C20 - Acids, Bases and pH Flashcards Preview

Year 13 - Chemistry > C20 - Acids, Bases and pH > Flashcards

Flashcards in C20 - Acids, Bases and pH Deck (33)
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1
Q

What’s a Brønsted-Lowry acid?

A

A proton donor (H+)

2
Q

What’s a Brønsted-Lowry base?

A

A proton acceptor (H+)

3
Q

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

A

By the arrows used.

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

4
Q

What’s a conjugate acid-base pair?

A

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.

5
Q

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

HCl + OH- H2O + Cl-

A

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-.

6
Q

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

HCl + H2O H3O+ + Cl-

A

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-.

7
Q

What’s a monobasic acid?

A

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

8
Q

What’s a dibasic acid?

A

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

9
Q

What’s a tribasic acid?

A

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

10
Q

What do acids react with? (4)

What do they form?

A

Metals (salt and hydrogen)

Carbonates (salt, water and CO2)

Metal oxides (salt and water)

Hydroxides / alkalis (salt and water)

11
Q

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

A

Full:

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

Ionic:

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

12
Q

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

A

Full:

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

Ionic:

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

13
Q

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

A

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.

14
Q

How is pH calculated?

A

pH = -log[H+ (aq)]

15
Q

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

A

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

16
Q

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

A

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

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

17
Q

What is the ‘strength’ of an acid?

A

The extent of dissociation of an acid.

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

18
Q

What is Ka?

A

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.

19
Q

How is Ka calculated?

A

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

20
Q

How is the acid dissociation constant calculated?

A

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

21
Q

What is pKa?

A

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

22
Q

How is pKa calculated?

A

pKa = -logKa

23
Q

How can Ka be calculated using pKa?

A

Ka = 10^-pKa

24
Q

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

A

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

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

25
Q

What happens to weak acids (HA) in equilibrium?

A

HA H+ + A-

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

26
Q

What assumptions are made abut the Ka equation:

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

What are the limitations?

A

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.

27
Q

What is Kw?

A

The ionic product of water.

28
Q

How is Kw calculated?

A

Kw = [H+][OH-]

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

29
Q

How is the ionic product of water calculated?

A

Kw = [H+][OH-]

30
Q

How can the pH of a strong base be calculated?

A

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

31
Q

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

A

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

32
Q

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

A

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

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

33
Q

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

A

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