Buffer Solutions

Question 1

What is a buffer solution?

Answer 1

A system that minimises pH changes when small amounts of acid or a base are added.

Question 2

What do buffer solutions contain?

Answer 2

Two components to remove added acid or alkali - a weak acid and its conjugate base.

Question 3

What is the purpose of a weak acid (HA)321 in a buffer solution?

Answer 3

It removes added alkali.

Question 4

What is the purpose of the conjugate base (A-) in a buffer solution?

Answer 4

It removes added acid.

Question 5

What happens to the 2 components of the buffer solution when alkalis and acids are added to a buffer?

Answer 5

They react and will eventually be used up. As soon as one component has all reacted, the solution loses its buffering ability towards the added acid or alkali.

Question 6

What happens to the pH as the buffer works?

Answer 6

The pH only changes by a smalla amount - you should not asume that the pH stays completely constant.

Question 7

What does a buffer solution based on a weak acid need?

Answer 7

A weak acid and its conjugate base.

Question 8

What are the 2 methods for preparing a buffer based on a weak acid?

Answer 8

1. Preparation from a weak acid and its salt.
2. Preparation by partial neutralisation of the weak acid.

Question 9

What is the method for preparing a buffer based on a weak acid from a weak acid and its salt? (CH3COOH)? (2)?

Answer 9

1. When CH3COOH is added to water, the acid partially dissociates and the amount of ethanoate ions in solution os very small. The CH3COOH is the source of the weak acid component of the buffer solution.
   - CH3COOH = H+ + CH3COO-. (component 1)
2. The salts of the weak acids are ionic compounds and provide a source of the conjugate base component. When added to water, the salt completely dissolves. Dissociation into ions is complete and so the salt is the source of the conjugate base component of the buffer solution.
   - CH3COONa + aq = CH3COO- + Na+. (component 2)

Question 10

What is the method for preparing a buffer based on the partial neutralisation of the weak acid (CH3COOH)? (4)?

Answer 10

1. Add aqueous solution of an alkali (NaOH) to an excess of the weak acid (CH3COOH).
2. The weak acid (CH3COOH) is partially neutralised by the alkali, forming the conjugate base.
3. Some of the weak acid is left over unreacted.
4. The resulting solutions contains a mixture of the salt of the weak acid and any unreacted weak acid.

Question 11

What happens in the ethanoic acid equilibrium?

Answer 11

The equilibrium position lies well towards ethanoic acid. When CH3COO- ions are added to CH3COOH, the equilibrium position shifts even further to the left, reducing the already small concentration of H+ ions, and leaving a solution containing mainly two components, CH3COOH and CH3COO-.

Question 12

What do CH3COOH and CH3COO- act as? Achieved?

Answer 12

Two reservoirs that are able to act independently to remove added acid and alkali. This is achieved by shifting the buffer’s equilibrium system either to the right or the left.

Question 13

What controls the pH in an acid buffer solution?

Answer 13

The conjugate acid-base pair (HA/A-).

Question 14

What can the control of pH be explained by?

Answer 14

The shifts in the equilibrium position using le Chatalier’s principle.

Question 15

How does a conjugate base remove added acid? Equation?

Answer 15

On addition of an acid, H+:
- The [H+] increases.
- H+ ions react with the conjugate base, A-.
- The equilibrium position shifts to the left, removing most of the H+ ions.

HA = H+ + A-
<—————–
added acid

Question 16

How does a weak acid remove added alkali? Equation?

Answer 16

On addition of an alkali, OH-:
- The [OH-] increases.
- The small concentration of H+ ions reacts with the OH- ions:
H+ + OH- = H2O.
- HA dissociates, shifting the equilibrium position to the right to restore most of H+ ions.

HA = H+ + A-
—————–>
added alkali

Question 17

When is a buffer most effective at removing either added acid or alkali?

Answer 17

When there are equal concentrations of the weak acid and its conjugate base.

Question 18

What components are used in a buffer solution when [HA] = [A-]?

Answer 18

• The pH of the buffer solution is the same as the pKa value of HA.
• The operating pH is typically over about two pH units, centred at the pH of the pKa value.
• The ratio of the concentrations of the weak acid and its conjugate base can then be adjusted to fine-tune the pH of the buffer solution.

Question 19

What does the pH of a buffer solution depend upon? (2)?

Answer 19

1. The pKa (or Ka) value of the weak acid.
2. The ratio of the concentrations of the weak acid, HA, and its conjugate base, A-.

Question 20

What is the equation for Ka of a weak acid?

Answer 20

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

Question 21

When you calculate the pH of a weak acid, what approximation is made? What is different for a buffer solution?

Answer 21

[H+] = [A-], however, this is no longer true for a buffer solution as A- has been added as one of the components of the buffer.

Question 22

How do we know calculate the pH of a buffer solution?

Answer 22

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

Question 23

What happens to the Ka and the pKa when HA and A- are the same?

Answer 23

1. Ka = [H+]
2. pKa = pH.

Question 24

CALCULATING THE pH OF A BUFFER SOLUTION:
- Calculate the pH when the buffer solution contains 0.100 moldm-3 CH3COOH and 0.300 moldm-3 CH3COONa.
- Ka (CH3COOH) = 1.74 x 10^-5 moldm-3.

Answer 24

1. Calculate [H+] from Ka, [HA], and [A-]:

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

1.74 x 10^-5 x (0.100/0.300)

5.80 x 10^-6 moldm-3.

2. Use your calculator to find the pH:

pH = -log([H+] = -log (5.80 x 10^-6) = 5.24.

Question 25

CALCULATING THE pH OF A BUFFER SOLUTION MADE FROM A WEAK ACID AND ITS SALT: - 1.50cm^3 of 1.00 moldm-3 HCOOH is mixed with 100cm^3 0.750 moldm-3 HCOONa. - Calculate the pH of the buffer solution formed. - Ka (HCOOH) = 1.78 x 10^-4 moldm-3.

Answer 25

1. Calculate the amounts, in mol, of HCOOH and HCOO- in the buffer solution: n(HCOOH) = (1.00 x 150) / 1000 = 0.150 mol. n(HCOO-) = (0.750 x 100) / 1000 = 0.0750 mol. 2. Calculate the concentrations of HCOOH and HCOO- in the buffer solution: Total volume of buffer solution = 250cm^3. [HCOOH] = (1000 x 0.150) / 250 = 0.600 moldm-3. [HCOO-] = (1000 x 0.0750) / 250 = 0.300 moldm-3. 3. Calculate [H+] from Ka, [HA] and [A-]: [H+] = Ka x ([HCOOH] / [HCOO-]) = 1.78 x 10^-4 x (0.600/0.300). = 3.56 x 10^-4 moldm-3. 4. Use your calculator to find the pH: pH = -log [H+] = -log (3.56 x 10^-4) = 3.45.

Question 26

CALCULATING THE pH OF A BUFFER SOLUTION MADE BY PARTIAL NEUTRALISATION: - 100cm3 of 0.750moldm^-3 NaOH is added to 150cm3 of 1.50moldm^3 HCOOH. - Calculate the pH of the buffer solution formed. - Ka (HCOOH) = 1.78 x 10^-4 moldm^-3.

Answer 26

1. Calculate the amount, in mol, of HCOO- in the buffer solution: The partial neutralisation is: HCOOH + NaOH = HCOONa + H2O. n(NaOH) added = (0.750 x 100) / 1000 = 0.0750 mol. n(HCOO-) formed = 0.0750 mol. 2. Calculate the amount, in mol, of HCOOH in the buffer solution. n(HCOOH) used = (1.50 x 150) / 1000 = 0.225 mol. n(HCOOH) remaining = n(HCOOH) used - n(NaOH) added = 0.225 - 0.0750 = 1.50 mol. 3. These are the same amounts as obtained from step 1 of the previous worked example. The remainder of the calculation is identical to give the pH of 3.45.