21.6 Buffers

Question 1

What are buffers

Answer 1

Solutions that can resist changes to PH when small amounts of acid or alkali are added

So keep concentration of H+ ions and OH- ions in the solution almost unchanged

Question 2

What is equation for weak acidic dissociation

Answer 2

Dissociation of a weak acid is an equilibrium reaction

Eg HA <—-> A- + H+
We can see from this equation that
[H+] = [A-]
The values of these are very small because most HA isn’t dissociated

Question 3

What happens if you add some alkali to weak acid buffer

Answer 3

The OH- ions from alkali will react with HA to make water molecules and A-

HA + OH- —> H2O + A-

This shows that solution is buffered because all the OH- ions are reacted, and products are water and acid.

So it won’t become alkali solution, PH remains the same

Question 4

What happens if you add acid to weak acid buffer

Answer 4

. If H+ ions are added, equilibrium shifts left from equation
HA <—> A- + H+
So more HA is produced as A- combine with H+ ions to make it

Since [A-] is small, the supply of it soon runs out so there is no more A- to combine with H+

This means solution is not a buffer to acid as many H+ ions won’t be reacted with

Question 5

How can you make acidic buffer more resistant against adding acid

Answer 5

. Add more A- ions to solution, but the way to do this is adding a soluble salt of HA which fully ionises
Eg Na+A-

This works by increasing supply of A- so more H+ ions can be used up

So this means there is now a way where both H+ and OH- ions can be removed

Question 6

What is an acidic buffer made from

Answer 6

. Mixture of a weak acid and its soluble salt.
It will maintain a PH of below 7

Question 7

What is the role of the weak acid part of buffer

What is role of soluble salt component of buffer

Answer 7

. Its role is to act as a source of HA to react with and remove OH-

HA + OH- <–> H2O + A-

. Salt component provides A- ions to react with H+ ions

A- + H+ –> HA

Question 8

What is half neutralisation and what assumption do we make here

Answer 8

. A useful way to make a weak acid and its salt is to add some NaOH to neutralise half the weak acid

At this point,
PH = Pka

This is because it is equally efficient at resisting a change in PH whether acid or alkali is added

Question 9

What is the job for basic buffers
What are they made from

Answer 9

Keep PH above 7

Made from mixture of weak base and its soluble salt

Question 10

How does weak base ammonia work to buffer solution

How does salt NH4Cl act as a basic buffer when added to aqueous ammonia

Answer 10

. The aqueous ammonia NH3 is the weak base which can react with and remove H+ ions

NH3 + H+ —> NH4+

. The NH4+ ion is from the salt and removes any added OH-

NH4+ + OH- —> NH3 + H2O

Question 11

How are buffers used in blood

Answer 11

. PH of blood needs to be kept at 7.4

H+ + HCO3- —> CO2 + H2O

Addition of extra H+ ions moves equilibrium to the right which forms water
Equilibrium can also move left to make more H+ ions when needed

Question 12

What equation do you use when calculating PH of buffer

Answer 12

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

Question 13

Question: A buffer consists of 0.1moldm-3 sodium ethanoate and 0.1moldm-3 ethanoic acid

Ka for ethanoic acid: 1.7 X 10^-5 and pka is 4.77

Answer 13

sub in values to get [H+]

so [A-] is from sodium ethanoate which is [0.1]

[ HA ] is from ethanoic acid which is [0.1]
[H+] and -log10 H+ = 4.77

Question 14

What happens when you have equal concentrations of acid and salt in buffer

Answer 14

PH is equal to pKa of acid used
So same situation as at half neutralisation point

Question 15

Calculate PH of buffer when 500cm3 of 0.4moldm-3 of NaOH is added to 500cm3 of 1moldm-3 HA

Ka= 6.25 X 10-5

Answer 15

Some weak acid is neutralised by NaOH, leaving solution of A- and HA

Moles of HA is 0.5 mol
Moles NaOH = Moles of OH- so 0.2mol

HA + NaOH —> NaA + H2O

So adding NaOH will decrease moles of HA making it 0.3 mol
And it will increase moles of NaA making it 0.2 moles

So now 1000cm3 of solution in total so use this to get concentrations of them

Then sub into calculation to get [H+] is 9.375 X 10^-5

So PH is 4.03

Question 16

What happens if you start with 0.1moldm-3 of a buffer solution of ethanoic acid at volume 1dm-3.

And you add HCl at concentration 1moldm-3 to this solution previously made

Answer 16

All the H+ ions added will react with ethanoate ions (A-) to form molecules of ethanoic acid (HA)

. Before adding acid: Moles of ethanoic was 0.1 mol
. Moles of sodium ethanoate was 0.1 mol

Then moles of HCl is 0.010