Topic 12 - Acid-Base Equilibria Flashcards

Question

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

Answer 1

It acts as the base.

Answer 2

HA(aq) + H₂O(l) {-} H₃O⁺(aq) + A⁻(aq) * Pair 1 = HA and A⁻ * Pair 2 = H₂O and H₃O⁺

Answer 3

HA(aq) + H₂O(l) {-} H₃O⁺(aq) + A⁻(aq) * HA = Acid * H₂O = Base * H₃O⁺ = Acid * A⁻ = Base

Answer 4

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

Answer 5

* HA is the conjugate acid of A⁻. | * A⁻ is the conjugate base of HA.

Answer 6

HA {-} H⁺ + A⁻ | H⁺ + H₂O {-} H₃O⁺

Answer 7

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

Answer 8

* HCl and Cl⁻ | * H₂O and H₃O⁺

Answer 9

When an acid or a base dissolves.

Answer 10

* B and BH⁺ | * H₂O and OH⁻

Answer 11

It acts as an acid.

Answer 12

B + H₂O {-} OH⁻ + BH⁺ * B = Base * H₂O = Acid * OH⁻ = Base * BH⁺ = Acid

Answer 13

Acid + Base -> Salt + Water

Answer 14

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

Answer 15

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

Answer 16

Exothermic

Answer 17

* 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.

Answer 18

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

Answer 19

* 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.

Answer 20

• Enthalpy of reaction between H⁺ and OH⁻ ions

Answer 21

A measure of the hydrogen ion concentration.

Answer 22

Logarithmic

Answer 23

pH = -log₁₀[H⁺]

Answer 24

Concentration of...

Answer 25

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

Answer 26

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

Answer 27

One that releases 2 protons into solution when it dissociates.

Answer 28

* Since it fully dissociates, [H] = Acid concentration | * pH = -log₁₀[H⁺] = -log₁₀(Acid concentration)

Answer 29

pH = -log₁₀(Acid concentration)

Answer 30

* [H⁺] = 0.050 | * pH = -log₁₀(0.050) = 1.30

Answer 31

[H⁺] = 10^-pH

Answer 32

[H⁺] = 10^-2.45 = 3.5 x 10^-3 mol/dm³

Answer 33

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.

Answer 34

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

Answer 35

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

Answer 36

HA(aq) {-} H⁺(aq) + A⁻(aq)

Answer 37

Ka = [H⁺][A⁻] / [HA]

Answer 38

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

Answer 39

* Ka = [H⁺]² / [HA] | * Because dissociation of the acid produces the same amount of H⁺ and A⁻, so [H⁺] = [A⁻].

Answer 40

* [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.

Answer 41

The temperature.

Answer 42

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

Answer 43

* 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

Answer 44

* [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

Answer 45

It can act as either.

Answer 46

By donating a proton.

Answer 47

By accepting a proton.

Answer 48

* Hydroxonium ions (H₃O⁺) | * Hydroxide ions (OH⁻)

Answer 49

• H₂O(l) + H₂O(l) {-}H₃O⁺(aq) + OH⁻(aq) OR MORE SIMPLY: • H₂O(l) {-} H⁺(aq) + OH⁻(aq) (This is essentially the dissociation of water.)

Answer 50

Kc = [H⁺][OH⁻] / [H₂O]

Answer 51

* Ionic product of water | * It is essentially a dissociation constant for water, except adjusted

Answer 52

* 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⁻]

Answer 53

Kw = [H⁺][OH⁻]

Answer 54

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

Answer 55

mol²/dm^-6

Answer 56

* [H⁺] = [OH⁻] * So Kw = [H⁺]² (NOTE: This is only for pure water.)

Answer 57

* Kw = [H⁺]² * Find [H⁺]. * pH = -log₁₀[H⁺]

Answer 58

1.0 x 10^-14 mol²/dm^6

Answer 59

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

Answer 60

* 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

Answer 61

-log₁₀ of ...

Answer 62

* -log₁₀Kw | * It is equal to 14 at STP

Answer 63

They are easier to work with.

Answer 64

* pKw = -log₁₀Kw | * Kw = 10^-pKw

Answer 65

-log₁₀Ka

Answer 66

* pKa = -log₁₀Ka | * Ka = 10^-pKa

Answer 67

pX = -log₁₀(X)

Answer 68

X = 10^-pX

Answer 69

pKa = -log₁₀(1.50 x 10^-7) = 6.824

Answer 70

Ka = 10^-4.32 = 4.8 x 10^-5 mol/dm³

Answer 71

The smaller the pKa, the stronger the acid.

Answer 72

* 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

Answer 73

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

Answer 74

* Probe | * Digital display

Answer 75

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

Answer 76

Pg 134 of revision guide

Answer 77

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]

Answer 78

* 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

Answer 79

* Diluting decreases [H⁺] | * This increases the pH

Answer 80

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

Answer 81

Increases it by 1.

Answer 82

Increases it by 0.5

Answer 83

* Strong acids -> pH increases by 1 | * Weak acids -> pH increases by 0.5

Answer 84

The concentration of an acid or base.

Answer 85

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.

Answer 86

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

Answer 87

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

Answer 88

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

Answer 89

They involve equimolar monoprotic acids and bases.

Answer 90

Pg 136 of revision guide + online.

Answer 91

* 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

Answer 92

Equivalence line

Answer 93

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

Answer 94

[H⁺] = [OH⁻]

Answer 95

* 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

Answer 96

Equivalence point

Answer 97

* Look at the straight part of the steep line and find the mid-point. * This is the equivalence point.

Answer 98

They are vertically flipped. | NOTE: Check for little curved parts at the start of some of the lines.

Answer 99

At the equivalence point.

Answer 100

* Look at the steep part of the curve | * See which indicators have a colour change within this region

Answer 101

* Methyl orange | * Phenolphthalein

Answer 102

* Acid = Red * Base = Yellow * Change -> 3.1 - 4.4

Answer 103

* Acid = Colourless * Base = Pink * Change -> 8.3 - 10

Answer 104

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.

Answer 105

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.

Answer 106

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.

Answer 107

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

Answer 108

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

Answer 109

Finding the Ka of a weak acid.

Answer 110

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

Answer 111

* 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.

Answer 112

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

Answer 113

* 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)

Answer 114

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

Answer 115

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

Answer 116

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

Answer 117

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

Answer 118

* -log₁₀[OH⁻] | * It is like pH, except it shows the concentration of OH⁻ ions rather than H⁺ ions.

Answer 119

* They add up to pKw. | * pKw = 14 = pH + pOH

Answer 120

pH + pOH = 14

Answer 121

-log₁₀Kb

Answer 122

* Kb = [OH⁻]² / [B] * [OH⁻]² = Kb x [B] * Find [OH⁻]. * pOH = -log₁₀[OH⁻] * pH = pKw - pOH * pH = 14 - pOH

Answer 123

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

Answer 124

No, but it does make changes very slight though.

Answer 125

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

Answer 126

* A buffer with a pH less than 7. | * Made by setting up an equilibrium between a weak acid and it’s conjugate base.

Answer 127

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

Answer 128

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

Answer 129

CH₃COOH(aq) {-} H⁺(aq) + CH₃COO⁻(aq)

Answer 130

* 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⁻

Answer 131

* 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⁻

Answer 132

* Lot of: Undissociated weak acid + Conjugate base | * Less of: H⁺ ions

Answer 133

* 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.

Answer 134

* 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.

Answer 135

* A buffer with a pH above 7 | * It is made from a weak base and one of its salts

Answer 136

Weak base and one of its salts

Answer 137

Yes, the idea is the same.

Answer 138

* 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

Answer 139

NH₄ {-} H⁺ + NH₃

Answer 140

* Lot of: Weak base + Conjugate acid | * Less of: H⁺ ions

Answer 141

* 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.

Answer 142

* 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.

Answer 143

* 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

Answer 144

* Weak acid with strong base | * Strong acid with weak base

Answer 145

When adding something strong to something weak. (i.e. Strong acid to weak base OR Strong base to weak acid) Note: Check this.

Answer 146

Maintaining blood pH.

Answer 147

Carbonic acid-hydrogencarbonate buffer system

Answer 148

H₂CO₃(aq) {-} H⁺(aq) + HCO₃⁻(aq)

Answer 149

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

Answer 150

* Kidneys | * These excrete an excess in the urine

Answer 151

* H₂CO₃(aq) {-} H⁺(aq) + HCO₃⁻(aq) | * H₂CO₃(aq) {-} H₂O(l) + CO₂ (controlled by breathing)

Answer 152

* 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

Answer 153

* 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

Answer 154

* 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

Answer 155

* Henderson-Hasselbalch equation * pH = pKa + log₁₀([A⁻]/[HA]) Note: This isn’t necessary. The standard Ka equation can be used.

Answer 156

* 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

Answer 157

Just write the equation for the Ka of the acid and rearrange this. Then sub-in values. (See pg 140 of revision guide)

Answer 158

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 ```

Answer 159

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