O - Acids, Bases, pH and buffers

Question 1

Give the Bronsted-Lowry definition of an acid

Answer 1

Proton donors

Question 2

Give the Bronsted-Lowry definition of a base

Answer 2

Proton acceptors

Question 3

Give the equation for the reaction between an acid and water

Answer 3

HA (aq) + H₂O (l) → H₃O⁺ (aq) + A⁻ (aq)

Question 4

Give the equation for the reaction between a base and water

Answer 4

B (aq) + H₂O (l) → BH⁺ (aq) + OH⁻ (aq)

Question 5

Define a strong acid and give an example

Answer 5

Strong acids ionise almost completely in water (nearly all H⁺ released)
Example: Hydrochloric acid
HCl (g) + H₂O (l) → H⁺ (aq) + Cl⁻ (aq)
(this reaction is reversible but the equilibrium is well to the right)

Question 6

Define a strong base and give an example

Answer 6

Strong bases ionise almost completely in water
Example: Sodium Hydroxide
NaOH (s) + H₂O (l) → Na⁺ (aq) + OH⁻ (aq)
(this reaction is reversible but the equilibrium is well to the right)

Question 7

Define a weak acid and give an example

Answer 7

Weak acids only ionise very slightly in water, so small numbers of H⁺ ions are formed. An equilibrium is set up which is well over to the left
Example: Ethanoic acid
CH₃OOH (aq) + H₂O (l) ⇌ CH₃OO⁻ (aq) + H⁺ (aq)

Question 8

Define a weak base and give an example

Answer 8

Weak bases only ionise slightly in water. An equilibrium is set up which is well over to the left
Example: Ammonia
NH₃ (aq) + H₂O (l) ⇌ NH₄⁺ (aq) + OH⁻ (aq)

Question 9

Describe how acids and bases react

Answer 9

Acids donate their protons to the base

HA (aq) + B (aq) ⇌ BH⁺ (aq) + A⁻ (aq)

Question 10

Describe what happens when you add an acid to water

Answer 10

The water acts as a base and accepts the protons:

HA (aq) + H₂O (l) → H₃O⁺ (aq) + A⁻ (aq)

Question 11

Define conjugate pairs

Answer 11

A pair made between an acid and a base either side of the equilibrium which transfer an Hydrogen ion between them

Question 12

State the conjugate pairs in the following reaction:

HCl + H₂O ⇌ H₃O⁺ + Cl⁻

Answer 12

• HCl and Cl⁻ form a conjugate pair where HCl is the acid

- H₂O and H₃O⁺ form a conjugate pair where H₃O⁺ is the acid

Question 13

State the conjugate pairs in the following reaction:
B + H₂O ⇌ BH⁺ + OH⁻
where B is a base

Answer 13

• B and BH⁺ form a conjugate pair where BH⁺ is the acid

- H₂O and OH⁻ form conjugate pair where H₂O is the acid

Question 14

Describe how water can behave as both an acid and a base

Answer 14

It can act as an acid by donating a proton or as a base by accepting a proton
Both hydroxonium ions and hydroxide ions will always be present at the same time due to the equilibrium:
H₂O ⇌ H⁺ + OH⁻

Question 15

Describe why the equilibrium lies to the left in the following reaction:
H₂O ⇌ H⁺ + OH⁻
Give an expression for the equilibrium constant Kc

Answer 15

Because water only dissociates a small amount

Kc = [H⁺][OH⁻] / [H₂O]

Question 16

Define Ionic product of water

Answer 16

The Ionic Product of Water, Kw, is the equilibrium constant for the reaction in which water undergoes an acid-base reaction with itself.
Kw = Kc x [H₂O] = [H⁺][OH⁻]

Question 17

What makes a solution neutral?

Answer 17

An equal concentration of H⁺ and OH⁻ ions

[H⁺] = [OH⁻]

Question 18

Why is the pH logarithmic?

Answer 18

Because the concentration of hydrogen ions can vary enormously

Question 19

Give the equation to find the pH of a solution

Answer 19

pH = -log₁₀[H⁺]

Question 20

Name 2 strong monoprotic acids

Answer 20

Hydrochloric Acid (HCl)
Nitric Acid (HNO₃)

Question 21

Describe the relationship between the concentration of hydrogen ions and the acid concentration for a strong monoprotic acid

Answer 21

Each mole of acid produces 1 mole of hydrogen ions

[HA] = [H⁺]

Question 22

Describe how do you find the pH of a strong base

Answer 22

[H⁺] is linked to [OH⁻] by the ionic product of water
Since Kw = [H⁺][OH⁻]
[H⁺] = Kw / [OH⁻]
And thus the pH of a base is:
pH = -log₁₀[H⁺] = -log₁₀(Kw / [OH⁻])

Question 23

Give the equation for the equilibrium constant Ka for the reaction:
HA ⇌ H⁺ + A⁻

Answer 23

Ka = [H⁺]² / [HA]

Question 24

Define pKa

Answer 24

pKa is the negative base-10 logarithm of the acid dissociation constant of a solution
pKa = -log₁₀Ka
Ka = 10⁻ᵖᴷᵃ
The lower the pKa, the stronger the acid

Question 25

What is Ka?

Answer 25

The acid dissociation constant

Question 26

Give 5 examples for the use of buffers

Answer 26

1) Shampoo
2) Biological washing powders
3) Regulation of cell pH
4) Regulation of blood pH
5) Food products to control pH change due to bacteria and fungi

Question 27

State the equation for the process of coupling of a buffer in the blood to the respiratory system

Answer 27

2H₂O + CO₂ ⇌ H₂CO₃ + H₂O ⇌ H₃O⁺ + HCO₃⁻