1.12- Acid and base equilibria

Question 1

What are the bronsted lowry definitions of acids and bases

Answer 1

A bronsted lowry acid is defined as a substance that can donate a proton

A bronsted lowry base is defined as a substance that can accept a proton

Question 2

What is the formula for calculating pH

Answer 2

pH = -log[H+]

Question 3

How do you calculate the pH of a strong acid

Answer 3

.Strong acids completely dissociate
. The concentration of hydrogen ions in a monoprotic strong acid will be the same as the concentration of the acid
. pH values are always given 2 d.p

Question 4

What is the ionic product of water

Answer 4

Kw = [H+][OH-]

. You can get this by rearraging the equilibrium expression for Kc
. If you rearange the equation and then as H2O is much bigger that the concentrations of the ions you can assume the value of it is constant and then you can make a new constant called Kw

Question 5

What is the Kw value for all aqeuous solutions at 25c

Answer 5

1X10^-14 mol2dm-3

Question 6

How do you find the pH of pure water

Answer 6

Pure water/neutral solutions are neutral because the [H+] = [OH-]
This means that Kw = [H+]^2

Question 7

What happens to the pH of water at higher temperatures

Answer 7

At different temperatures water has a diffrent pH, at higher temperatures the solution is more acidic, this is because the dissociation of water is endothermic so increasing the temperature would push the equilbirum to right giving a bigger concentration of H+ ions and a lower pH

Question 8

How do you calculate the pH for a strong base

Answer 8

.Strong bases completely dissociates into theier ions
. The concentration of the OH- ions is the same as the concentration of the Base
.You can work out the concentration of the H+ using the concentraion of the OH- and the Kw value

Question 9

How do you calculate the pH of a weak acid

Answer 9

Weak acids only slightly dissociate when dissolved in water, giving in equilibrium mixture

When you use Kc on a weak acid then you get the expression

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

The larger the Ka the stronger the acid

Question 10

What is pKa

Answer 10

pKa = - log Ka

Ka = 10^-pKa

Question 11

What are 2 other assumptions that are made to simplify Ka

Answer 11

. We assume the [H+] = [A-] because they dissociate to a 1:1 ratio

. As the amount of dissociation is small, assume that the intial concentration of the undissociated acid has remained constant

.[HA] eqm = [HA] intial

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

Question 12

How do you work out the pH of the solution if too much alkali has been added

Answer 12

.Work out the new concentration of the excess OH- ions

. [OH-] = Moles excess OH-/ Total volume dm3

. Use kw to find the concentration of the [H+] and then the pH

Question 13

How do you work out the pH of a partially neutralised acid

Answer 13

. Work out the concentration of the excess [H+] ions and then use these to find the pH

Question 14

What is different about strong diprotic acids and bases compared to strong monoprotic acids and bases

Answer 14

For a diprotic acid the concentration of H+ ions is equal to 2x the amount of concentration of the acid

For a diprotic base the concentration of OH- is equal to 2x the amount of concentration of the base

Question 15

How do you work out the pH of a neutralisation were weak alkali is added in excess or where weak acid is added in excess

Answer 15

.Work out the new concentration of excess HA
. [HA] = intitial moles of HA - moles OH-/Total volume dm3
. Work out the conentration of the salt formed [A-]
.[A-] = moles of the OH- added/ total volume dm3
. Rearange Ka to find the concentration of H+ ions, and then use this to find the pH

Question 16

How do you work out the pH of a weak acid at half equivalence

Answer 16

,When a weak acid has reacted with exactly half the neutralisation volume of alkali you can simplify Ka further
. You can make the assumption that [HA] = [A-]

. [H+] = Ka

. pH = pKa

Question 17

How do you work out the pH of a diluted acid and a diluted base

Answer 17

. The pH of a diluted strong acid-

[H+] new = [H+] old x (old volume/ new volume)

. The pH of a diluted strong base-

. [OH-] = [OH-] old x (old volume/ new volume)
. Use ka to find the concentration of the H+

Question 18

Give the definition of a buffer solution

Answer 18

A buffer solution is one where the pH does not change signigicantly if small amounts of acid or alkali are added to it

Question 19

What is a buffer solution made up of

Answer 19

.An acidic buffer solution is made from a weak acid and a salt of that weak acid

. A basic buffer solution is made from a weak base and a salt of that weak base

Question 20

How does a buffer solution work, outline this using ethanoic acid

Answer 20

In an ethanoic acid buffer solution there is a much higher concentration of the salt ion that in the pure acid, the buffer contains a reservor of HA and A- ions

If small amounts of acid are added to the buffer then the equilibrium will shift to the left removing nearly all of the H+ ions added

As there is a large concentration of the salt ion in the buffer, the ratio of the ethanoic acid and the ions almost stays constant so the pH stays fairly constant

If small amounts of alkali is added to the buffer then the OH- ions will react will the H+ ions to form water , the equilibrium will then shift to the right to produce more H+ ions

Some of the ethanoic acid molecules are changed to ethanoate ions but are as there is a large concentration of the salt ion in the buffer the ration almost remains constant so the pH remains fairly constant

Question 21

How do you work out the pH of a buffer solution

Answer 21

You still use the Ka expression for weak acids, but here you assume that the concentration of the [A-] is due to the added salt only

We can also assume that the initial concentration of the acid has remained constant because the amount that has been dissociated or reacted is small

Question 22

How do you work out the pH of a solution of a partially neutralised acid

Answer 22

For a partially neutralised acid you find out both the moles of the ethanoic acid and then the base. To find the moles of the acid in excess you take the moles of the base away from the acid

You can then work out the concentration of the acid and the base and then use the Ka experssion to work out the concentration of the hydrogen ions

Question 23

How do you calculate the change in the pH of a buffer solution in which there has been a small addition of alkali

Answer 23

If a small amount of alkali is added to a buffer solution then the moles of the buffer acid would reduce by the number of moles of alkali added and the moles of salt would increase by the same amount so a new calculation of the pH can be done with the new values

Question 24

How do you calculate the change in the pH of a buffer solution if there has been a small addition of acid

Answer 24

If a small amount of acid is added to a buffer then the moles of the buffer salt would reduce by the number of moles of acid added and the moles of buffer acid would increase by the same amount so a new calculation of pH can be done with these new values

Question 25

What happens when you dilute a buffer solution

Answer 25

Diluting a buffer solution with water will not change it’s pH, this is because in the buffer equation the ration of the HA and A- ions will stay constant as both concentrations of salt and acid would be diluted by the same proportion

Question 26

How do you construct a pH curve

Answer 26

. Transfer 25cm3 of acid to a conical flask with a volumetric pipette
. Measure the intial pH of the acid with a pH meter
. Add alkali in small amounts noting the volume added
. Stir the mixture to equalise the pH
. Measure and record the pH to 1dp
.Repeat steps 3-5 but when approaching the endpoint add in smaller volumes of alkali
.Add until the alkali is in excess

.First you must calibrate the meter by measureing known pH of a buffer solution, pH meters can lose accuracy on storage, most pH meters a calibrated by putting probe in a set buffer and then presssing a calibration button for that pH , sometimes this is repeated with a second buffer at a different pH

Question 27

What are the features for a titration curve that includes a strong acid and a strong base

Answer 27

. There would be a long steep part from around 3-9
. The pH at the equivalence point will be 7
. The equivalence point lies at the mid point of the extrapolated vertical portion of the curve

Question 28

What are the key points when sketching a titration curve

Answer 28

.Intial and final pH
.Volume at neutralisation
. General shape(pH at neutralisation)
.

Question 29

What are the key features of a titration curve that involves a weak acid and a weak base, and give the key features for working out the pH

Answer 29

At the start the pH rises quickly and then it wil leve off, the flattened part is called the buffer region and is formed because a buffer solution is made, the equivalence point for this reaction is greater that pH 7, and the steep part of the curve occurs around 7-9

At the half neutralisation volume, the concentration of the HA ions and the A- ions will be the same

So Ka = [H+] and pKa = pH

If we know the Ka then it is also possible to find the pH of the solution at half equivalence point, if a pH curve is plotted then the pH of a weak acid at the half neutralisation point will be equal to the pKa

Question 30

What are the features for a titration curve involving a strong acid and a weak base

Answer 30

Steep part of the curve occurs less that 7 pH, around 4-7, the equivalance point will also be less than 7

Question 31

What are the feature of the titration curve involving a weak acid and a weak base

Answer 31

. There will be no steep part of the curve, the curve will just slowly go up in pH depending on what is being added in excess

Question 32

How do you choose an indicator for a titration

Answer 32

You choose an indicator based on what pH the end colour change of the indictor, the pH must lie in between the end point that coincides with the equivalence point for the titration

Question 33

How do indicators work

Answer 33

. You have an equilibrium reaction, where Hln and ln are 2 different colours

Hln > ln- + H+

. If you apply le chateliers principle to this then in an acidic solution there will be more H+ ions which will then push the equilibrium to the reactants so this means that Hln colour will be the colour when acidic

. In an alkaline solution there will be OH- ions that will then react with the H+ and this will then shift the reaction towards the products, this means that the colour will now shift the ln colour

Question 34

What area of the titration curve will an indicator work

Answer 34

An indicator will work if the pH range of the indicator lies on the steep part of the titration curve, in this case the indicator will change colour rapidly and the colour change will correspond to the neutralisation point