Chapter 21 - Buffers and neutralisation Flashcards

1
Q

What is a buffer solution?

A

A system that minimises PH changes when small amounts of an acid or base are added

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
2
Q

What 2 components does a buffer solution contain?

A

Weak acid and it’s conjugate base

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
3
Q

How can a weak acid buffer solution be prepared from a weak acid and its salt?

A
  • When the weak acid (e.g. ethanoic acid ) is added to water, the acid partially dissociates and the number of ions is very small
  • The salt (e.g. sodium ethanoate) completely dissolve into ions when added to water
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
4
Q

How can a weak acid buffer solution be prepared by partial neutralisation?

A

Adding an aqueous solution of an alkali (e.g. NaOH(aq)) to an excess of the weak acid is partially neutralised forming the conjugate base. Some of the weak acid is left over unreacted. Solution contains mixture of the salt of the weak acid and any unreacted weak acid

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
5
Q

What controls the PH in an acid buffer solution?

A

Conjugate acid-base pair (HA(aq)/A-(aq)) - using le chatelier’s principle

HA(aq) H+(aq) + A-(aq)

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
6
Q

What factors determine the PH of a buffer solution?

A
  • Ka/ acid strength
  • Temperature
  • Concentrations/ratio of weak acid and conjugate base/salt
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
7
Q

In a buffer solution, what happens on addition of an acid?

A

HA(aq) H+(aq) + A-(aq)

  1. ) [H+(aq)] increases
  2. ) H+(aq) ions react with the conjugate base, A-(aq)
  3. ) Equilibrium position shifts left, removing most of the H+(aq) ions - increasing PH again
How well did you know this?
1
Not at all
2
3
4
5
Perfectly
8
Q

In a buffer solution, what happens when an alkali is added?

A
  1. ) [OH-(aq)] increases
  2. ) small concentrations of H+(aq) ions react with the OH-(aq) ions:

H+(aq) + OH-(aq) -> H2O(l)

3.) HA dissociates, shifting the equilibrium position to the right to restore most of H+(aq) ions - decreasing PH again

HA(aq) H+(aq) + A-(aq)

–> added alkali

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
9
Q

When is a buffer solution most effective?

A

Equal concentrations of the weak acid and it’s conjugate base

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
10
Q

Why is a buffer solution most effective at equal concentrations of the weak acid and its conjugate base?

A
  • PH of the buffer solution is the same as the pKa value of HA
  • Operating PH is typically over about 2 PH units, centred at the PH of the pKa value

Used to find operating PH range and ratio of conc of weak acid and conjugate base are adjusted to fine-tune the PH of the buffer solution

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
11
Q

How do you calculate PH of a buffer solution?

A

[H+(aq)] = Ka x [HA(aq]/[A-(aq)]

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
12
Q

How is PH calculated if [A] = [Ha]?

A

Ka = [H+]

pKa = PH

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
13
Q

Calculate the PH when the buffer solution contains 0.1 moldm-3 CH3COOH and 0.3 moldm-3 CH3COONa.

Ka (CH3COOH) = 1.74 x 10-5 moldm-3

A

[H+(aq)] = Ka x [HA(aq)]/[A-(aq)]

[H+(aq)] = Ka x [CH3COOH(aq)]/[CH3COO-(aq)]

[H+(aq)] = 1.74 x 10-5 x 0.1/0.3

= 5.8 x 10-6 moldm-3

PH = -log[H+(aq)] = -log(5.8 x 10-6) = 5.24

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
14
Q

150 cm3 of 0.1 moldm-3 HCOOH is mixed with 100 cm3 0.750 moldm-3 HCOONa. Calculate the PH of the buffer solution formed

Ka (HCOOH) = 1.78 x 10-4 moldm-3

(Calculating the PH of a buffer solution made from a weak acid and its salt)

A

calculate the mol of HCOOH and HCOO-:

n(HCOOH) = 1.00 x 150/1000= 0.150 mol

n(HCOO-) = 0.75 x 100/1000 = 0.075 mol

calculate concentrations of HCOOH and HCOO- in the buffer solution:

total volume = 250cm3

[HCOOH(aq)] = 1000 x 0.150/250 = 0.6 moldm-3

[HCOO-(aq)] = 1000 x 0.075/250 = 0.3 moldm-3

Calculate [H+(aq)] using buffer equation:

[H+(aq)] = Ka x [HCOOH(aq)]/[HCOO-(aq)]

= 1.78 x 10-4 x 0.6/0.3

= 3.56 x 10-4 moldm-3

PH = -log(3.56 x 10-4) = 3.45

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
15
Q

100 cm3 of 0.75 moldm-3 NaOH(aq) is added to 150 cm3 of 1.5 moldm3 HCOOH. Calculate the PH of the buffer solution formed.

Ka (HCOOH) = 1.78 x 10-4moldm-3

(calculating the PH of a buffer solution made by partial neutralisation)

A

write equation: HCOOH(aq) + NaOH(aq) → HCOONa + H2O

Calculate moles of acid and base:

n(NaOH) = 0.75 x 100/1000 = 0.075 mol

n(HCOOH) = 1.5 x 150/1000 = 0.225 mol

Acid in excess, so all the base reacts. 0.075 moles of NaOH at start of reaction, so 0.075 moles of moles of salt formed. As 1 mole of acid reacts with 1 mole of base to form 1 mole of salt:

n(HCOOH) remaining = 0.225 - 0.075 = 0.150 mol

Calculate concentration of acid and salt in buffer solution:

[HCOOH] = 1000 X 0.150/250 = 0.6 moldm-3

[HCOO-] = 1000 X 0.075/250 = 0.3 moldm-3

Calculate [H+(aq)] using buffer equation:

[H+(aq)] = Ka x [HCOOH(aq)]/[HCOO-(aq)]

= 1.78 x 10-4 x 0.6/0.3

= 3.56 x 10-4 moldm-3

PH = -log(3.56 x 10-4) = 3.45

How well did you know this?
1
Not at all
2
3
4
5
Perfectly
16
Q

Where does blood PH need to be maintained?

A

PH 7.35-7.45

17
Q

What happens if the blood PH falls outside this range?

A

Below 7.35 - acidosis causing fatigue, shortness of breath and in extreme cases - shock or death

Above 7.45 - alkalosis: causes muscle spasms, light-headedness, and nausea

18
Q

What buffer system controls the PH in the blood?

A

Carbonic acid - hydrogen carbonate

19
Q

What is the equilibrium for the buffer system in blood?

A

H2CO3(aq) H+(aq) + HCO3-(aq)

20
Q

When acid is added to blood, what happens?

A
  1. ) [H+] increases
  2. ) H+(aq) ions react with conjugate base, HCO3-(aq)
  3. ) equilibrium shifts to the left, removing most of the H+(aq) ions

H2CO3(aq) H+(aq) + HCO3-(aq)

21
Q

When alkali is added to blood, what happens?

A
  1. ) [OH-(aq)] increases
  2. ) Small concentration of H+(aq) ions reacts with the OH-(aq) ions:

H+(aq) + OH-(aq) -> H2O (l)

3.) H2CO3 dissociates, shifting equilibrium to the right to restore most of H+ ions

H2CO3(aq) ⇔ H+(aq) + HCO3-(aq)

→ added alkali

22
Q

What is the Henderson-Hasselbalch equation?

A

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

  • PH of buffer solution
  • Shows how pKa and the base/acid ratio controls the PH
23
Q

How is a buffer solution able to resist change to PH on dilution?

A

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

[HA] and [A-] altered to the same extent - maintaining [H+] and PH

24
Q

What is a PH meter?

A

electrode dipped into a solution and connected to a meter that displays the PH reading

25
Q

How do you use a PH meter to monitor the PH as an aqueus base is added to an acid solution?

A
  1. ) Using a pipette, add a measured volume of acid to a conical flask
  2. ) Place the electrode of the PH meter in the flask
  3. ) Add the aqueous base to the burette and add to the acid in the conical flask, 1cm3 at a time.
  4. ) After each addition, swirl the contents. Record the PH and the total volume of the aqueous base added.
  5. ) Repeat step 3 and 4 until PH starts the change more rapidly. Then add the aqueous base dropwise for each reading until the PH changes less rapidly.
  6. ) Now add the aqueous base 1cm3 at a time again until an excess has been added and the PH has been basic, with little change, for several additions
26
Q

What are the key features of PH-titration curve?

A
27
Q

What is the equivalence point?

A

When the volume of one solution is reached where it exactly reacts with the volume of the other solution. (matching the stoichiometry of the reaction)

28
Q

What is an acid-base indicator?

A

Weak acid (HA) that has a distinctively different colour from it’s conjugate base (A-).

29
Q

What is the end point of a titration?

A
  • Point of titration where the indicator contains equal concentrations of HA and A-, so colour will be between the 2 extremes.
  • Ideally the same as equilibrium point
30
Q

Explain the colour changes in the indicator methyl orange when a basic solution is added

A
  • OH-(aq) ions react with H+ in the indicator. (H+(aq) + OH-(aq) -> H2O(l))
  • Weak acid, HA, dissociates shifting the equilibirum position to the right. Colour changes first to orange at the end point and finally to yellow as the equilibrium position shifts to right

HA(aq) ⇔ A-(aq) + H+(aq)

31
Q

Explain the colour changes int the indicator methyl orange when an acid is added to a basic solution

A
  • H+(aq) ions react with conjugate base, A-(aq)
  • The equilibrium position shifts to the left
  • Colour changes first to orange at the end point and finally to red when position has shifted to the left

HA(aq) ⇔ A-(aq) + H+(aq)

32
Q

In a titration, when is an indicator suitable?

A

Choose an indicator that has a colour change which coincides with the vertical section of the tiration curve

33
Q

What is the PH range of methyl orange?

A

3.1 - 4.4

34
Q

What is the PH range of phenolphthalein?

A

8.3 -10

35
Q

Draw a PH-titration curve for strong acid - strong base titration

A
  • Both phenolphathalein and methyl orange suitable
36
Q

Draw the PH titration-curve of a weak acid-strong base titration

A
  • Phenolphthalein suitable
37
Q

Draw the PH titration-curve of a strong acid-weak base titration

A
  • Methyl orange suitable
38
Q

Draw the PH titration-curve of a weak acid-weak base titration

A
  • No indicator suitable
39
Q

What colour does phenolpthalein turn in presence of an acid, a base and at the endpoint?

A

Acid - colourless

end point - pink

base- purple