O 2 - Equilibria (Acid-Base)

Question 1

What is a Bronsted-Lowry acid and base?

Answer 1

Bronsted-Lowry acid: proton donor
Bronsted-Lowry base: proton acceptor

Question 2

What is the proton donor and acceptor in this reaction:
HNO2(s) + H2)(l) –> H3O+(aq) + NO2-(aq)

Answer 2

Proton donor: HNO2(aq)
Proton acceptor: H2O(l)

Question 3

What is a conjugate acid-base pair?

Answer 3

A conjugate acid-base pair contains two species that can be easily converted by transferring a proton

     Acid 1       Base 2             Acid 2         Base 1 E.g.  HCl(aq)  +  H2O(l)  <=>  H3O+(aq)  +  Cl-(aq) HCl and Cl- are conjugate acid-base pairs H2O and H3O+ are conjugate acid-base pairs

Question 4

What is a strong acid?

Answer 4

A strong acid is an acid that completely dissociates in solution:
HCl(aq) <=> H+(aq) + Cl-(aq)

Question 4

What is a strong base?

Answer 4

A strong base is a base that completely dissociates in solution:
NaOH(aq) <=> Na+(aq) + OH-(aq)

Question 4

What is a weak acid?

Answer 4

A weak acid is an acid that only partially dissociates in solution:
CH3COOH(aq) <=> CH3COO-(aq) + H+(aq)

Question 5

What is the acid dissociation constant?

Answer 5

It measures, quantitatively, the strength of acid in solution
Ka = [H+][A-] / [HA]

Question 6

What is pKa?

Answer 6

pKa = -logKa
pKa values are much more manageable than Ka values and make it easier to compare the strengths of solutions

Question 7

What is pH?

Answer 7

It is an easier way of measuring hydrogen ion concentration
There’s a large range of [H+] values with negative powers of 10. The negative logarithm of [H+] gives a more manageable scale of 1 to 14 rather than 10^-1 to 10^-14

Question 8

How can you measure the pH of a solution?

Answer 8

With a pH probe or using pH scales with a suitable indicator

Question 9

How do you calculate the pH of a strong acid?

Answer 9

HA <=> H+ + A-
Strong acid: therefore the concentration of acid = concentration of H+ ions
pH = -log[H+]

Question 10

How can you calculate the pH of a strong base?

Answer 10

XOH <=> X+ + OH-
Strong base: therefore concentration of base = concentration of OH- ions
Kw = [H+][OH-] so [H+] = Kw / [OH-]
pH = -log[H+]

Question 11

How do you calculate the pH of a weak acid?

Answer 11

HA <=> H+ + A-
Write Ka expression: Ka = [H+][A-] / [HA]
We can assume [H+] = [A-] so: Ka = [H+]^2 / [HA]
Rearrange to make [H+] the subject: [H+] = (root) Ka[HA]
pH = -log[H+]

Question 12

What is a buffer?

Answer 12

A system that minimises pH changes on addition of small amounts of an acid or base

Question 13

How do buffers work?

Answer 13

E.g. CH3COOH(aq) <=> CH3COO-(aq) + H+(aq)
- Upon addition of acid: More H+ ions are present in the solution and so combine with CH3COO-(aq) to form CH3COOH. The reverse reaction is favoured and the position of equilibrium shifts to the left
- Upon addition of base: More OH- ions are present in the solution and so combine with H+ to form H2O. The forward reaction is favoured and the position of equilibrium shifts to the right

Question 14

What is a weak acid buffer?

Answer 14

A mixture of a weak acid and its conjugate base (usually in the form of one of its salts i.e. CH3COO-Na+)

Question 15

How do you calculate the pH of a weak acid buffer solution?

Answer 15

E.g. CH3COOH(aq) <=> CH3COO-(aq) + H+(aq)
Write Ka expression: Ka = [H+][CH3COO-] / [CH3COOH]
Make [H+] the subject: [H+] = Ka[CH3COOH] / [CH3COO-]
Calculate [H+] and then substitute into pH = -log[H+]

Question 16

What assumptions are made in this weak acid buffer calculation?

Answer 16

[Salt] = [A-]
[Acid] = [HA] as only slightly depleted

Question 17

How do you calculate the pH of a buffer solution made by partial neutralisation?

Answer 17

E.g. HCOOH(aq) + NaOH(aq) –> HCOO-Na+(aq) + H2O(l)
Here, the acid is in excess so not all acid is converted into the salt, leaving some leftover. All NaOH reacts however
Calculate the number of moles of each compound reacted and at equilibrium and hence their concentrations
Substitute into the Ka expression and rearrange for [H+]
pH = -log[H+]