O ijklm: Acid-base equilibria

Question 1

Bronsted-Lowry theory

Define:

• Acids
• Bases

Answer 1

• Acids = proton donors
• Bases = proton acceptors

Question 2

Write the formula for the conjugate base of ethanoic acid.

Answer 2

CH₃COO^-

Question 3

Write the formula for the conjugate base of:

• Hydrochloric acid
• Chloric(VII) acid
• Sulfuric acid
• Hydrogen sulfide
• Water

Answer 3

• HCl → Cl^-
• HClO₄ →ClO₄^-
• H₂SO₄ →HSO₄^-
• H₂S → SH^-
• H₂O → OH^-

Question 4

Water may act as an acid or base. Write 2 equations, including hydrochloric acid and ammonia, to demonstrate this.

Answer 4

Acting as a base: accepting a proton from HCl

H₂O_(l) + HCl_(aq) → H₃O⁺_(aq) + Cl^-_(aq)

Acting as an acid: donating a proton to ammonia

H₂O_(l) + NH_3(aq) → OH^-_(aq) + NH₄⁺_(aq)

Question 5

• What equation is used to calculate pH?
• How can [H⁺] be calculated from a pH value?

Answer 5

pH = -log₁₀[H⁺]

[H⁺] = 10^-pH

Question 6

What is the [H⁺] in water?

Answer 6

pH = 7

-log[H⁺] = 7

[H⁺] = 10^-7 mol dm^-3

Question 7

What can be said of [H⁺] in the case of a solution with pH < 0?

Answer 7

pH < 0

-log[H⁺] < 0

log[H⁺] > 0

[H⁺] > 10⁰

[H⁺] > 1 mol dm³

Question 8

Compare strong and weak acids. Use equations to demonstrate the comparison.

Answer 8

Strong acids dissociate protons completely in aqueous solution:

HA + H₂O → H₃O⁺ + A^-

HA → H⁺ + A^-

Weak acids do not dissociate protons completely in aqueous solution:

HA + H₂O ⇌ H₃O⁺ + A^-

HA ⇌ H⁺ + A^-

The further to the left the eq. position lies, the weaker the acid

Question 9

Write equations to demonstrate why solutions of carbon dioxide are acidic.

Answer 9

CO_{2 (aq)} + H₂O ⇌ H⁺_(aq) + HCO₃^-_(aq)

HCO₃^-_(aq) ⇌ H⁺_(aq) + CO₃^2-_(aq)

Weak acidity; equilibrium lies to the left.

Question 10

• Write an equation for the equilibrium constant of a weak acid, used to find [H⁺].
• Name this constant

Answer 10

HA ⇌ H⁺ + A^-

K_a = [H⁺] [A^-] / [HA]

K_a, acidity constant / acid dissociation constant

Question 11

To find [H⁺] for a weak acid by calculating K_a, what 2 assumptions are made for simplicity?

Answer 11

1: [H⁺] = [A^-]

Negate small number of protons added by ionisation of water: H₂O ⇌​ H⁺ + OH^-

2: [HA_eq] = [HA_initial]

Negate fraction of HA which have dissociated H⁺ - small since weak acid

Question 12

To find [H⁺] for a weak acid by calculating K_a, the following assumptions are made. When are they not good assumptions?

1: [H⁺] = [A^-] Negate small number of protons added by ionisation of water: H₂O ⇌​ H⁺ + OH^-

2: [HA_eq] = [HA_initial] Negate fraction of HA which have dissociated H⁺

Answer 12

1. For very dilute solutions
2. For stronger acids

Question 13

Calculate the pH of a solution of methanoic acid made up by adding 0.25 mol to 100 cm³ of water. K_a at 298 K = 1.6 x 10^-4 mol dm^-3.

Answer 13

K_a = [H⁺]² / [HA]

[H⁺]² / (0.25/0.1) = 1.6 x 10^-4

[H⁺] = √(1.6 x 10^-4 x 2.5)

pH = -log(√(1.6 x 10^-4 x 2.5)) = 1.7

Question 14

How do values of K_a compare for strong and weak acids?

Answer 14

For stronger acids, [HA] is lower and [H⁺] is greater, so K_a is greater.

Question 15

Since values of K_a for weak acids can be small, a log scale pK_a is often used. How is pK_a calculated?

Answer 15

pK_a = -log₁₀(K_a)

Question 16

The weaker the acid, the __ the value of pK_a

Answer 16

Greater

Question 17

Define an:

• Acidic solution
• Alkaline solution

Answer 17

• Acidic: [H⁺] > [OH^-]
• Alkaline: [OH^-] > [H⁺]

Question 18

What is a strong base?

Answer 18

One which produces OH^- ions completely in aqueous solution.

Question 19

1. What is the ionisation product of water?
2. How is it represented symbolically?

Answer 19

1. [H⁺] [OH^-], i.e. the product of the products of ionisation of water
2. K_w

Question 20

Give the value and unit of the ionisation product of water, K_w, at 298 K.

Answer 20

K_w = 1 x 10^-14 mol²dm^-6

Question 21

Explain why the expression for K_w is [H⁺] [OH^-].

Answer 21

K_a = [H⁺] [OH^-] / [H₂O]

Water is in excess so has no effect on equilibrium. Therefore, it is omitted:

K_w = [H⁺] [OH^-]

[H₂O] is ignored because it’s so large that it makes values small, and has a practically identical effect on all values

Question 22

Calculate the pH of 0.1 mol dm^-3 NaOH_(aq).

Answer 22

Strong base; dissociates completely. Neglect small amount of OH^- formed from ionisation of water. So:

[OH^-] = 0.1 mol dm^-3

K_w = [H⁺] [OH^-]

1 x 10^-14 = 0.1 [H⁺]

pH = -log(1 x 10^-14 / 0.1) = 13

Question 23

Calculate the pH of a 0.100 mol dm^-3 barium hydroxide solution at 298 K.

Answer 23

Ba(OH)₂ → Ba²⁺ + 2OH^-

Strong base; dissociates completely. Neglect small amount of OH^- formed from ionisation of water. So:

[OH^-] = 0.2 mol dm^-3

K_W = [H⁺] [OH^-]

1 x 10^-14 = 0.2 [H⁺]

pH = -log(1 x 10^-14 / 0.2) = 13.3

Don’t square [OH^-] due to the 2OH^-; this is not done for K_w as it is for K_c and K_sp

Question 24

Answer 24

a) Pink. H⁺ is removed by reaction with OH^-, so equilibrium position shifts to right to restore the constant K_a, so [In^-] increases.
b) K_a = [H⁺] [In^-] / [HIn]
c) pK_a = 9.3 so K_a = 10^-9.3

At end point, [In^-] = [HIn] so K_a = [H⁺] = 10^-9.3

pH = -log(10^-9.3) = 9.3

Question 25

Write the formula for a hydrogencarbonate ion.

Answer 25

HCO₃^-

Question 26

What is a buffer solution?

Answer 26

A solution which resists pH changes for small additions of acid or alkali.

Question 27

What do buffer solutions usually comprise of?

Answer 27

* Weak acid and one of its salts (e.g. CH₃COOH + CH₃COONa) * Weak base and one of its salts (e.g. NH₃ + NH₄Cl)

Question 28

The action of a buffer solution depends on the weak acid equilibrium: HA(aq) ⇌ H⁺(aq) + A^-(aq) What 2 assumptions are required to explain how buffers counteract small pH changes?

Answer 28

* Almost no weak acid (HA) molecules dissociate * (And therefore) all A^- ions come from salt; negate small amount from weak acid (HA)

Question 29

Explain how the pH of a buffer solution is reestablished when acid is added.

Answer 29

HA(aq) ⇌ H⁺(aq) + A^-(aq) * Conjugate base suppresses ionisation of acid: H⁺ ions added react with A^-, forming HA + water * H⁺ ions removed from solution * Equilibrium shifts to left * pH reestablished

Question 30

Explain how the pH of a buffer solution is reestablished when alkali is added.

Answer 30

HA(aq) ⇌ H⁺(aq) + A^-(aq) * OH^- ions added react with and remove H⁺ ions, forming H₂O * Weak acid HA ionises further, regenerating H⁺ * Equilibrium moves to right * pH reestablished

Question 31

Explain how the expression for K_a can be used to work out the pH of a buffer solution.

Answer 31

K_a = [H⁺] [A^-] / [HA] Assume that: * Almost no weak acid molecules dissociate * (And therefore) All A^- ions come from salt K_a = [H⁺] [salt] / [acid] [H⁺] = K_a [acid] / [salt] pH = -log(K_a [acid] / [salt])

Question 32

What factors does the pH of a buffer solution depend upon?

Answer 32

* The value of K_a * The ratio of acid to salt ## Footnote *[H⁺] = K_a x [acid] / [salt]*

Question 33

Explain why the pH of a buffer solution is not affected by dilution.

Answer 33

pH = -log[H⁺] = -log (K_a x [acid] / [salt]) * K_a is a constant, so it is unaffected by dilution (overall change in concentration does not alter equilibrium position) * Dilution reduces concentrations of acid + salt by same proportion, so ratio of concentrations is unchanged * These factors are what affect [H⁺]; pH depends on H⁺ so is unchanged

Question 34

Calculate the pH of a buffer solution containing 0.1 mol dm^-3 ethanoic acid and 0.2 mol dm^-3 sodium ethanoate. K_a for ethanoic acid is 1.7 x 10^-5 mol dm^-3 at 298 K.

Answer 34

K_a = [H⁺] [A^-] / [HA] = [H⁺] x [salt] / [acid] [H⁺] = K_a x [acid] / [salt] pH = -log(K_a x [acid] / [salt]) = -log(1.7 x 10^-5 x 0.1 / 0.2) = 5.1

Question 35

A mass of a solid base is added to an acid of known concentration to make up a buffer solution. The mass of base, concentration of acid and volume of solution are used to work out the pH. What assumption is made in doing this?

Answer 35

The total volume of the solution is not altered when the base is added.

Question 36

What is the relationship between K_a, K_b and K_w? *K_b is the weak base ionisation constant.*

Answer 36

K_w = K_b x K_a

Question 37

Suggest why alkaline solutions dissolve CO₂ more effectively than water.

Answer 37

CO₂ is weakly acidic: CO₂(g) ⇌ CO₂(aq) CO₂(aq) + H₂O ⇌ H⁺(aq) + HCO₃^-(aq) HCO₃^-(aq) ⇌ H⁺(aq) + CO₃^2-(aq) Overall: CO₂(g) + H₂O ⇌ CO₃^2-(aq) + 2H⁺(aq) * OH^- ions from alkali react with H⁺ ions * [H⁺] decreases * Equilibrium position moves right to restore K_a since it is a constant * More CO₂ gas dissolves

Question 38

Sodium ethanoate is used in buffer solutions. Calculate the mass of sodium ethanoate that would need to be added to 1.0 mol dm^-3 ethanoic acid in order to make 250 cm³ of buffer solution of pH 4.80. K_a of ethanoic acid is 1.74 x 10^-5 mol dm^-3.

Answer 38

[H⁺] required = 10^-4.8 K_a = [H⁺] [ethanoate] / [acid] 1.74 x 10^-5 = 10^-4.8 [ethanoate] / 1 [ethanoate] = 1.74 x 10^-5 / 10^-4.8 = 1.098 Mol CH₃COONa = conc x vol = 1.098 x 0.25 = 0.274 mol Mass CH₃COONa = mol x Mr = 0.274 x 82 = 23 g

Question 39

Answer 39

*Acid: NaH₂PO₄ Salt: Na₂HPO₄* [H⁺] = 10^-7 mol dm^-3 K_a = [H⁺] [salt] / [acid] → [acid] = [H⁺] [salt] / K_a Conc acid = (10^-7 x 0.1) / (6.2 x 10^-8) = 0.161 mol dm^-3 Mol acid = conc x vol = 0.161 x 1 = 0.161 Mass acid = mol x Mr = 0.161 x 120 = 19 g

Question 40

A student titrates sodium hydroxide solution against hydrochloric acid and records a titre of 7.5 cm³. They suggest that less than 7.5 cm³ of sodium hydroxide solution would be needed to neutralise H₂PO₄^- of the same concentration, since it is a weak acid. Discuss the student's suggestion.

Answer 40

* For a weak or strong acid of a given concentration, the total number of protons which will eventually be dissociated, and react with the base, is the same * It takes longer for a weak acid to reach the end-point, but the same volume of base is required

Question 41

Which statement is true of buffer solutions? 1. A buffer does not change in pH when small amounts of acid or alkali are added. 2. An acidic buffer is a mixture of a weak acid and a solution of its salt. 3. All buffers are acidic. 4. The pH of a buffer depends only on the ratio of its components’ concentrations.

Answer 41

2 * 1 incorrect since pH of buffers _does_ vary - just less than pH of things that aren't buffers* * 3 incorrect since they can be basic* * 4 incorrect since also dependent on K_a*

Question 42

Answer 42

B