  • Find [H⁺].
  • pH = -log₁₀[H⁺]
91
Q

What is the value of Kw (at 25*C and 298K)?

A

1.0 x 10^-14 mol²/dm^6

92
Q

Describe how you can work out the pH of a strong base when given the concentration of OH⁻ ions.

A

1) Put all of the values into Kw = [H⁺][OH⁻].
2) Rearrange to find [H⁺].
3) pH = -log₁₀[H⁺]

93
Q

Find the pH of 0.10 mol/dm³ NaOH at 298K, given that Kw at 298K is 1.0 x 10^-14 mol²/dm^6.

A
  • Kw = [H⁺][OH⁻]
  • 1.0 x 10^-14 = [H⁺] x 0.10
  • [H⁺] = 1.0 x 10^-14 / 0.10 = 1.0 x 10^-13 mol/dm³
  • pH = -log₁₀[H⁺]
  • pH = 13.00
94
Q

IMPORTANT: What does a ‘p’ in front of value mean?

A

-log₁₀ of …

95
Q

What is pKw?

A
  • -log₁₀Kw

* It is equal to 14 at STP

96
Q

What is pKw equal to under standard conditions?

A

14

97
Q

What is the advantage of ‘p’ values e.g. pKw?

A

They are easier to work with.

98
Q

Describe the two relationships between pKw and Kw.

A
  • pKw = -log₁₀Kw

* Kw = 10^-pKw

99
Q

What is pKa?

A

-log₁₀Ka

100
Q

Describe the two relationships between pKa and Ka.

A
  • pKa = -log₁₀Ka

* Ka = 10^-pKa

101
Q

How can you convert from X to pX?

A

pX = -log₁₀(X)

102
Q

How can you convert from pX to X?

A

X = 10^-pX

103
Q

If an acid has a Ka value of 1.50 x 10^-7 mol/dm³, what is it’s pKa?

A

pKa = -log₁₀(1.50 x 10^-7) = 6.824

104
Q

What is the Ka value of an acid if its pKa is 4.32?

A

Ka = 10^-4.32 = 4.8 x 10^-5 mol/dm³

105
Q

How does pKa relate to acid strength?

A

The smaller the pKa, the stronger the acid.

106
Q

Calculate the pH of 0.0500 mol/dm³ methanoic acid (HCOOH). Methanoic acid has a pKa of 3.75 at this temperature.

A
  • Ka = 10^-pKa = 10^-3.75 = 1.77 x 10^-4
  • Ka = [H⁺]² / [HCOOH]
  • [H⁺]² = Ka x [HCOOH] = 1.77 x 10^-4 x 0.0500 = 8.91 x 10^-6
  • [H⁺] = 2.98 x 10^-3
  • pH = -log₁₀(2.98 x 10^-3) = 2.53
107
Q

What device can be used to measure pH?

A

pH meter

108
Q

What is a pH meter?

A

An electronic gadget used to measure the pH of a solution.

109
Q

What are the parts of a pH meter?

A
  • Probe

* Digital display

110
Q

How can you use a pH meter?

A

TO CALIBRATE:
• Place the bulb into deionised water and allow the reading to settle
• Do the same with a standard solution of pH 4 and another of pH 10.
• Make sure to rinse with deionised water between each reading
TO USE:
• Allow reading to settle each time
• Rinse between each reading

111
Q

Remember to practice identifying the type of substance using its pH.

A

Pg 134 of revision guide

112
Q

How can you work out the Ka of a weak acid using the mass dissolved and the pH of the resulting solution?

A

1) Work out the number of moles of the substance using Moles = Mass / Mr.
2) Then work out concentration using Conc = Moles x 1000 / Volume
3) Use pH to work out [H⁺] using [H⁺] = 10^-pH
4) Work out Ka using Ka = [H⁺]² / [HA]

113
Q

1.31g of ethanoic acid (CH₃COOH) are dissolves in 250cm³ of distilled water to create a solution of ethanoic acid. The solution has a pH of 2.84. Calculate the acid dissociation constant for ethanoic acid.

A
  • Moles = Mass / Mr = 1.31 / 60 = 0.02183 mol
  • Conc = Moles x 1000 / Volume = 0.02183 x 1000 / 250 = 0.0873 mol/dm³
  • [H⁺] = 10^-pH = 10^-2.84 = 0.00144
  • Ka = [H⁺]² / [HA] = 0.00144² / 0.0873 = 2.4 x 10^-5
114
Q

How does diluting an acid change the pH? Why?

A
  • Diluting decreases [H⁺]

* This increases the pH

115
Q

Which is affected by dilution more, strong acids or weak acids?

A

The pH of strong acids changes more, because no equilibrium exists.

116
Q

How does diluting a strong acid by a factor of 10 change its pH?

A

Increases it by 1.

117
Q

How does diluting a weak acid by a factor of 10 change its pH?

A

Increases it by 0.5

118
Q

Describe how diluting by a factor of 10 affects the pH of strong and weak acids.

A
  • Strong acids -> pH increases by 1

* Weak acids -> pH increases by 0.5

119
Q

What can a titration by used to find?

A

The concentration of an acid or base.

120
Q

Describe how to carry out a titration.

A

1) Measure out some base using a pipettes and put it in a flask.
2) Rinse a burette with some of your standard solution of acid.
3) Do a rough titration to get an idea of where the end point is. Do this by reading off the initial reading and then adding the acid to the base until a permanent colour change is seen. Swirl regularly.
4) Record the final reading from the burette.
5) Now do an accurate titration, adding the acid dropwise once you are within 2cm³ of the rough endpoint.
6) Repeat the titration until you get results within 0.1cm³ of each other.
7) Calculate a mean titre from these.

121
Q

What is a titration curve?

A

A graph of pH against volume of acid or base added to an acid or base.

122
Q

What can a titration curve be used for?

A

Working out exactly how much base is needed to neutralise a quantity of acid or vice versa.

123
Q

Do all titration curves look the same.

A

No, it depends on whether the acid/base is strong or weak.

124
Q

What is an assumption for the standard titration curves?

A

They involve equimolar monoprotic acids and bases.

125
Q

What is another name for a titration curve?

A

pH curve

126
Q

Remember to revise the shapes of all titration curves for adding a base and for adding an acid.

A

Pg 136 of revision guide + online.

127
Q

Explain how you can work out the general shape of a titration curve.

A
  • Think about the starting substance in the solution (acid/base) and its strength -> This will give the starting pH, so you know the y-intercept
  • Think about the substance being added (acid/base) and it’s strength -> This will give you the end pH and so the point at which the curve ends
  • The steepness of the sharp part is dependant on the strength of the reactants: strong-strong is very steep, strong-weak or weak-strong is steep, weak-weak is gentle
128
Q

What is the steep part of a titration curve called?

A

Equivalence line

129
Q

What is the equivalence point?

A
  • The point at the centre of the equivalence line at which all of the acid/base is just neutralised.
  • [H⁺] = [OH⁻]
130
Q

What can be said about [H⁺] and [OH⁻] at the equivalence point?

A

[H⁺] = [OH⁻]

131
Q

Explain the general shape of a titration curve for a base being added to an acid.

A
  • Starts at a low pH -> Due to the acid
  • Very slow change of pH at first -> Addition of a base has little effect on pH, since there are so many H⁺ ions
  • Equivalence line is steep -> Addition of base has great effect on pH, since there are few H⁺ ions
  • Curve ends at a high pH due to the base
132
Q

What is the point in the middle of the steep part of a titration curve called?

A

Equivalence point

133
Q

How can you find the equivalence point on a titration curve?

A
  • Look at the straight part of the steep line and find the mid-point.
  • This is the equivalence point.
134
Q

How so the graphs titration curves for adding a base to an acid differ from adding an acid to a base?

A

They are vertically flipped.

NOTE: Check for little curved parts at the start of some of the lines.

135
Q

At what point on a titration curve can the acid/base be said to be neutralised?

A

At the equivalence point.

136
Q

How can a titration curve be used to decide which indicator to use for a neutralisation?

A
  • Look at the steep part of the curve

* See which indicators have a colour change within this region

137
Q

What are two indicators commonly used in acid-base titrations?

A
  • Methyl orange

* Phenolphthalein

138
Q

What is the colour of methyl orange in acid and base? What is the approximate pH of colour change?

A
  • Acid = Red
  • Base = Yellow
  • Change -> 3.1 - 4.4
139
Q

What is the colour of phenolphthalein in acid and base? What is the approximate pH of colour change?

A
  • Acid = Colourless
  • Base = Pink
  • Change -> 8.3 - 10
140
Q

Looking at methyl orange and phenolphthalein, which can be used as an indicator for a strong acid/strong base neutralisation reaction?

A

Both, because:
• Methyl orange changes at a pH about 3.1 - 4.4, while phenolphthalein changes at about 8.3 - 10.
• The steep part of the titration curve covers both of these ranges.

141
Q

Looking at methyl orange and phenolphthalein, which can be used as an indicator for a strong acid/weak base neutralisation reaction?

A

Only methyl orange, because:
• Methyl orange changes at a pH about 3.1 - 4.4, while phenolphthalein changes at about 8.3 - 10.
• The steep part of the titration curve covers low pHs, so only methyl orange is within this range.

142
Q

Looking at methyl orange and phenolphthalein, which can be used as an indicator for a weak acid/strong base neutralisation reaction?

A

Only phenolphthalein, because:
• Methyl orange changes at a pH about 3.1 - 4.4, while phenolphthalein changes at about 8.3 - 10.
• The steep part of the titration curve covers high pHs, so only phenolphthalein is within this range.

143
Q

Looking at methyl orange and phenolphthalein, which can be used as an indicator for a weak acid/weak base neutralisation reaction?

A

Neither, because:
• There is no sharp pH change.
• No indicators would work, so it’s better to use a pH meter.

144
Q

Can an indicator be used to determine the endpoint of a weak acid/weak base titration?

A

No, because there is no sharp change, so a pH meter should be used instead.

145
Q

Aside from working out the equivalence point, what can titration curves be used for?

A

Finding the Ka of a weak acid.

146
Q

What titration curve can be used to find the pKa of a weak acid?

A

It must be a strong base being added to a weak acid.

147
Q

What is the half-equivalence point?

A
  • The point at which half of the acid has been neutralised in a titration.
  • It’s when half of the equivalence volume has been added.
148
Q

How are the equivalence point and half-equivalence point related?

A

The half-equivalence point is when half of the volume of base needed to reach the equivalence point has been added.

149
Q

Describe how and why the pKa of a weak acid can be found from a titration curve of a strong base being added to the weak acid.

A
  • A weak acid dissociates like this: HA {-} H⁺ + A⁻
  • So Ka = [H⁺][A⁻] / [HA]
  • At the half equivalence point, [HA] = [A⁻] so the Ka expression simplifies to:
  • Ka = [H⁺]
  • Therefore pKa = pH
  • So the pKa can be found by looking at the pH at the half-equivalence point (half the volume of the equivalence point)
150
Q

At the half equivalence point, the … equals the …

A

pH, pKa

151
Q

What is a pH chart?

A
  • A diagram that shows the colour of an indicator at different pH values.
  • It is usually shown as a continuous spectrum.
152
Q

How can a pH chart be used?

A

After adding an indicator to a solution, look at the colour and the pH chart to see what the pH is.

153
Q

What is Kb?

A

The dissociation constant for a weak base (just like Ka for acids).

154
Q

How is Kb different from Ka?

A

It is the same, except every [H⁺] is replaced by [OH⁻].

155
Q

What is pOH?

A
  • -log₁₀[OH⁻]

* It is like pH, except it shows the concentration of OH⁻ ions rather than H⁺ ions.

156
Q

How are pH and pOH related?

A
  • They add up to pKw.

* pKw = 14 = pH + pOH

157
Q

pH + pOH =

A

pH + pOH = 14

158
Q

What is pKb?

A

-log₁₀Kb

159
Q

How can the pH of a weak base be found when given the concentration and Kb?

A
  • Kb = [OH⁻]² / [B]
  • [OH⁻]² = Kb x [B]
  • Find [OH⁻].
  • pOH = -log₁₀[OH⁻]
  • pH = pKw - pOH
  • pH = 14 - pOH
160
Q

What is a buffer?

A

A solution that minimises changes in pH when small amounts of acid or base are added.

161
Q

Do a buffer stop pH changing completely?

A

No, but it does make changes very slight though.

162
Q

When can buffers work?

A

With a small amount of acid or base - putting too much in means they won’t be able to cope.

163
Q

What is an acidic buffer?

A
  • A buffer with a pH less than 7.

* Made by setting up an equilibrium between a weak acid and it’s conjugate base.

164
Q

What does an acidic buffer contain?

A

Equilibrium between:
• Weak acid
• It’s conjugate base

165
Q

How can an acidic buffer be prepared?

A

1) Mix a weak acid with the salt of its conjugate base
• The salt fully dissociates into ions when it dissolves
• The acid only partly dissociates
• An equilibrium is set up between the weak acid and its conjugate base
OR
2) Mix an excess of weak acid with a strong base
• All the base reacts with the acid.
• The weak acid was in excess so there’s still some left in solution once all the base has reacted. This slightly dissociates.
• An equilibrium is set up between the weak acid and its conjugate base

166
Q

Give the equation for the equilibrium in an acidic buffer based on ethanoic acid.

A

CH₃COOH(aq) {-} H⁺(aq) + CH₃COO⁻(aq)

167
Q

Describe how an equilibrium is set up in a buffer prepared by mixing a weak acid with the salt of a conjugate base.

A
  • The salt fully dissociates into ions when it dissolves
  • e.g. CH₃COO⁻Na⁺ {-} CH₃COO⁻ + Na⁺
  • The acid only partly dissociates
  • e.g. CH₃COOH {-} H⁺ + CH₃COO⁻
  • An equilibrium is set up between the weak acid and its conjugate base
  • e.g. CH₃COOH {-} H⁺ + CH₃COO⁻
168
Q

Describe how an equilibrium is set up in a buffer prepared by mixing an excess of weak acid with a strong base.

A
  • All the base reacts with the acid.
  • e.g. CH₃COOH + OH⁻ {-} CH₃COO⁻ + H₂O
  • The weak acid was in excess so there’s still some left in solution once all the base has reacted. This slightly dissociates.
  • e.g. CH₃COOH {-} H⁺ + CH₃COO⁻
  • An equilibrium is set up between the weak acid and its conjugate base
  • e.g. CH₃COOH {-} H⁺ + CH₃COO⁻
169
Q

In the equilibrium in an acidic buffer, what is there more and less of?

A
  • Lot of: Undissociated weak acid + Conjugate base

* Less of: H⁺ ions

170
Q

Describe how an acidic buffer copes when the solution becomes more acidic.

A
  • This sort of equilibrium exists: CH₃COOH {-} H⁺ + CH₃COO⁻
  • If the amount of H⁺ increases, the extra H⁺ ions combine with the conjugate base on the right of the equation to produce the undissociated acid.
  • This reduces the amount of H⁺ ions, so the pH returns close to the original.
171
Q

Describe how an acidic buffer copes when the solution becomes more alkaline.

A
  • This sort of equilibrium exists: CH₃COOH {-} H⁺ + CH₃COO⁻
  • If the amount of OH⁻ increases, the extra OH⁻ ions combined with the H⁺ ions to form H₂O.
  • This reduces the amount of H⁺ ions, so the acid dissociates more to form more H⁺ ions, shifting the equilibrium to the right.
  • So the pH returns close to the original.
172
Q

What is an alkaline buffer?

A
  • A buffer with a pH above 7

* It is made from a weak base and one of its salts

173
Q

What is an alkaline buffer made from?

A

Weak base and one of its salts

174
Q

Do acidic and alkaline buffers work in the same way?

A

Yes, the idea is the same.

175
Q

How can an alkaline buffer be prepared?

A
  • A mixture of a weak base and one of its salts is prepared
  • e.g. NH₃ and NH₄Cl
  • The salt fully dissociates
  • e.g. NH4Cl -> NH₄⁺ + Cl⁻
  • An equilibrium is set up between the conjugate acid and weak base
  • e.g. NH₄⁺ {-} H⁺ + NH₃
176
Q

Give the equation for the equilibrium in an alkaline buffer based on ammonia.

A

NH₄ {-} H⁺ + NH₃

177
Q

In the equilibrium in an alkaline buffer, what is there more and less of?

A
  • Lot of: Weak base + Conjugate acid

* Less of: H⁺ ions

178
Q

Describe how an alkaline buffer copes when the solution becomes more acidic.

A
  • This sort of equilibrium exists: NH₄⁺ {-} H⁺ + NH₃
  • If the amount of H⁺ increases, the extra H⁺ ions combine with the weak base on the right of the equation to produce the conjugate acid.
  • This reduces the amount of H⁺ ions, so the pH returns close to the original.
179
Q

Describe how an acidic buffer copes when the solution becomes more alkaline.

A
  • This sort of equilibrium exists: NH₄⁺ {-} H⁺ + NH₃
  • If the amount of OH⁻ increases, the extra OH⁻ ions combined with the H⁺ ions to form H₂O.
  • This reduces the amount of H⁺ ions, so the conjugate acid dissociates more to form more H⁺ ions, shifting the equilibrium to the right.
  • So the pH returns close to the original.
180
Q

Explain the distinctive shape of a titration curve for sodium hydroxide being added to ethanoic acid.

A
  • At first, there is a very small sharp increase -> The base is strong and contains many OH⁻ ions to react with the H⁺ ions
  • The curve then levels off to be almost horizontal -> A buffer solution of sodium ethanoate in ethanoic acid is formed which resists further changes in pH
  • Large increase -> Eventually all the ethanoic acid is used up and the equivalence point is reached
  • Graph levels off -> Maximum pH is reached
181
Q

What titration curves do buffers explain?

A
  • Weak acid with strong base

* Strong acid with weak base

182
Q

In which titration curves is a small, sharp increase/decrease seen at the very start?

A

When adding something strong to something weak.
(i.e. Strong acid to weak base OR Strong base to weak acid)

Note: Check this.

183
Q

In the human body, how are buffer solutions important?

A

Maintaining blood pH.

184
Q

What pH does the blood need to be kept at?

A

7.4

185
Q

What system controls blood pH?

A

Carbonic acid-hydrogencarbonate buffer system

186
Q

Give the equation for the buffer system that controls the blood pH.

A

H₂CO₃(aq) {-} H⁺(aq) + HCO₃⁻(aq)

187
Q

What controls the levels of H₂CO₃ in the blood?

A

• Breathing
• Breathing out CO₂ reduces the H₂CO₃ concentration by shifting this equilibrium:
H₂CO₃(aq) {-} H₂O(l) + CO₂

188
Q

What controls the levels of HCO₃⁻ in the blood?

A
  • Kidneys

* These excrete an excess in the urine

189
Q

Give the two equilibria important in the control of blood concentration.

A
  • H₂CO₃(aq) {-} H⁺(aq) + HCO₃⁻(aq)

* H₂CO₃(aq) {-} H₂O(l) + CO₂ (controlled by breathing)

190
Q

Describe how you can calculate the pH of an acidic buffer solution.

A
  • Write the equation for the Ka of the weak acid
  • Rearrange this to give [H⁺]
  • Plug in the data given, assuming the salt of the conjugate base fully dissociates and that the concentration of the weak base is unchanged
  • Convert [H⁺] to pH
191
Q

What are the two assumptions used in calculating the pH of a buffer solution?

A
  • The salt of the conjugate base fully dissociates -> So the concentration of A⁻ is the same as the initial concentration of the salt
  • HA only slightly dissociates -> So the equilibrium concentration is the same as the initial concentration
192
Q

At a certain temperature, a buffer solution contains 0.40 mol/dm³ methanoic acid, HCOOH, and 0.60 mol/dm³ sodium methanoate, HCOO⁻Na⁺. At this temperature, Ka for methanoic acid = 1.8 x 10⁻⁴ mol/dm³. What is the pH of the buffer?

A
  • Ka = [H⁺][HCOO⁻] / [HCOOH]
  • [H⁺] = Ka x [HCOOH] / [HCOO⁻]
  • [H⁺] = 1.8 x 10⁻⁴ x 0.40 / 0.60 = 1.2 x 10⁻⁴ mol/dm³
  • pH = -log₁₀[H⁺] = -log₁₀[1.2 x 10⁻⁴] = 3.92
193
Q

What equation is used to work out the concentration of salt and acid or base that is needed to make a buffer of a specific pH?

A
  • Henderson-Hasselbalch equation
  • pH = pKa + log₁₀([A⁻]/[HA])

Note: This isn’t necessary. The standard Ka equation can be used.

194
Q

What assumptions does the Henderson-Hasselbalch equation use?

A
  • The salt of the conjugate base fully dissociates -> So the concentration of A⁻ is the same as the initial concentration of the salt
  • HA only slightly dissociates -> So the equilibrium concentration is the same as the initial concentration
195
Q

What is the alternative to using the Henderson-Hasselbalch equation for calculating the concentration of salt and acid or base that is needed to make a buffer of a specific pH?

A

Just write the equation for the Ka of the acid and rearrange this. Then sub-in values.

(See pg 140 of revision guide)

196
Q

A buffer is made using ethanoic acid (CH₃COOH) and an ethanoic acid salt (CH₃COO⁻Na⁺). 1.20 mol/dm³ of the ethanoic acid salt is used. What concentration of ethanoic acid is required so that the buffer has a pH of 4.9? Under these conditions, Ka of ethanoic acid = 1.75 x 10⁻⁵.

A

Assume that the concentration of the salt is unchanged at equilibrium and it is fully dissociated, and that the acid is only slightly dissociated, so that it is at the same concentration at equilibrium as at the start.

METHOD 1: Henderson-Hasselbalch
• pKa = -log₁₀(1.75 x 10⁻⁵) = 4.756...
• pH = pKa + log₁₀([A⁻]/[HA⁻])
• 4.9 = 4.756... + log₁₀([CH₃COO⁻]/[CH₃COOH])
• log₁₀([CH₃COO⁻]/[CH₃COOH]) = 0.143
• [CH₃COO⁻]/[CH₃COOH] = 10⁰*¹⁴³ = 1.39
• 1.20/[CH₃COOH] = 1.39
• [CH₃COOH] = 1.20 / 1.39 = 0.86 mol/dm³
METHOD 2: Classic
• pH = 4.9
• [H⁺] = 10⁻⁴*⁹
• Ka = [H⁺][CH₃COO⁻] / [CH₃COOH]
• [CH₃COOH] = [H⁺][CH₃COO⁻] / Ka
• [CH₃COOH] = 10⁻⁴*⁹ x 1.20 / (1.75 x 10⁻⁵) = 0.86 mol/dm³
197
Q

When doing buffer questions calculations, what is a good approach?

A
  • Just go back to the basic equations and fill everything you can in
  • Remember about assumptions, like minimal dissociation or full dissociation

A LEVEL - Chemistry (Inorganic) (21 decks)

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