Buffer solutions (Acid)

Question 1

a buffer solution is

Answer 1

a solution that minimises the change in pH when a small amount of either acid or base is added

Question 2

the two most common ways of making a buffer solution are

Answer 2

• to mix a weak acid with its conjugate base

- to mix a weak base with its conjugate acid

Question 3

in a buffer solution made from a weak acid with its conjugate base, the salt of the weak acid has to be……………in water which is why………… and …………………salts are commonly used

Answer 3

salt must be soluble in water

therefore sodium and potassium are used

Question 4

a simple example of a weak acid with its conjugate base buffer solution is

Answer 4

ethanoic acid and sodium ethanoate

Question 5

in the mixture made up of ethanoic acid and sodium ethanoate, the acid is……………dissociated and the salt is………………dissociated

Answer 5

acid is partially dissociated

salt is fully dissociated

Question 6

the equations which show the relative dissociations of the acid and salt making up the ethanoic acid/sodium ethanoate buffer solution are

Answer 6

CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)

CH3COONa(aq) → CH3COO-(aq) + Na+(aq)

Question 7

the ethanoic acid/sodium ethanoate buffer solution produces a soltion with a pH of

Answer 7

less than 7

Question 8

if we mix equal volumes of 1.00 mol dm-3 ethanoic acid with 1.00 mol dm-3 sodium ethanoate, we can assume that the extent of dissociation of the acid is …………therefore meaning the concentration of CH3COOH at equilibriuum is ….mol dm-3

Answer 8

the extent of dissociation of CH3COOH is negligible, so its concentration at equilibrium will be 0.5 mol dm-3, half of 1.00 mol dm-3 because it has been diluted by its equal volume by the addition of sodium ethanoate

Question 9

write the equation for Ka for the dissociation of CH3COOH

Answer 9

[CH3COO-(aq)][H+(aq)]
Ka= ———————————- = 0.0000174 mol dm-3
[CH3COOH(aq)]

Question 10

if Ka = [CH3COO-(aq)][H+(aq)] / [CH3COOH(aq)] = 0.0000174 mol dm-3, then the pH of the buffer solution will be:

Answer 10

pH = -lg [H+(aq)]
[H+(aq)] = 0.0000174 x 0.5 / 0.5 = 1.74 x 10-5
pH= -lg10 [0.0000174] 
pH= 4.76

Question 11

in the following equation, the position of equilibrium lies to the…..
CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)

Answer 11

left, meaning more CH3COOH is present

Question 12

when adding ethanoate ions (CH3COO-) through the addition of sodium ethanoate, the position of equilibrium in the following equation shifts to the……………
CH3COOH(aq) ⇌ CH3COO-(aq) + H+(aq)

Answer 12

left even more

Question 13

the buffer solution will therefore contain the following (after the addition of CH3COONa, shifting the position of equilibirum further to the left)

Answer 13

• lots of un-ionised ethanoic acid (CH3COOH)
• lots of ethanoate ions from the sodium ethanoate
• enough H+ ions to make the solution acidic

Question 14

when adding an acid to this buffer solution, most of the new H+ ions are removed because ………..

Answer 14

of the abundance of ethanoate ions which react with the H+ ions to form ethanoic acid. as the position of equilibrium is well to the left, this means the ethanoic acid will not dissociate fully, meaning most H+ ions are removed in this way.
the pH will only fall a little bit due to the equilibria involved

Question 15

when adding a base to this buffer solution, the two ways in which the OH- ions can be removed are:

Answer 15

• that the hydroxide ions react with a ethanoic acid molecule to form CH3COO- and H2O
• the hydroxide ions react with H+ ions to make water. as this happens, the equilibrium shifts to the right to replace the H+ ions until most of the OH- are removed

the pH does increase slightly as not all the OH- ions are removed, and the H2O will ionise to a certain degree, producing OH- and H+ ions