Acids, bases and buffers cheat sheet.pdf

Acids, bases and buffers calculations

The basics [H+] = the concentration of protons

[HA] = concentration of the acid

[A-] = concentration of the conjugate base

You will either be ending you calculations with this:

pH = -log[H+]

Or starting them with this:

[H+] = 10-pH

Remember pH should always be given to 2 decimal places

pH calculations can be confusing because it often looks like you're not given all the information you need. The reason for this is that we make a lot of assumptions aboutthe behaviour of acids and bases. These are in pink in this document). Learn these and you won't struggle.

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Calculations for strong Acids

Assumption:

Strong acids fully dissociate in aqueous solution.

e.g. HCl H+ + Cl-

This means that [H+] = [HA]

How to calculate pH of a monobasic/monoprotic strong acid:

You will be given [HA]

pH = -log[HA]

To calculate [HA] of a strong acid:

You will be given pH

[HA] = 10-pH

How to calculate the pH of a diprotic/dibasic strong acid:

The acids still fully dissociate but the mole ratio is 1 HA : 2 H+

For example: H2SO4 2H+ + SO42-

Therefore the calculation you need to do is:

1. [H+] = 2 x [HA]

2. pH = -log[H+]

How to calculate [HA] of diprotic/dibasic strong acid:

1. [H+] = 10-pH

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


Calculations for weak acids

Assumption:

Weak acids do not fully dissociate in solution.

This means that [H+] DOES NOT = [HA]

So pH cannot be calculated in the same way as for strong acids.

Weak acids dissociate in an equilibrium reaction.

e.g. CH3COOH CH3COO- + H+

Ka is the acid dissociation constant and the Ka equation describes this dissociation:

Ka = [H + ][A - ]

[HA]

(if you are asked for the expression for Ka this is the one to give. We will be using asimplified version of it in the calculations but the simplified version isn't enough for

the marks.)

The bigger Ka the more dissociated the acid is.

The dissociation of acids is an endothermic process so Ka increases withtemperature.

How to calculate Ka

You will be given [H+] and [HA]

You will not be given [A-] for weak acid calculations.

There is an assumption that explains this:

For weak acids [H+] = [A-]

This is because they are both produced by the dissociation of the acid and they areformed in a 1:1 ratio and you don't need [A-]


The Ka equation can therefore be simplified to:

Ka = [H + ] 2

[HA]

This simplified equation can be used to calculate Ka but remember never togive this as the expression for Ka in a stand-alone question.

This equation uses another important assumption:

You will have probably noticed that Ka is basically the same equation as Kc(i.e. products/reactants). In Kc calculations we have to use the equilibrium

concentrations for everything. But for calculations involving Ka we can use theinitial concentration of Ka (i.e. the concentration you are given). This is

because the dissociation is extremely small.

Remember this assumption:

[HA] after dissociation = initial [HA]

How to calculate the pH of weak acids

We will also use the Ka equation for this but it will need to be rearranged tomake [H+] the subject. You will be given Ka and [HA]:

1. [H+] = ka[HA]

2. pH = -log[H+]

How to calculate the concentration of weak acids

We will also use the Ka equation for this but it will need to be rearranged to make[HA] the subject. You will be given pH and Ka:

1. [H+] = 10-pH

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


pKa

pKa is related to ka in the same way that pH is related to [H+]

pka = -logKa

ka = 10-pKa

Calculations involving pKa

You might just be asked to calculate the pKa of a weak acid. In this case justcalculate Ka as normal and then do pka = -logKa.

Another possibility is that you will be given pKa instead of pH doing a weak acid or buffer calculation.

All you need to do in this case is do ka = 10-pKa and then do the rest of thecalculation as normal.


Calculations for bases

Remember: pH = log[H+]

This still applies to bases, though bases do not dissociate to release protons.

Bases accept protons from water so if we want to calculate the pH of a base weneed to find out the concentration of protons left in the water after the base has been

added.

To do this we need to use Kw (the ionic product of water)

Kw = [H+][OH-]

Pure water dissociates to produce [H+] and [OH-]:

H2O H+ + OH-

At room temperature (298K) the pH of pure water is 7. This means that [H+] ofpure water is 1.0 x 10-7 (because [H]+ = 10-pH) and [OH-] is therefore also 1.0 x

10-7 because H2O dissociates in a 1:1 ratio.

This means that at room temperature Kw = [10-7][10-7] = 1.0 x 10-14 mol2 dm-6

Kw is dependent on temperature as the temperature increases the value ofKw increases. So if you are doing calculations involving a temperature other

than 298K you will be given/asked to calculate a different Kw.

Assumption for base calculations:

Strong bases fully dissociate in aqueous solution.

e.g. NaOH Na+ + OH-

So [OH-] = concentration of the base.

But watch out for the mole ratio:

e.g. Ca(OH)2 Ca2+ + 2OH-

In this case [OH-] = 2 x the concentration of the base.


How to calculate the pH of a strong base:

You will be given the concentration of the base. You may be given Kw (if not just use1.0 x 10-14 mol2 dm-6).

Use the Kw equation but rearrange it to make [H+] the subject:

1. [H+] = Kw/[OH-]

(where [OH-] the concentration of the base. Remember to watch out for the moleratio see the assumptions above)

2. pH = -log[H+]

To calculate the concentration of a base:

You will be given pH. You may be given Kw (if not just use 1.0 x 10-14 mol2 dm-6).

1. [H+] = 10-pH.

Use the Kw equation but rearrange it to make [OH-] the subject

2. [OH-] = Kw/[H+]

If your base is the type that dissociates to give 1 OH- you're done but if your basedissociates to give 2 OH- (like Mg(OH)2) you need to do 2 x [OH-] to get the

concentration of the base (see the assumption above).


Other calculations involving Ka

You may be asked to work out the pH of a solution of pure water. You might thinkthat the pH of pure water should always be 7 but this is only the case at a

temperature of 298K.

The pH of pure water depends on the concentration of H+, as usual. Theconcentration of H+ depends on how dissociated the water is. The dissociation of

water is an endothermic process so the higher the temperature the more dissociatedthe water, the higher the H+ and the lower the pH.

How to calculate the pH of pure water

Assumption:

In pure water [H+] = [OH-]

You will be given the relevant Kw for the particular temperature.

1. Use the Kw equation Kw = [H+][OH-]

2. Because [H+] = [OH-] we can find [H+] by doing

[H+] = Kw

3. Do pH = -log[H+]


Buffers

Acidic Buffers

A buffer solution is a mixture weak acid and its conjugate base.

The conjugate base is provided by a salt of the weak acid.

e.g. a buffer solution might be made from ethanoic acid and sodium ethanoate.

The composition of acidic buffers

The weak acid will not be fully dissociated in solution:

[HA] [H+] + [A-]

So the acid will provide lots of undissociated HA, a little A- and a little H+

The salt will be fully dissociated in solution:

NaA Na + A-

The salt will therefore provide lots of A- (and some Na+/K+/random metal ion, whichwe are not interested in)

Overall, the buffer contains:

Lots of HA

Lots of A-

A bit of H+

You need to know how buffers work. We will look at this in a few pages time.


Standard buffer calculations

All buffer calculations use the Ka equation that you learned for weak acids:

Ka = [H+][A-]

[HA]

Assumption:

In a buffer mixture [H+] DOES NOT EQUAL [A-]. This is because the A-comes from both the weak acid and the salt. You will be given the

concentration (or mass - I will tell you what to do in each case) whichrepresents the A-.

Not sure which is [HA]? It's the one that you have a Ka for.

E.g. NH4Cl doesn't look like an acid, but it is and it ca1n be used to make a buffer. Inthis case you would be a given a Ka for [NH4+]

How to calculate the pH of a buffer solution

You will most likely be given [HA] and the concentration of the salt (this is [A-]) and Ka.

1. Calculate [H+]. You will need to rearrange the Ka equation to make [H+] thesubject:

[H+] = ka[HA]

[A-]

2. Do pH = -log[H+]

1


If you are given the mass of the salt: You need to convert this to concentration bydoing moles = mass/Mr then doing concentration = moles/volume(in dm3). You won't

have a volume for the salt so use the volume of the acid as you are dissolving thesalt in the acid.

To calculate the concentration of [HA]

You will be given: pH, Ka and [A-]

1.do [H+] = 10-pH

2. Then rearrange the Ka equation to make [HA] the subject and use it to calculate[HA]:

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

Ka

How to calculate the concentration of [A-]

You will be given: pH, Ka and [HA]

1. do [H+] = 10-pH

2. Then rearrange the Ka equation to make [HA] the subject and use it to calculate[A-]:

[A-] = Ka[HA]

[H+]

How to calculate the mass of [A-]


1. do [H+] = 10-pH


[A-] = Ka[HA]

[H+]

3. Calculate the moles of A- by doing moles = concentration x volume(in dm3).


(This volume will most likely be the volume of the acid, since A- will be a solid thatyou are adding to the acid.)

4. Calculate the mass by doing moles x Mr of the salt.


Calculations for buffers made from a weak acid andsodium Hydroxide

If you add NaOH to a weak acid they will react to make the salt. But, the acid needsto be in EXCESS otherwise you will just have load of salt and water, not a buffer.

Example: CH3COOH + NaOH CH3COONa + H2O

How to calculate the pH of a buffer made from a weak acid and NaOH

You will be given: Ka, [HA] and volume of the weak acid, EITHER the concentrationand volume of NaOH or mass of NaOH.

This calculation is quite long so just follow the steps:

1. Calculate the moles of the weak acid: moles = concentration x volume

2. Calculate the moles of the NaOH: EITHER moles = concentration x volumeor moles = mass/Mr

3. Calculate the excess moles of the acid:

excess moles HA = moles of HA moles of NaOH

4. Deduce the moles of the salt (the A-): Moles of NaOH = moles of A-

5. You can now calculate [H+] by using the Ka equation:

[H+] = ka[HA]

[A-]

Where [HA] is the moles of the weak acid and [A-] is the moles of the salt.

6. Get the pH by doing pH = -log[H+]

IMPORTANT: Remember [HA] and [A-] are, strictly speaking, concentrationsbut in this example we have used moles instead of concentration. This is

because the units of [HA] and [A-] cancel out anyway, so when we use the Kaequation we don't need to bother converting to volume.


Other calculations where NaOH is used to make a buffer:

How to calculate the concentration of NaOH needed to make a buffer of a particularpH:

The concentration of A- is the same as the concentration of NaOH so you just need to do exactly what you would do to calculate [A-]:


1. do [H+] = 10-pH


[A-] = Ka[HA]

[H+]

How to calculate the mass of NaOH

Again, you just need to do exactly what you would do to calculate the mass of A-,since the moles of NaOH = moles of A-


1. do [H+] = 10-pH


[A-] = Ka[HA]

[H+]

3. Calculate the moles of A- by doing moles = concentration x volume(in dm3).

(This volume will most likely be the volume of the acid, since A- will be a solid thatyou are adding to the acid.)

4. Calculate the mass by doing moles x Mr of the salt.


More awkward buffer questions

You might be asked to calculate the pH of a buffer made from equalconcentrations/proportions of the acid and the salt. All you will be given is the Ka.

To calculate pH you would first need to get [H+] by using the Ka equation:

[H+] = ka[HA]

[A-]

But, as [HA] and [A-] are the equal, they will cancel out, leaving you with [H+] = ka.

Therefore to get you just need to do pH = -log Ka

You might also notice that in this situation the pH is equal to the pKa, sincepKa = -logKa. This observation is important.

How to calculate pH when you are given proportions of acid and salt, the ratio ofacid and salt etc.

For example: Calculate the pH of a buffer solution made from 2 parts Ethanoic acidand 1 part Sodium Ethanoate. Ka = 1.7 x 10-5 mol dm-3.

To calculate pH you would first need to get [H+] by using the Ka equation:

[H+] = ka[HA]

[A-]

All you have for [HA] and [A-] is the parts. These can be used instead ofconcentrations as can ratios, proportions, anything else you are given for [HA] and

[A-]. This is because the units cancel out, so it doesn't matter if you use moles,concentrations, whatever, in this equation.

So, for this example, you would do [H+] = 1.75 x 10-5 x [2]

[1]

Then do pH = -log[H+]


How to calculate the ratio of [A - ] to [HA] needed to make a buffer solution of aparticular pH

You will be given: the pH and the Ka of the acid.

1. Work out [H+] by doing [H+] = 10-pH

Then you need to rearrange the Ka equation to make [A-]/[HA] the subject.

This leaves you with [A-]/[HA] = [H+]/Ka

So if you do [H+]/Ka the answer will the the ratio of [A-]:[HA]

For example: Calculate the ratio of Sodium ethanoate to ethanoic acid needed tomake a buffer solution of pH 4.63. Ka = 1.75 x 10-5 mol dm-3

[H+] = 10-4.63 = 2.34 x 10-5

So [A-]/[HA] = 2.344 x 10-5/1.75 x 10-5

= 1.43

This is your ratio so this is your answer.

You may also get asked to convert this ratio to concentrations. Remember withdecimal ratios, the number you get represents the variable on top of the division (in

this case the [A-] and the bottom of the division [HA] is 1 by default)

So the concentration of [A-] needs to be 1.43 mol dm-3 and the concentration of [HA]will be 1 mol dm-3


Calculate changes to the pH when an acid or alkali isadded to a buffer solution.

To do these calculations you need to make sure you understand how buffers work.

The whole point of a buffer solution is to resist changes in pH so that the solution towhich the buffer is added will stay at around the characteristic pH of the buffer

solution.

The buffer keeps the pH roughly constant by mopping up any added protons(to prevent a reduction in pH) and reacting with any base added (to prevent a

pH increase).

How it does it:

Remember Le Chateliers principle: If the conditions of a dyamic equilibrium arechanged the position of equilibrium will shift to oppose the change

You buffer is made up of a weak acid and its salt, which can be described by thisequation:

[HA] [H+] +[ A-]

If acid is added to a solution:

[H+] will increase. As there is lots of A- present, this can react with the H+ to reduce itsconcentration. The H+ will therefore be removed from the solution and the pH will not

decrease.

What happens to the buffer:

The concentration of A- decreases

(all of the H+ will react with A- so A- will decrease by however many moles of H+ youadded)

The concentration of HA increases, as the A- and H+ combine to make HA

(As the H+ and A- make the HA, the amount of HA will increase by however manymoles of H+ you added)


How to calculate changes to the pH if you add an acid.

1. Calculate the moles of the acid you are adding. You may actually be given themoles but if not you will have to do moles = concentration x volume or moles =

mass/Mr depending on what you are given.

2. Calculated the moles of the weak acid that makes up the buffer and calculate themoles of the salt. moles = concentration x volume

3. Remember, the moles of A- will decrease because it will react with the [H+] so youneed to calculate how many moles of [A-] are left by doing:

New moles of A- = initial moles of A- - moles of added acid

4. The moles of HA will increase as more is produced so you need to calculate howmany moles of HA you will end up with in total by doing:

New moles of HA = initial moles of HA + moles of added acid

5. You can now calculate the new pH by using the Ka equation, rearranged to make[H+] the subject:

[H+] = ka[H]

[A-]

6. Where Ka is the Ka of the weak acid that makes up the buffer, [HA] is the newmoles of HA and [A-] is the new moles of A-.

Then to calculate pH you will need to do pH = -log[H+]


If alkali is added to a solution

The added OH- will react with the H+ in the buffer. This decreases the concentrationof the H+. HA then dissociates to replace the lost H+ and the equilibrium position

shifts to the right and prevents a decrease in the pH. This also makes some more A-

What happens to the buffer:

The concentration of HA decreases

(by however many moles of alkali was added, since all the alkali will react)

The concentration of A- increases

(by however many moles of alkali were added, since the alkali reacts with the HA toproduce A-)

How to calculate changes to the pH if you add an alkali.

1. Calculate the moles of the alkali you are adding. You may actually be giventhe moles but if not you will have to do moles = concentration x volume or

moles = mass/Mr depending on what you are given.

2. Calculated the moles of the weak acid that makes up the buffer andcalculate the moles of the salt.

3. Remember, the moles of A- will increase as more is formed when HAdissociates so you need to work out how much A- you will end up with in total:

New moles of A- = initial moles of A- + moles of added acid

4. The moles of HA will decrease as it dissociates to replace the H+ so youneed to find out how much you have left by doing

New moles of HA = initial moles of HA - moles of added acid

5. You can now calculate the new pH by using the Ka equation, rearranged tomake [H+] the subject

[H+] = ka[HA]

[A-]

Where Ka is the Ka of the weak acid that makes up the buffer, [HA] is the new molesof HA and [A-] is the new moles of A-.

6. Then to calculate pH you will need to do pH = -log[H+]


Calculations involving neutralisation reactions

In this type of calculation you will be asked to work out the pH of the solution after astrong acid is mixed with a base. How you will do this depends on whether the acid

or the base is in excess.

What you will be given:

[HA], the volume of the acid and either the concentration and volume of the base orthe mass of the base.

For monoprotic/monobasic acids (e.g HCl):

1. Calculate the moles of both the acid and the alkali (moles = concentration xvolume or moles = mass/Mr)

2. Deduce which reagent is in excess. Whichever has the greatest number of molesis in excess. The other reagent is considered to be the limiting reagent.

3. Calculate the number of excess moles by doing:

excess moles = moles of excess reagent - moles of limiting reagent.

4. Convert the moles of the excess reagent to concentration by doing:concentration = moles/total volume. (The total volume is the volume of the acid +

the volume of the base)

5. The next steps depend on whether the acid or base was in excess. If the acidwas the excess reagent, follow the steps for determining the pH of a strong acid. If

the base was in excess follow the steps for determining the pH of a base.


For diprotic/dibasic acids (e.g. H2SO4)

The steps above still apply but you have to be careful with the mole ratio - this willmake it harder to see which is in excess. This is the one that everyone struggles with

so I will show you a full worked example:

Example question: Calculate the pH when a 0.0200 dm3 of a 0.250 mol dm-3solution of H2SO4 is mixed with 0.0200 cm3 of a 0.190 cm3 solution ok KOH.

H2SO4 + 2KOH --> K2SO4 + H2O

Following the steps in the previous example we need to calculate the moles of eachreagent first (using moles = concentration x volume):

Moles of H2SO4 = 0.0200 x 0.250

= 0.005 moles

Moles of KOH = 0.0200 x 0.190

= 0.0038 moles

Then, we need to deduce which is in excess and by how much. This is where is is alittle bit more awkward than in it is for monoprotic acids. We can see that the H2SO4

has the greater moles but the 1:2 molar ratio between the H2SO4 and the KOHmeans we cannot just do moles of H2SO4 - moles of KOH in order to find out the

excess moles.

This is what we need to do:

Even though there is a 1:2 molar ratio between the H2SO4 and the KOH the molarratio between H+ and OH- is still 1:1 (this is always the case in neutralisation

reactions). This is because the ionic equation for the reaction simplifies to H+ + OH---> H2O

So, we can determine the moles of H+ and then compare that to the moles of KOH inorder to work out what is in excess.

H2SO4 is a strong acid so one mole of H2SO4 dissociates to give 2 moles of H+:

H2SO4 --> 2H+ + SO42-

We already know that the moles of H2SO4 is 0.005, so this means the moles of H+ is0.01.


We can now follow the same steps as for a monoprotic acid, but we need to makesure we use the moles of H+ rather than the moles of the acid. These are the steps:

1. Deduce which reagent is in excess. Whichever has the greatest number of molesis in excess. The other reagent is considered to be the limiting reagent.

2. Calculate the number of excess moles by doing:

excess moles = moles of excess reagent - moles of limiting reagent.

3. Convert the moles of the excess reagent to concentration by doing:concentration = moles/total volume. (The total volume is the volume of the acid +

the volume of the base)

4. The next steps depend on whether the acid or base was in excess. If the acidwas the excess reagent, follow the steps for determining the pH of a strong acid. If

the base was in excess follow the steps for determining the pH of a base.

So, for our example: the excess moles will be the moles of H+ - moles of KOH

excess moles = 0.01 - 0.0038

= 0.0062 moles of H+

We now need to convert back to concentration. Remember that the volume needs tobe the total volume which in this case is 0.0200 + 0.0200 = 0.040 dm3.

The concentration of H+ is therefore 0.0062/0.040 = 0.155 mol dm-3

Since the acid was in excess we can now just do pH= -log[H+]

= 0.81

To recap: the difference between this calculation and the monoprotic acid calculationis that we need to calculate the moles of H+ (moles of acid x 2) and use this rather

than the concentration of the acid when we are working out the excess moles.

This is important: if you get a question which asks you to work out the pHwhen a base is added to a weak acid you will probably find that the acid is in

excess and you need to treat it like a weak acid calculation. If you have areaction involving a weak acid and the base is in excess you can treat is as a

neutralisation calculation and follow these instructions.


Acids, bases and buffers cheat sheet.pdf

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