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Edexcel Chemistry A level > Topic 12 Acid - Base equilibria > Flashcards

Topic 12 Acid - Base equilibria Flashcards

1
Q

What are bronsted lowry acids and bases?

A

Bronsted lowry acids are proton donors. When we mix acids with water, H+ ions are released.
H+ ions don’t exist on their own in water. They form hydroxonium ions (H3O+). It is these ions that make a solution acidic however for simplicity you will normally just see H+ in equations
HA(aq) + H2O(l) —–> H3O+(aq) + A-

Bronsted Lowry bases are proton acceptors. When we mix bases with water, they react with H+ ions to form hydroxide ions.
B(aq) +H2O(l) —-> BH+ + OH-

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2
Q

Difference between Strong and weak acids and bases (examples)

A

When acids and bases react with water they form a reversible reaction.
HA + H2O <—–> H3O+ + A-
or
B + H2O <——> BH+ + OH-

Strong bases dissociate almost completely. Weak bases dissociate partially.
Strong acids dissociate almost completely. Weak acids dissociate poorly

Weak acid - CH3COOH (Ethanoic acid) and other carboxylic acids. Backwards reaction is favoured so not many H+ produced
Strong acid - HCl, H2SO4, HNO3. Forwards reaction is favoured so lots of OH- ions are produced.
Strong base - NaOH, KOH. Forward reaction is favoured strongly. Lots of OH- ions produced.
Weak Base - NH3. Backwards reaction favoured so not many OH- ions produced.

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3
Q

What are conjugate pairs?

A

A conjugate pair is linked by a proton.
Any species that has gained a proton is the CONJUGATE ACID. and a species that has lost a proton is the CONJUGATE BASE.
Conjugate acid - HA
Conjugate base A-
Conjugate acid - BH+
Conjugate base - B

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4
Q

What happens in acid - base reactions?

A

When acids and bases react with each other, protons are exchanged.

In this generic example, the acid HA donates a proton to the base B. Positive and Negative ions are produced.
HA + B<—–> BH+ + A-

Water behaves like a base when acid is added to it.
HA + H2O <—–> H3O+ + A-

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5
Q

What is formed in an acid base reaction?

A

Acids react with bases to form salts which are PH neutral

The H+ ions produced from acids react with OH- ions produced from alkalis, we make water which is neutral. [H+]=[OH-].
Salts are made from the metal from the base and the non-metal from the acid.

Lithium chloride:
HCl + LiOH —-> LiCl + H2O

Potassium nitrate:
HNO3 + KOH —-> KNO3 + H2O

Ammonium sulfate:
Ammonia reacts with acids to make Ammonium salts but NO water
2NH3 + H2SO4 —–> (NH4)2SO4

Be careful with ammonia. Ammonia doesn’t produce OH- ions directly. It reacts with water first and accepts a proton to produce ammonium ions (NH4+) and OH- ions

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6
Q

What is enthalpy change of neutralisation?

A

We get a change in enthalpy when neutralisation occurs

Standard enthalpy change of neutralisation:
The enthalpy change when acid and base solutions react together, under standard conditions, to produce 1 mole of water.

Neutralisation reactions are always exothermic so enthalpy changes of neutralisation are always negative

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7
Q

Enthalpy change of neutralisation for weak acids and bases

A

These dissociate weakly and the OH- and H+ ions are used up quickly in a neutralisation reaction as there are only a small number in the solution.
Because of this the acid and base are constantly dissociating to replace the H+ and OH- ions that have reacted

There are 2 types of enthalpy involved in neutralisation:
1. Enthalpy of dissociation
2. Enthalpy is when OH- and H+ ions react

The enthalpy change of dissociation varies depending on the acid or the base being used. This means the enthalpy of neutralisation of weak acids and bases varies too

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8
Q

Enthalpy change of neutralisation for strong acids and bases

A

These dissociate fully to produce OH- and H+ ions in solution

In neutralisation reactions involving strong acids and bases, there is only 1 type involved. Enthalpy when OH- and H+ ions react.
Unlike neutralisation reactions in weak acids and bases, there is no enthalpy of dissociation with strong acids and bases as they fully dissociate

As a result, the standard enthalpy of neutralisation is similar for all reactions of strong acids and bases

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9
Q

How do you calculate PH?

A

Ph is a logarithmic scale that measures the concentration of + ions in solution

PH = -log10[H+]
- need to find out concentration of H+ ions
[H+] = 10 ^ -PH

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10
Q

What are polyprotic acids?

A

This is when acids can donate more than 1 proton

HNO3. 1 mole of nitric acid is monoprotic. It will produce 1 mole of H+ ions
H2SO4. 1 mole of sulfuric acid is diprotic. It will produce 2 moles of H+ ions
H3PO4. 1 mole of phosphoric acid is triprotic. It will produce 3 moles of H+ ions

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11
Q

How to calculate the PH of strong acids?

A

We assume they fully dissociate

Monoprotic acids dissociate to produce 1 H+ ion for every acid molecule. This means the concentration of the acid = the concentration of the H+ ions
For example, the PH of 0.25moldm-3 of HCl is -
[H+] = [Acid]
PH = -log10 0.25
PH = 0.6

Diprotic acids dissociate to produce two H+ ions for every acid molecule. This means the concentration of the acid = 2 x the conncentration of the H+ ions
For example, the PH of 0.25moldm-3 of sulfuric acid is -
2[H+] = [Acid]
0.25 of Acid produces 0.5 of H+ ions
PH = -log10 0.5
PH = 0.3

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12
Q

How to calculate the PH of strong bases?

A

We assume they fully dissociate

Most strong bases dissociate to reduce 1 OH- ion for every base molecule. This means the concentration of the base = the concentration of the OH- ions.

To calculate the PH of a base we still need to [H+]. To get this we need to use the ionic product of water (Kw).

Kw = [H+][OH-]
Rearrange to calculate [H+]. Then put it in the PH = -log10 [H+]

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13
Q

How to calculate the PH or Ka or concentration of a weak acid?

A

Weak acids only dissociate slightly in aqueous solutions so we have to use another constant to help work out their PH values.

We can’t assume [H+] = [Acid]

Assumption1:
Only a small amount of the weak acid (HA) dissociates so we can assume that
[HA] equilibrium = [HA] star

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

Assumption 2:
The dissociation of acid is greater than the dissociation of water present in the solution. We can assume all the H+ ions come from the acid.
[H+] = [A-]

Ka = [H+]^2 / [HA]

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14
Q

What is the ionic product of water? (Kw)

A

Water exists in equilibrium with its ions. In other words, a glass of water doesn’t just contain water molecules.

Water dissociates into hydroxide ions and hydroxonium ions as illustrated in this equation -
2H2O <—–> H3O+ + OH-. A simplified version is H2O <—–> H+ + OH-
Water dissociates into its ions very weakly. In fact, there is so little OH- and H+ ions compared with H2O molecules that we assume that the concentration of water has a constant value.

Kw = [H+][OH-]

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15
Q

What are some important points related to the value of Kw?

A

The value of Kw is the same in a solution at a given temperature
Kw = 1 x 10^-14 mol2dm-6
This value changes if the temperature changes.
Pure water has an equal concentration of H+ and OH- ions. In other words, [H+] = [OH-]
When referring to pure water, Kw = [H+]^2

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16
Q

What is pKw?

A

It is used to display Kw values on a smaller scale which make it easier to use.

pKw = - log10 Kw

Kw = 10 ^ -pKw

17
Q

How to calculate pKa?

A

It is another way of measuring the stength of an acid similar to PH. The lower the value, the stronger the acid

pKa = -log10 Ka
Ka = 10 ^ -pKa

Calculate the pKa value for an acid with a Ka value of 7.52 x 10^-3
pKa = -log10 Ka
= -log10 7.52 x 10^-3
= 2.12

Calculate the PH of 0.025moldm-3 of ethanoic acid at 298K which has a pKa value of 4.75 at 298K

  1. Calculate Ka
  2. Calc [H+] from Ka expression
  3. Calculate PH
18
Q

How do you measure pH experimentally?

A

pH meters measure the pH of a solution however they must be calibrated correctly to provide reliable readings

The pH probe must be placed in distilled water first. The meter should be reading pH 7. Adjust if it is not.
Then we repeat this process with standard solutions at PH 4 and PH 10. Make sure you rinse with distilled water before each PH solution.

The PH meter should be calibrated for use. Remember - wash the probe with distilled ater after each test

19
Q

Understanding PH

A

PH indicates whether a substance is an acid or base and if it’s weak or strong

Here we will compare equimolar (same number of moles) substances

Strong acid:
1 moldm-3 HCl (PH 0)
[H+] = 10^-PH
[H+] = 1
HCl has to be a strong acid as its H+ = 1

Weak acid:
1 mol of CHOOH (PH 2.4)
H+ = 10 ^ -PH
H+ = 3.98 x 10^-3
CHOOH has to be a weak acid as H+ = 0.00398

Neutral:
1 mol of NaCl (PH 7)
[H+] = 10 ^ -PH
[H+] = 1 x 10^-7
NaCl has to be neutral as it’s [H+] = 1 x 10 ^-7
This means [H+] = [OH-]
and Kw = 1x10^-14

Weak base:
1 mol of NH3 (PH 10.6)
[H+] = 10^-PH
[H+] = 2.4 X 10^-11
NH3 has to be a weak base as it’s [H+] = 2.4 x 10^-11
This means [OH-] = 4.2 x 10^-4

Strong base
1 mol of NaOH (PH 14)
[H+] = 10^-PH
[H+] = 0
NaOH has to be a strong base as it’s [H+] = 1 x 10^-14
This means [OH-] = 1

20
Q

How to calculate the Ka from PH and mass?

A

Calculate the Ka of a solution made from 1.2g of methanoic acid (HCOOH) dissolved in 250cm3 of water with a PH of 3.14

  1. Calculate moles of methanoic acid
    n = m / M
    = 1.2 / 46
    = 0.026 moles
  2. Calculate the concentration of methanoic acid solution
    c = n/v
    = 0.026 / 0.25
    = 0.104 moldm3
  3. Calculate [H+] using the pH equation
    pH = =log10[H+]
    [H+] = 10 ^ -PH
    [H+] = 10 ^ -3.14
    [H+] = 7.24 x 10^-4 moldm-3
  4. Use Ka expression
    Ka = [H+]^2 / [HCOOH]
    = (7.24 X 10^-4)^2 / 0.104
    = 5.04 X 10^-6
21
Q

Acid dilution and PH

A

Strong acids:
If we diliute a strong acid by a factor of 10, then PH will increase by 1
Acid conc = PH of HCl at 298K
1 = 0
0.1 = 1
0.01 = 2
0.001 = 3

Weak acids:
If we dilute a weak acid by a factor of 10 then PH will increase by 0.5
Acid conc = PH of CH3COOH at 298K
1 = 2.5
0.1 = 3
0.01 = 3.5
0.001 = 4

22
Q

How to carry out a titration?

A
  1. Have acid or alkali in burette with a known concentration
  2. Have an acid or alkali with an unknown concentration but a known volume in the conical flask. Add a few drops of indicator too
  3. Add the chemical in the burette to the conical flask until the indicator changes colour. This is known as the endpoint. Add drop by drop near endpoint
  4. Read how much chemical was added from the burette to neutralise the chemical in the conical flask. Read the bottom of the meniscus. Always read at eye level
  5. Record your results to 2 d.p and repeat until you get 2 results that are concordant (within 0.10cm3 of each other)
23
Q

What do different titration curves look like?

A

Strong acid / Strong base:
Graph starts at PH 1 as there is excess strong acid
Ends at around PH 13 as we now have excess strong base

Strong acid / Weak base:
Graph starts at PH 1 as there is excess strong acid
Ends at around PH 9 as we now have excess weak base

Weak acid / Strong base:
Graph starts at around PH 5 as there is excess weak acid
Ends at around PH 13 as we now have excess strong base

Weak acid / Weak base:
Graph starts at around PH 5 as there is excess weak acid
Ends at around PH 9 as we now have excess base

24
Q

What are some features about titration curves?

A

The equivalence point or end point - at this point the acid has been neutralised by the base. The sharp vertical rise shows a rapid change in PH.

The half neutralisation point - This is the point half way between 0 and the equivalence point. It can be used to calculate pKa of a weak acid by taking the PH at this point.
At this point [HA] = [A-]
Ka = [H+][A-] / [HA]
Ka = [H+]
-logKa = -log [H+]
pKa = PH

25
Q

Indicators in titration curves

A

A suitable indicator must change colour entirely within the vertical part of the titration curve for it to be effective.

Methyl orange colour change = PH 3 - 4.5
It is RED at low PH and YELLOW at high PH values
It can be used for strong acid / weak base and strong acid / strong base

Phenolphthalein colour change = PH 8.2 - 10
It is COLOURLESS at low PH and PINK at high PH
It can be used for Weak acid / strong base

Weak acid / weak base have no sharp change. No indicator is suitable so we have to use a PH meter

26
Q

What is a buffer?

A

A buffer is a chemical that can resist the change in PH when small amounts of acid or base are added

Buffers don’t stop the change in PH, just resist it.
There are 2 types of buffers:
Acidic and Basic

27
Q

What is an acidic buffer and how does it work?

CH3COOH (weak acid) and Sodium ethanoate CH3COO-Na+ (its salt)

A

They resist the change in PH in order to keep the solution below PH 7. They are made from a WEAK acid and a salt of its conjugate base.

An example of an acidic buffer is using ethanoic acid CH3COOH (weak acid) and Sodium ethanoate CH3COO-Na+ (its salt)
In any buffer solution, there are 2 equilibrium equations at plau. The 2 equations coexist in the same beaker

CH3COOH <—-> CH3COO- + H+
[high] [low] [high]
Weak acid dissociates weakly so equilibrium lies well over to the left
CH3COO-Na+ <——> CH3COO- + Na+
[low] [high] [high]
Salts dissociate strongly (fully) so equilibrium lies well over to the right.

28
Q

What happens when we add an acid (H+) to this buffer?

CH3COOH (weak acid) and Sodium ethanoate CH3COO-Na+ (its salt)

A

The H+ ions react with the CH3COO- ions in solution
There is a high concentration of these from the salt
More CH3COOH is produced which means equilibrium shifts to the left

29
Q

What happens when we add a base (OH-) to this buffer?
CH3COOH (weak acid) and Sodium ethanoate CH3COO-Na+ (its salt)

A

The OH- ions react with the H+ ions in solution
There is a low concentration of these however they can be reproduced from a high concentration of CH3COOH to counteract the change (Le Chatelier’s principle)
Equilibrium shifts to the right to replace the reacted H+ ions

30
Q

What else can acidic buffers be made from?

A

They can be made from an excess weak acid and a strong base.

An example of an acidic buffer is using Ethanoic acid CH3COOH (weak acid) and Sodium hydroxide NaOH.
All of the base reacts with the acid in the following way.
CH3COOH + OH- —-> CH3COO- + H2O

Here we have an excess of ethanoic acid so we still have some left after al the OH- ions have reacted. In our beaker, we have a mixture of weak acid, its salt and water.
Because there is still some weak acid remaining it partially dissociates in the following way
CH3COOH <—–> CH3COO- + H+

31
Q

What are basic buffers and how do they work?

NH3 (weak base) and NH4+Cl- (it’s salt)

A

They resist the change in PH in order to keep the solution above PH7. They are made from a weak base and its salt.

An example of a basic buffer is using NH3 (weak base) and NH4+Cl- (it’s salt)
NH3 + H2O <—–> NH4+ + OH-
[High] [Low] [Low]
Weak base produces very little OH- ions so equilibrium lies well over to the left
NH4+Cl- <——> NH4+ + Cl-
[Low] [High] [High]
Salts dissociate strongly (fully) so equilibrium lies well over to the right

32
Q

What happens when we add a base (OH-) to this buffer?

NH3 (weak base) and NH4+Cl- (it’s salt)

A

The OH- ions react with the NH4+ ions in solution
There is a high concentration of these from the salt
More NH3 and H2O is produced which means equilibrium shifts to the left.

33
Q

What happens when we add an acid (H+) to this buffer?

NH3 (weak base) and NH4+Cl- (it’s salt)

A

The H+ ions react with the OH- ions in the solution
There is a low concentration of these however they can be reproduced from a high concentration of NH3 and H2O to counteract the change.
Equilibrium shifts to the right to replace the reacted OH- ions

34
Q

What is the link between titration curves and buffers?

A

Initially, PH changes quickly as there are plenty of OH- ions from the strong base to react with H+ ions from the weak acid

The curve has then become more level. A buffer solution exists between the weak acid and the salt of a conjugate base. This resists changes in PH as more base is added and so the curve is more level.

When all of the weak acid has been used up, the equivalence point is reached and the curve rises steeply. The buffer system doesn’t exist now

35
Q

What are the uses of buffers?

A

A good example would be blood.

It is vital to make sure blood is maintained as close to PH 7.4 as possible. Our body system relies on this so a buffer is present in our blood to help. Carbon dioxide plays a big role here.

A Carbonic acid (H2CO3) - hydrogen carbonate (HCO3-) system exists
H2CO3 <——> H+ + HCO3-
and
H2CO3 <—–> H2O + CO2

Carbonic acid is (H2CO3) controlled by respiration in your cells. When we breathe out CO2 the level of carbonic acid reduces as equilibrium shifts right to attempt to replace them.

HCO3- is controlled via your kidneys. Excess is removed here

36
Q

How do you calculate the PH of a buffer?

A

You need to know the Ka value and the concentration of the weak acid and its salt

Calc the PH of a buffer that contains 0.0235 mol of methanoic acid and 0.0184 mol of sodium methanoate. The value of Ka at 25degrees for methanoic acid is 0.000178moldm-3

  1. Write the equation and the Ka expression
    HCOOH <—–> H+ + HCOO-
    Ka = [H+][HCOO-] / [HCOOH]
    (we must use equilibrium concentrations here not initial concentrations
  2. Rearrange the expression to get [H+]
    [H+] = Ka x [HCOOH] / [HCOO-]
  3. Calculate [H+]
    [H+] = 0.0000178 x 0.0235 / 0.0184
    [H+] = 2.27 x 10^-4
37
Q

How do you calculate the concentration in a buffer?

A

PH = pKa + log10 ([A-] / [HA])

This equation assumes that [HA] = [HA] start
[A-] = [A-] start

Remember pKa = -log10Ka

Edexcel Chemistry A level (22 decks)

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