C20 - Acids, Bases and pH

Question 1

What’s a Brønsted-Lowry acid?

Answer 1

A proton donor (H+)

Question 2

What’s a Brønsted-Lowry base?

Answer 2

A proton acceptor (H+)

Question 3

How can weak and strong acids be shown within an equation?

Answer 3

By the arrows used.

Equilibrium / reversible reaction signs suggest a weak acid.
A single arrow suggests it is strong.

Question 4

What’s a conjugate acid-base pair?

Answer 4

Two species that can be interconverted by transfer of proton e.g. HCl and Cl-.

In the forward reaction HCl releases a proton to form its conjugate base, Cl-.

In the reverse direction, Cl- accepts a proton to form its conjugate acid, HCl.

Question 5

What are the conjugate acid-base pairs in the reaction:

HCl + OH- H2O + Cl-

Answer 5

HCl + OH- H2O + Cl-

Acid 1 Base 2 Acid 2 Base 1

HCl donates a proton which is accepted by OH-.
In reverse, H2O donates a proton which is accepted by Cl-.

Question 6

What are the conjugate acid-base pairs in the reaction:

HCl + H2O H3O+ + Cl-

Answer 6

HCl + H2O H3O+ + Cl-

Acid 1 Base 2 Acid 2 Base 1

HCl donates a proton which is accepted by H2O.
H3O+ donates a proton in the reverse which is accepted by Cl-.

Question 7

What’s a monobasic acid?

Answer 7

An acid with one hydrogen atom, able to donate 1 proton.

Question 8

What’s a dibasic acid?

Answer 8

An acid with two hydrogen atoms, able to donate 2 protons.

Question 9

What’s a tribasic acid?

Answer 9

An acid with three hydrogen atoms, able to donate 3 protons.

Question 10

What do acids react with? (4)

What do they form?

Answer 10

Metals (salt and hydrogen)

Carbonates (salt, water and CO2)

Metal oxides (salt and water)

Hydroxides / alkalis (salt and water)

Question 11

What is the full and ionic equation for the reaction between HCl and Mg?

Answer 11

Full:

2HCl (aq) + Mg (s) -> MgCl2 (aq) + H2 (g)

Ionic:

2H+ (aq) + Mg (s) -> Mg 2+ (aq) + H2 (g)

Question 12

What is the full and ionic equation for the reaction between H2SO4 and Mg?

Answer 12

Full:

H2SO4 (aq) + Mg (s) -> MgSO4 (aq) + H2 (g)

Ionic:

2H+ (aq) + Mg (s) -> Mg 2+ (aq) + H2 (g)

Question 13

What is the mathematical relationship between pH and [H+ (aq)] ?

Answer 13

pH = -log[H+ (aq)]

It’s inverse is:
[H+] = 10^(-pH)

Being logarithmic, each pH has a scale difference of 10; so a pH of 1 has 10 times the concentration of a pH 2.

Question 14

How is pH calculated?

Answer 14

pH = -log[H+ (aq)]

Question 15

How is the concentration of protons / H+ ions within a substance calculated (using pH)?

Answer 15

[H+ aq] = 10 ^(-pH)

Question 16

What is the concentration of protons in a substance of pH 2 and pH 8?

Answer 16

pH 2: 10^-2 = 0.01 moldm-3

pH 8: 10^-8 = 0.00000001 moldm-3

Question 17

What is the ‘strength’ of an acid?

Answer 17

The extent of dissociation of an acid.

The stronger the acid, the greater the dissociation of H+ ions.
Weak acids only partially dissociate.

Question 18

What is Ka?

Answer 18

The acid dissociation constant.
It shows how acidic a solution is - the greater the value, the further the equilibrium is to the right and the stronger the acid.

Question 19

How is Ka calculated?

Answer 19

Ka = [H+][A-] / [HA]

Question 20

How is the acid dissociation constant calculated?

Answer 20

Ka = [H+][A-] / [HA]

Question 21

What is pKa?

Answer 21

A negative log value of Ka to compare pH using simpler numbers than the large numbers for Ka.

Question 22

How is pKa calculated?

Answer 22

pKa = -logKa

Question 23

How can Ka be calculated using pKa?

Answer 23

Ka = 10^-pKa

Question 24

What are the values of Ka and pKa like for strong and weak acids?

Answer 24

For strong acids, Ka is large and pKa is a small value.

For weak acids, Ka is small and pKa is large.

Question 25

What happens to weak acids (HA) in equilibrium?

Answer 25

HA H+ + A-

[H+] depends on the the concentration of the acid of acid [HA] and acid dissociation Ka.

Question 26

What assumptions are made abut the Ka equation:

Ka = [H+][A-] / [HA]

What are the limitations?

Answer 26

Assumptions:
1) [H+] = [A-]
H20 H+ + OH-
Water also releases H+ ions but this is ignored.

2) [HA] at the start = [HA] at equilibrium
We assume that the dissociation of H+ ions is so small that the concentration of acid does not change.

Limitations:
1) Only works with strong-weak acids. With weak-weak acids, the dissociation is significant.

2) Only works with weak-weak acids. With strong-weak acids, more has been dissociated so difference in concentration will be greater between the start and equilibrium.

Question 27

What is Kw?

Answer 27

The ionic product of water.

Question 28

How is Kw calculated?

Answer 28

Kw = [H+][OH-]

At 25’C, Kw = 1*10^-14 and the pH of water is 7.

Question 29

How is the ionic product of water calculated?

Answer 29

Kw = [H+][OH-]

Question 30

How can the pH of a strong base be calculated?

Answer 30

Using the concentration of the base and ionic product of water (Kw).

Question 31

How would the concentrations of H+ and OH- ions in a solution of oH 3.25 at 25°C be calculated?

Answer 31

1) Use calc to find [H+].
10^-3.25 = [H+] = 5.62*10^-4

2) Use [H+] and Kw to calculate [OH-].
[OH-] = Kw/[H+] = 110^-14 / 5.6210^-4

= 1.78*10^-11 moldm-3

Question 32

What is the pH of a solution with [OH-] of 2*10^-2 moldm-3 at 25°C?

Answer 32

1) Calculate [H+] from Kw and [OH-].
[H+] = 110^-14 / 210^-2
= 5*10^-13

2) Calculate pH
pH = -log[H+]
= 12.30

Question 33

What is the pH of a solution of 0.075 moldm-3 NaOH at 25°C?

Answer 33

1) Convert NaOH into OH-

NaOH is a strong monoacidic alkali which completely dissociates.
[OH-] = [NaOH] = 0.075 moldm-3

2) Use Kw and [OH-] to find [H+]

[H+] = 1.33*10^-13 moldm-3

3)Calculate pH
pH = -log[H+]
=12.88