Acid Base Equilibria. Flashcards
1
Q
What is the Bronsted lowry definition of an acid and also a base.
A
An acid is a substance that can donate a proton. A Base is a substance that can accept a proton.
2
Q
What is an alkali defined as.
A
A water soluble base that produces OH- ions in aqeous solution.
3
Q
What is different about strong acids and weak acids.
A
Strong acids will fully dissociate in solution to form the maximum number of H+ ions. Weak acids will only partially dissociate into H+ ions.
4
Q
What is special about the {H+} of a strong monoprotic acid.
A
The concentration of [H+] is the same as the concenration of the acid, [HA}. Disociate in a 1:1 ratio.
5
Q
How must acids and bases react.
A
They must recact in pairs, one acid and one base. Each acid is linked to a conjugate base on the other side of the equation always.
6
Q
How can you easily identify conjugate acid and base pairs.
A
Need to think about what has potential to donate a H+ ion. Potential is the key concept. Potential is in terms of could it take a H+ from its conjugate pair to form the form the substance that the other compound was originaly.
7
Q
Give the conjugate acid and base pairs in
1) Hydrogen chloride plus ammonia going into ammonium chloride.
A
HCL is acid Cl- is base ,consider the ions sepertely if there is only one ionic product to give two needed products.
NH3 is a base and NH4+ is an acid.
8
Q
Give the conjugate acid and base pairs in HCl + H2O going into H3O+ + CL-
And name the ion that is produced.
A
HCl is a an acid and Cl- is the conjugate base. Pair
H20 is the base and H30+ is the acid. Pair
It is known as the hydroxonium ion.
9
Q
What acuracy should you give pH to.
A
Always 2 decimal places.
10
Q
Define pH.
A
Give this exact definition nothing else
pH= -log10 [H+(aq)]
11
Q
How do you find the concentration of H+ ions in Moldm-3 given the pH. How is the concentration of H+ in solution represented.
A
Take the antilog. Just do 10 to the power of the negative pH. The square brackets just means concentration of the entity in moldm-3.
12
Q
What equilibrium is established in all aqeous solutions an pure water.
A
H20(l) reversable reaction into H+(aq) and OH- (aq)
13
Q
Why does the H+ ion have strange chemical properties compared to other possitive ions.
A
It has no electrons just the hydrogen nucleus. Its diameter is much less than any other chemical entity, this extreamly small diameter means that it has an intense electric field.
14
Q
In reality how is H+ found in solution and how is H+ actualy represented for the sake of simplicity.
A
H+ in reality is actually always bonded to at least one water molocule and so is found as H3O+ however it is just represented as H+ for simplicities sake.
15
Q
Since H+ has no electrons of its own what is the only type of species that the H+ ion is able to bond with.
A
It is only able to dative covalent bond with species that have a lone pair of electrons.
16
Q
What Kc Expression can be created from the equilibrium established in water and aqeous solutions.
A
Kc = [H+(aq)} {OH-(aq)}
{H20(l)}
17
Q
Considering the equilibrium that is established in water how can this be rearanged to give soemthing usefull and state the asumptions you made to get there.
A
The equilibrium established in water and aqeous solutions can be rearanged to give Kc x [H2O} = {H+} {OH-}
You asume that since {H2O] is much larger than the ions you asume that since so little of it dissociates it remains constant. H20 has stong covalent bonds so little disocaites.
You can hence ignore this from th expression but since you have done so it can be called Kc beacsue you just got rid of a term.
It is instead called the ionic product of water- Kw.
18
Q
Write an expression for the ionic product of water Kw and state why Kw is practically usefull in calculation.
A
Kw= {H+(aq)} {OH-(aq)] Must know off by heart.
If you need to convert between {H} and {OH} you just take Kw as 10 to the power of negative 14 (unless another value given for another temp) and re-arange. Divide Kw by either.
Note that the values must be concentrations in moldm-3 to go into the expression.
19
Q
What is the value for Kw including units, at 298K, 25 celcius.
A
10 to the -14 Mol2dm-6
20
Q
Why is pure water considered a neutural solution.
A
It is considered a neutural solution since a neutural solution is defined as one where {H+(aq)} = {OH-(aq)]
21
Q
Define a neutural solution.
A
One where {H+(aq)} = {OH-(aq)]
22
Q
Since Kw is a constant at a given temperature what must happen to {OH-}eqm if [H+] eqm is increased and vice versa.
A
If the concentration of one of thease ions increases the the concentration of the other must decrease proportionally in order to keep Kw constant.
23
Q
What can be said for the Kw expresion in all neutural soloutions such as water.
A
Kw = [H+} squared (very different from modified ka)
24
Q
Write a proof that the pH of pure water will be 7 at 25 degrees celcius.
A
SEE PIC square root ka and sub into pH expression.
25
How will the pH of pure water change with increased temperature.
The disociation of water is an endothermic process. In an exam you would write the equation for this. Since energy is required to break luiqid water bonds to form aqeous ions the forward reaction is endothermic. Increasing the temperature would shift the pH to the RHS favoring a forward reaction to resist the temperature increase, and hence the degree of dissociation will increase and {H+} will increase.
So pH will decrease.
26
How is pH and neutrality related. Can water be neutural if the pH doesnt =7.
Need to understand that pH is a measure of [H+} and is -log10[H+(aq)] numerical value. A neutural solutuion is one where {H+} = {OH-} and so NEUTURAL SOLUTIONS DONT ALWAYS HAVE pH=7. It is possible for water to still be neutural with a pH that isnt 7 if by definition {H+} = {OH-}.
27
Why is a logarithmic scale used for pH even though it is more complex than just stating the number of H+ ions.
The concentration of H+ ions in most solutions is actualy very small. So would give very small values out. The pH scale will give almost all possitive numbers since the log of anything less than one will be negative and so by X that by -1 you get possitive non awkward values, doing away with tiny values like 10 to the negative 3. normally between 0-14 , but can be outside this. Decimals are possible.
28
State cases where the pH of a solution would be less than 0 and also greater than 14.
Would be less than 0 , negtive , if the solution has a [H+] of greater than 1 moldm-3. Since you would esentialy be raising 10 to a power than would get you a possitive number so this would have to be a possitive fraction. When this is timesed by -1 it will turn negative.
A 1M H+ solution will have a Ph of zero.
A 1M OH- solution will have a PH 14
Going over 1 either will mathaematically breaks such boundries.
29
What does a decrease in 1 mean for the change in [H+].
It has increased by a factor of 10.
30
Give a detailed method for constructing a Titration Curve.
1.Transfer 25cm3 of acid to a conical flask with a volumetric pipette2 .Measure initial pH of the acid with a pH meter3 .Add alkali in small amounts(2cm3) noting the volume added 4.Stir mixture to equalise the pH5. Measureand record the pH to 1d.p.6. Repeat steps 3-5 but when approaching end point add in smaller volumes of alkali 7 .Add until alkali in excess.
31
State one way you could improve acuracy when experimentally constructing a pH curve.
By keeping a constant temperature.
32
State how you would calibrate a pH meter and why it is neccasary to do so.
Calibrate meter first by measuring known pH of a buffer solution. This is necessary because pH meters can lose accuracy on storage. Most pH probes are calibrated by putting probe in a set buffer (often pH 4) and pressing a calibration button/setting for that pH.Sometimes this is repeated with a second buffer at a different pH.
33
If you where given a tritration curve or needed to draw one how would you find the neutralisation volume and wht data will be relevent for this.
You can find the neutralisation volume by way of simple titration calculation. You need to find the volume of whatever is added from the beurete. If they give you a value then just consider this as the amount you have available, carry out a titration calculation as normal and find your own volume by considering it as unknown as you are concerned with the amount required to neuurtalise (recact with all the moles) of what is in the conical flask.
34
What is known about strong bases. Show this via an equation.
Strong bases completely dissociate into their ions. NaOH into Na+ + OH-
35
Regarding most strong base calculations how would you find the pH of an isolated strong base.
For bases we are normally given the concentration of the hydroxide ion. To work out the pH we need to work out [H+(aq)] using the Kw expression taking the Kw to= 10 to the minus 14 if at 25 celcius.
36
Calculatethe pH of the strong base 0.1mol dm-3 NaOH , assume complete dissociation.
Answer is 13.00. Solution on example 3 in notes.
37
What would you need to do for a strong base calculation that involved 2OH ions in the molocule such as Ba(OH)2.
Take the concentration of the base in M and multiply by 2 to find the concentration of OH- since it will dissociate in 1:2 ratio , each molocule giving way to 2 Ions when in aqeous solution.
38
Calculate the pH of 20grams per dm3 of NaOH
13.7
39
Calculate the pH of solution formed when 50cm cubed of 0.25M KOH is made up to the mark of 250cm3 by adding water.
12.7
40
When diluting an acid or an alkali what formula should you use and what from this formula is needed to go into any expression.
pH of diluted strong acid method
[ H+]new= [H+] old x old volume
---------------------
new volume
Same method if using OH- but you will need to put the [OH-] new into the kw expression to get conversion for {H+}.
You can only put the new diluted concentrations into expressions to calculate pH value.
41
Calculate the new pH when 50.0 cm3 of 0.150 mol dm-3. HCl is mixed with 500 cm3 of water
1.87 .taken from example 13 on revision notes.
42
Calculate the pH of 0.2M of Ba(OH)2 .
13.6
43
What must you know about weak acids. As always use an equation to support your answer.
Weak acids only slightly dissociate when dissolved in water, giving an equilibrium mixture.
HA + H2O (l) REVERSABLE INTO H3O+(aq)+ A-(aq)
Can be simplified into more usefull.
HA(aq) REVERSABLE INTO H+(aq)+ A-(aq)
44
Write an expression for the weak acid dissociation constant, Ka. Do this in both general terms and also for propanoic acid giving the equation for the dissociation of propanoic acid.
```
General : Ka= [H+(aq)][A-(aq)]
[HA (aq)]
For weak propanoic acid:
CH3CH2CO2H(aq) Reversable into H+(aq) CH3CH2CO2-(aq)
So........
Ka= [H+(aq)][CH3CH2CO2-(aq)]
[CH3CH2CO2H(aq)]
```
45
What does a given value for Ka tell you about the strength of an acid and why.
The larger the Ka the stronger the acid. A stronger acid will dissociate to a greater extent this means that the magnitude of [H+(aq)][A-(aq)] will be greater and hence the top half of the ka "fraction" will be greater.
46
State what scenario would require you to modify your Ka expresion for the purpose of making calculation possible.
When given an isolated weak acid with too many unknowns. You will be given either the concentration of the isolated weak acid or its pH and asked to calculate the one you have not been given.
47
What two asumptions are made to modify the Ka expression and state what the new modified Ka expression actually is.
1)[H+(aq)]eqm= [A-(aq)]eqm because they have dissociated according to a 1:1 ratio.
2) As the amount of dissociation is small we assume that the initial concentration of the undissociatedacid has remained CONSTANT . So[HA(aq)]eqm= [HA(aq)]initial
Since its weak and will dissociate a small amount we assume that none has dissociated at all.
Ka = [H+(aq)]2
[HA(aq)]initial Where Ka is given and [H+(aq)] can be found from the antilog of pH.
48
Example 5: What is the pH of a solution of 0.01 mol dm-3ethanoic acid (ka is 1.7 x 10-5mol dm-3)?
pH = 3.38 (about right for a weak acid)
| Use the simplified ka expression.
49
Example 6: Calculate the concentration of propanoic acid with a pH of 3.52 (ka is 1.35 x 10-5mol dm-3)
[CH3CH2CO2H(aq)] = 6.75x 10-3mol dm-3
| Hint: use the simplified Ka expression.
50
What is the ralationship between Ka and pKa.
| What is the relationship the same as.
Sometimes Ka values are quoted as pKa values
pKa =-log Ka so Ka = 10 to the power-pKa
It is the same relationship as pH and [H+] where the terms that start with p have the same relationship to the ones that don't in both cases.
51
If asked in an exam for the Ka expression which one should you give and which one should you never give.
Should always give the full version without the assumptions , that simplified version should only be used for the purposes of isolated weak acid calculations.
Ka= [H+(aq)][A-(aq)] This is what you will give.
[HA (aq)]
If given a specific acid the use the relevent parts of that weak acid.
52
What is a similarity and a diffrenece in the method you would follow to find pH in a strong base strong acid neutraliation compared to what you would do with a strong base weak acid neutralisation.
In both cases you would need to find the initial moles added from the data in the question. You do convert moles to concentration by dividing by the total volume in both cases at some point.
However you need to acount for the fact the moles of the weak acid won't be the same as the moles of H+ in the solution since not all will dissociate. The weak acid method is significantly more complicated since instead of just subtracting one from the other to find if OH or H+ is in exess you need to consider the equation that happens when HA reacts with OH-.
The main difference is that a weak acid neutralisation will need an EQUILIBRIUM TABLE (where the raw initial moles will be put into the table to find exess stuff) based on the equation HA + OH- into A- + H2O is needed if exess Acid.
where the change in moles will be the limiting reagent
that gets used up leaving a decreased final moles of that which was in exess. This will find the neccasary unknowns to qualify for use of Ka expression to get {H+}.
However even the weak acid neutralisation is easy if it is found that you have exess amount of alkali since the degree of dissociation doesnt need to be considered. You can Bi-pass the equilibrium table. Use Kw using {OH-} as being the exess amount divided by total volume as with strong base strong acid.
53
Give the equation that explains what is going on in a weak acid strong base titration where the acid is in exess.
HA + OH- into A- + H2O
Base your equilibruim table on this.
And use this in explanations and calculations.
54
Example 7: 15cm3 of 0.5 mol dm-3 HCl is reacted with 35cm3 of 0.55 mol dm-3 NaOH. Calculate the pH of the resulting mixture.
= 13.37
Hint: find the exess moles of OH-
and use Kw
55
Give the equation that explains what goes on in a strong acid strong base titration.
H + + OH- into H2O
| Use this in explanations and calculations.
56
Example8 : 45cm3 of 1 mol dm-3 HCl is reacted with 30cm3 of 0.65 mol dm-3 NaOH. Calculate the pH of the resulting mixture.
= 0.47
| Hint: Find the moles of H+ that is in exess and divide by the total volume .
57
Example9: 35cm3 of 0.5 moldm-3 H2SO4 is reacted with 30 cm3 of 0.55 mol dm-3 NaOH.Calculate the pH of the resulting mixture.
=0.55
Find the moles of exess H+ by considering.
H+ + OH- into H2O
and the fact the acid will give off 2 moles of H+ for every acid mole.
58
Example 10: 15cm3 of 0.5mol dm-3 HCl is reacted with 35cm3 of 0.45 mol dm-3Ba(OH)2. Calculate thepH of the resulting mixture.
=13.68
| Both are strong so find the concentration of the exess in M.
59
Example11: 55cm3of 0.5 mol dm-3 CH3CO2H is reacted with 25cm3of 0.35 mol dm-3NaOH.Calculate the pH of the resulting mixture? ka is 1.7 x 10-5mol dm-3
=4.44
Hint: CH3CO2H+ NaOH into CH3CO2Na+ H2O should be stated.
Work out initial moles and see that by inspection the acid is in exess.
construct an equilibrium table and insert the raw initial moles you found to get A-.
60
If for a weak acid vs Base titration you find that the Alkali is in exess after you inspected the initial moles which method do you follow.
The exact same method as strong base strong acid where you just find [OH-] by doing exess divided by total volume and then using Kw as 10 to the -14.
61
What is significant , in terms of volume , of a weak acid vs base titration and explain mathematically why this is.
When a weak acid has been reacted with exactly half the neutralisation volume of alkali, the above calculation can be simplified considerably.
At half neutralisation we can make the assumption that [HA] = [A-] and will cancel from the Ka expression leaving.
so [H+(aq)] = ka and then by TLOBS
pH = pka at half neutralisation volume.
62
Calculate thepH of the resulting solution when 25cm3of 0.1M NaOH is added to50cm3 of 0.1M CH3COOH (ka1.7 x 10-5)
This whole process is greatly simplified by ...
| From the volumes and concentrations spot it is half neutralisation (or calculate)pH =pka=-log (1.7 x 10-5) = 4.77
63
Give the equations you would use the find the pH of a diluted acid. (Diluted with water for example.)
[H+] diluted = [H+] old x old volume
new volume
Bang streight into pH = -log [H+]
64
Give the equations you would use to find the pH of a base that has been diluted by say water.
[OH–] = [OH–] oldx old volume
| new volume
65
Example 13: Calculate the new pH when 50.0 cm3 of 0.150 mol dm-3HCl is mixed with 500 cm3of water.
= 1.87
| Hint : multiply the old H+ concentration by the old volume divided by the new volume and put this into the pH formula.
66
Give a perfect exam answer to "define a buffer soluton".
A Buffer solution is one where the pH doesnot change significantly if small amounts of acid or alkali are added to it.
Must include small amounts and not "significant" change.
67
Describe how an acid buffer would be made.
An acidic buffer solution is made from aweak acid and a salt of that weak acid( made from reacting the weak acid with a strong base).Example :ethanoicacid and sodium ethanoate - CH3CO2H(aq) and CH3CO2-Na+
68
Describe how a basic buffer would be made.
A basic buffer solution is made from a weak base and a salt of that weak base ( madef rom reacting the weak base with a strong acid).Example :ammonia and ammonium chloride NH3 and NH4+Cl
69
Explain how a buffer is any different from the weak acid or base that it is made from and therefore explain what it is that buffers contain that make them any different.
In a buffer solution there is a much higher concentration of the salt CH3CO2- ion than in the pure acid.
The buffer contains a reservoirof HA and A- ions.
70
Give the word perfect exam answer for how a buffer solutions works to resist change in pH when small amounts of alkali are added . MUST USE EQUATIONS.
State the equilibrium that is present in a buffer (note if question is specific to actual reactants then use thsoe).
This is your go to reversable reaction you should state at the start of every buffer explanation:
CH3CO2H (aq) REVERSIBLE CH3CO2- (aq)+ H+(aq)
or HA(aq) REVERSIBLE H+(aq)+ A-(aq) (general terms).
State how the equilibrium will shift in order to remove the thing that has been added by a non reversable equation.
OH- + H+(aq) NON REVERSABLE CH3CO2H(aq)
Then finaly explain that since a buffer has large HA concentration HA can ionise to regenerate lost H+
71
Give the word perfect exam answer for how a buffer works to resist change in pH when small amounts of alkali are added. MUST USE EQUATIONS.
Start with the reversable reaction that is present in a buffer. HA(aq) REVERSIBLE into H+(aq)+ A-(aq) (general terms).
Now consider the NON REVERSIBLE reaction that would occur if OH- ions from an alkali where added and how they would be removed.
H+ + OH- NON REVERSIBLE into H2O
Get removd by producing water which will decrease the concentration of H+ ions in solution.
Consider the original equilibrium it would shift to the right to produce more salt and H+ ions. The acid would dissociate and regenetare H+.
Finish explanation by reference to buffer equation and how Some ethanoic acid molecules are changed to ethanoate ions (as regeneration involves equilibrium shifting to right) but as there is a large concentration of the salt ion in the buffer the ratio [CH3CO2H]/ [CH3CO2-] stays almost constant, so the pH stays fairly constant.
72
When we rearange Ka to give the Buffer solution expression what asumption do we make in order to use it for buffer solutions. What is the buffer formula.
[H+(aq)]= Ka X [HA(aq)]
[A-(aq)]
Where we asume that all of the A- is from the salt and also that so little of the acid has dissociated that none has at al l- such that the initial concentration of the acid has remained constant.
73
Give the three possible ways in which a buffer solution can be formed and how the calculations will differ from each other .
1) making a buffer by adding a salt solution
2)making a buffer by adding a solid salt
3) buffer can also be made by partially neutralising a weak acid with alkali and therefore producing salt.
The first two are the same in that you calculate the moles and put it in the rearanged Ka expression (Buffer formula) where [A-} is just the salt moles. Of couse for 1 you would use m= c X v and for 2 divide by mr for moles.
Three is a whole different kettle of fish.
Need to write down the reaction that will actually occur which of course will be neurtalisation.
HA + OH - into A- and H20
To find the moles you will need to put this into an equilibrium table as with weak acid neutralisation since this is what it is ultimately.
This will give you the moles of both HA (exess) and of A- which you can pop into your buffer expression now
74
How to you differenciate weather you should follow the method for a simple buffer calculation where its made by mixing with conjugate base. Or the harder neutralisation method instead.
You need to look at the actual REACTANTS that are given.
If the reactant is just an alkali e.g NaOH its harder neutralisation.
If you can see that the reactant is in fact the conjugate base of the other the you can use the simple method without having to worry about partial neutralisation.
75
Why when entering the moles into a buffer solution do you not need to worry about converting the moles into concentrations by dividing by total volume.
Since [H+] = Ka [HA]
[A-]
you would be doing the same thing to both HA and A- and therefore the effect will just cancel out so the moles is just the same think.
76
Example14: Calculate the pH of a buffer made from 45cm3of 0.1 mol dm-3 ethanoic acid and 50cm3 of 0.15 mol dm-3 sodium ethanoate (Ka= 1.7 x 10-5)
4.99
| Work out both moles of HA and A- and then bang into the buffer formula since volumes will cancel out.
77
A buffer solution is made by adding 1.1g of sodium ethanoate into 100 cm3 of 0.4 mol dm-3 ethanoic acid. Calculate its pH. (Ka=1.7x10-5)
4.29
| Work out both moles of HA and A- and then bang into the buffer formula since volumes will cancel out.
78
Example 15 : 55 cm3 of 0.5 mol dm-3 CH3CO2H is reacted with 25cm3 of 0.35 mol dm-3NaOH. Calculate the pH of the resulting buffer solution.
ka is 1.7 x 10 -5
4.44
Write out the equation for the neutralisation reaction that will occur and form an "equilibrium table" for this reaction.
Use the table to get values for exess HA and also transfer over using the change value to give your A- moles. Put the moles that you worked out into the buffer expression , the total volume will cancel so dont worry about this.
79
Give an equation for what will happen if you add a small amout of alkali to an ethanoic acid based buffer for the purposes of calculation. Hence what would happen to the moles of buffer acid and buffer salt that where originally mixed to make the buffer.
State hoe this equation is different from what you would use to explain resistance to pG change in general.
CH3CO2H(aq)+OH- non reversable CH3CO2-(aq)+H2O(l)
Neutralisation: HA + OH- INTO A- + H20
If a small amount of alkali is added to a buffer 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 pH can be done with the new values.
Since pH stays the same and so [H+] stays the same the increase and decrease must be proportional.
For purposes of pH resistance explanation you would just use H+ + OH- --> H2O
80
Give an equation for what would happen if you added a small amount of acid to an ethanoic acid based buffer.
And state what would happen to the moles of buffer acid and the moles of buffer salt that where originally mixed to make the buffer solution.
CH3CO2-(aq)+ H+ NON REVERSIBLE CH3CO2H(aq)
Reform the acid: A- + H+ into HA
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 the new values.
81
How should you think about calculating the pH of a buffer when a little acid or alkali is added to it.
Ignore the current pH and carry out new calculation with new data.
Find out what the new values for the moles of the acid and the salt are.
Work out the new value for H+ indirectly by using the buffer formula with the new edited moles that you have found.
This is the case you need to take an indirect rout to find new H+ and so it can't be included in your change calculations.
Find new pH.
82
Example 17: 0.005 mol of NaOH is added to 500cm3 of a buffer where the concentration of ethanoic acid is 0.200 mol dm-3 and the concentration of sodium ethanoate is 0.250 mol dm-3.(Ka= 1.7 x 10-5)
Calculate the pH of the buffer solution after the NaOH has been added.
4.91
Consider the equation that actually occurs to remove what has been added in this case : HA + OH- into A- and H20 (remember in buffers will react with the acid not the H+ since there is little available)
Work out the moles of acid and salt in the initial buffer solution Moles ethanoicacid=conc x vol=0.200x0.500=0.100mol Moles sodium ethanoate = conc xvol=0.25x0.500=0.125mol Work out the moles of acid and salt in buffer after the addition of 0.005mol NaOH Moles ethanoic acid = 0.100-0.005 = 0.095 mol Moles sodium ethanoate =0.125 +0.005 = 0.130 mol
Put this streight into buffer formula to get the concentration of H+.
83
How do you calculate the pH after acid or alkali has been added to a bufer.
Edit the moles of buffer acid and buffer salt. Use the new values to go into the buffer formula.
84
Explain, mathematically, what will happen when you dilute a buffer solution with water.
Diluting a buffer solution will not change its pH.
This is as the buffer equation is as follows:
[H] = Ka [HA]
[A-]
In the buffer equation the ratio of {HA}
{A-}
Will stay constant as both the concentration of buffer salt and also buffer acid will be diluted by the same proportion.
85
When should you consider the moles of a solution to be constant and thus consider a calculation as only a dilution.
When water is added.
86
Draw a rough sketch of the titration curve of a strong acid strong base. State on your sketch where you would find the equivalance point. Say the neutralisation volume is 25cm3 and substances is NaOH and HCl.
Key points: pH at equivalence point = 7
Longsteeppart from around 3 to 9.
Start around pH 1 and finish around pH 13.
87
Draw a titration curve for ethanoic acid and NaOH. Label the neutralisation volume as V. And label the diagram with rough pH 's. Comment on the shape of this curve and why it is significant.
Should start around pH 3 and finish around pH 13 still.
The steep vertical section should be squashed upwards since the acid pH is higher on the y Axis. Vertical section should now be around 7-9 with the equivalance point being > 7.
Equivilance point > 7 is key .
At the start the pH rises quickly and then levels off. The flattened part is called the buffer region and is formed because a buffer solution is made.
88
What is the name of the significant volume that occurs before the neutrlasisation volume and with which type of substances will it occur.
The half neutralisation volume and it will occur with weak acid strong base titrations.
89
State what can be worked out from the half neutralisation point and give the mathematical explanation of why this is so.
Well as we know,
Ka= [H+(aq)][A-(aq)]
[HA (aq)]
At ½ the neutralisation volume the [HA] = [A-]
And so thease terms will cancel from the expression leaving So Ka= [H+] and By TLOBS
pKa = pH at the half neutralisation volume
Why do we care
This means that given the Ka we can work out the pH at half neutralisation volume and vice versa , can work out the Ka if we have the pH at this point.
`Therefore : If a pH curve is plotted then the pH of a weak acid at half neutralisation (½ V) will equal the pKa.
90
Sketch a titration curve for the second most boring case where a strong acid is titrated with a weak base. Let the neutralisation volume be say 25cm3. State and label on all of the significant features.
Here the vertical section is squashed downwards since the pH of the base is brought downwars since it is weaker. The vertical section should be around 4-7 with an equivelance point that has been lowered to a pH < 7 .
Equivalence point of < 7 is the key take away here.
91
What is an indicator and give the equilibrium that is established for an indicator in aqeous solution.
Indicators can be considered as weak acids. Where the acid must have a different colour to its conjugate base.
HIn (aq) REVERSIBLE INTO + In-(aq) H+(aq)
colour A colour B
92
HIn (aq) REVERSIBLE INTO In-(aq) + H+(aq)
What can be apllied to the indicator equilibrium established in aqeous solution to find the colour change when acid or alkali is added.
So for HIn (aq) REVERSIBLE INTO In-(aq) + H+(aq)
colour A colour B
We can apply LeChatelier to give us the colour In an acid solution the H+ions present will push this equilibrium towards the reactants.Therefore colour A is the acidic colour. In an alkaline solution the OH-ions will react and remove H+ ions causing the equilibrium to shift to the products. Colour B is the alkaline colour.
93
What is defined as the "end point" of a titration and give a summary of how this relates to choosing the correct indicator and should never be confusd for the equivalnece point. In your answer define both.
The end-point of a titration is defined as the point when the colour of the indicator changes colour . The end-point of a titration is reached when [HIn] =[In-]. This is the acids half neutralisation point howver since the colour of indicators is equilibrium depedent this is the point one side of the equilibrium is favored over the other and hence the colour chnages. THIS IS DEPENDENT ON THE INDICATOR
The equivilance point is DEPENDENT ON THE REAGENTS (acid and base reaction) and is reached when the reaction is complete and the acid and base have reated in there exact stoichometric ratio.
Here [H+] = [OH-} neither in excess.
To choose a correct indicator for a titration one should pick an indicator whose end-point coincides with the equivalence point for the titration. Therefore the equivalance point is not the same as the endpoint but it will hopefully be close.
94
Where must the pH range of an indicator lie for it to work and what will be seen if this is the case.
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 CORRSPOND TO THE NEUTRALISATION POINT.
95
When should phenolphthalein be used and state the colout change you would observe.
Coment on the pH range of this indicator.
Only use phenolphthalein in titrations with strong bases but not weak bases. SEE Diagram. Its pH range is rather high.
Colour change: colourless acid INTO pink alkali.
The pH range is rather high.
96
When should Methyl orange indicator be used and comment both on the colour change and also the relative pH range of this indicator.
SEE Diagram.
Use methyl orange with titrations with strong acids but not weak acids , Colour change: red acid INTO yellow alkali (orange end point).
Methyl Orange has a rather low pH range relative to other indicators so is good for strong acids that havent had their pH pushed up on the graph yet.
97
Give an expression for the Ka of an indicator and then state what this expresion is known as.
Give what this epression can be simplified down to at the point the indicator changes colour.
Since an indicator is just a weak acid Ka = [H+] [In-]
[HIn]
Instead of Ka since this is specific to an indicator then the expression can be called KIn .
At the point of colour change [Hln+] = [ In-} by definition.
And so like with all weak acids , and with this being teqnically the half neutralisation point of the indicator in terms of the established equilibrium being surpased to the otehr side, terms cancel leaving Ka = H+
Or more specifically KIn = H+ at end point, Then by TLOBS
PkInd= pH at the colour change end point of the titration.
98
State in very simple terms what a titration is and what an indicator is.
Where a substance of acutarely known concentration is added gradually to another solution of unknown concentration and until the chemical reaction is complete. An indicator is a weak acid that will change colour at the end point that is hopefully close to the equivalance point.
99
What is the simple goal when picking an indicator.
That it will change colour at a pH that is close to the equivalance point. You Want the half neutralisation volume of the indicator to be as close to the full neutralisation volume of the reaction as possible.
100
Calculate the pH of solution formed after the addition of 50cm3 of 0.15M NaOH to the original 25cm3 of strong acid that contains 1.5 X 10-3 moles of H+ ions .
Include the equation of the reaction that occurs in your answer and what type of reaction this is.
pH = 12.9
H+ + OH- ---> H20 strong acid strong base neutralisation.
Work out concentration of XS OH- and hence using Kw find pH.
101
Calculate the volume of water that must be added to 25cm3 of a strong acid to increase its pH from 0.5 up top 0.7. State what type of problem this is and also the two methods you could use. Take the H+ concentration to be 0.3 to 1dp for original pH of 0.5.
[H+] at pH = 0⋅7 is 0⋅2 mol dm–3
m1 v1 = m2 v2 ∴ 0⋅3 × 25 = 0⋅2 × v (1) Hence v = 37⋅5
Water added = 37⋅5 – 25 = 12⋅5
or can use idea that {H+new} = [H+old] X old volume
new volume
102
What are the units of Ka
mol dm –3
103
Explain the terms acid and conjugate base according to the bronzed Lowry theory.
Acid is a proton donor
The conjugate base to that acid is the substance formed when acid has lost proton / substance that
becomes an acid / potential to become an acid
by gaining a proton (not just proton acceptor).
104
Give the conjugate acid and base in the following reactions.
NH3+HBr→NH4+ +Br–
H2SO4 + HNO3 → HSO 4– + H2NO3+
(i) acid: HBr base: Br–
| (ii) acid: H2SO4 base: HSO4–
105
Give the expression for the equilibrium constant, Kc, for the reaction in (c)(i) and use this to derive the expression for the ionic product of water, Kw.
endothermic and attempt at reason (1)
more dissociation / ionization / H+ ions at higher temperature (1)
4
if (iii) not completed, allow endothermic with sensible reason
for 1 mark if answer to (iii) is pH>7, allow 1 mark for exothermic
with attempt at reason 2
106
Identify two components that could be used to make a buffer solution.
correct weak acid / co-base
or correct weak base / co-acid
Ethanoic acid and sodium ethanoate.
or mixture of aqueous ammonia and ammonium chloride.
107
Give a use of a buffer solution.
Blood is buffered to a pH of around 7.4.
| Shampoo is buffered so it is slightly alkaline.
108
Why is neutralising some of a weak acid with an alkali such as sodium hydroxide practically useful if half of the weak acid is neutralised , and state also the mathematical significance of this. Give some numerical values and conditions to how this buffer could be made.
This is a very useful buffer as it is equally efficient in resisting change in pH whether acid or alkali is added.
If something like 0.1 mol of ethanoic acid is mixed with 0.1 mol of sodium ethanoate with water making it up to the total volume, then there is an equal supply of HA and A-.
Therefore in a case such as this you will have a buffer with a pH that is = to the pKa of the weak acid at half neutralisation.
109
What expression are you going to be using in regards to most buffer calculations.
Ka of the weak acid at hand.
110
How could you calculate the original concentration of an acid given the titration curve and the concentration of the base that was added.
Remember the only things relevant here is moles and volume if you remember it is just a neutralisation reaction.
The neutralisation volume multiplied by concentration of the base will give the moles of acid.
Divide this value by the volume in dm3 , as you would in a normal neutralisation reaction , to get the original concentration of the acid.
111
To determine the volume of NaOH added when [HA] = [ A–] on a titration cure and hence the pH what should you so.
For thus you can juts read off the graph from half neutralisation volume and the corresponding pH however if you need to go on to find Ka then you would need to use this value for pH at this volume and use pH = pKa.
112
In a mixture of aqueous NH3 and and NH4+Cl- state how this acts as a buffer.
The aqueous ammonia will remove the added H+ by the reaction of NH3 + H+ ---> NH4+
The ammonium ion will remove the added OH- by the reaction of NH4+ + OH- ---> NH3 + H20 ( just like the reaction in a weak acid buffer to remove OH- )
113
SEE Picture: If a titration curve shows how pH changes when 0.12 mol of HCl is added to a 25cm3 solution of sodium carbonate there will be two and points.
Give the equations for both end points and state how you would know how to write such equations.
CO3 2- + H + --->. HCO2. end point 1. or
Na2CO3 + HCl → NaHCO3 + NaCl
HCO3- + HCl ---> H20 + CO2 end point 2. or
NaHCO3 +HCl→NaCl+CO2 +H2O
Need to consider the overall reaction and then think about the two possible neutralisations that could happen to get the overall final products of overall reaction.
Remember once you have completed one neutralisation the acid will continue to be added and so this means that the basic of the two products from neutralisation 1 will go on to react with the acid again.
114
How do you calculate the concentration initial or otherwise of an acid or base when just given a titration curve.
You will have the concentration of what is on the x-axis (what is being added from burette) and so you will be able to calculate the moles that have been added at a given neutralisation volume , if there is 2 end points you can use the volume from the second.
You will also have the volume of what is in the burette and so simply divide the moles you have calculated from the neutralisation volume by the volume in the beurtte.
115
Suggest and explain with reference to pKa value the pH range of phenolphthalein. Explain also in terms of the species present the colour change that would be seen if NaOH where being added from the burette.
At an indicators end point the pKa = pH since [In–] = [HIn] and so pH= pKa = 9.3 , need also to state that the colour change for an indicator is detectable over a pH range of 2 and so the range is 8.3-10.3
(colourless to) pink / red (1)
Following is established for an indicator
HIn(aq) Reversible H+(aq) + In– (aq)
colourless red
As OH- is added equilibrium will shift right to regenerate lost H+ but more impprtantly [In–] increases to the point where [In–] becomes ≥[HIn] and so the colour changes to that of the conjugate base of HIn
116
In a 0.25 M solution, a different acid HY is 95% dissociated. (i) Calculate the pH of this solution
.Calculate the value of Ka for the acid HY.
[H+] = 0.25 × 0.95 = 0.2375. pH = 0.62
```
[H+] = [Y–] = 0.2375
[HY] = 0.05 × 0.25 = 0.0125 need to count for 5 percent of the acid that did not dissociate.
```
Use isolated acid formula.
Ka = [H+] [Y–] / [HY]
= (0.2375)2 / 0.0125 (1) Ka = 4.51
117
Write a equation for the reaction of ammonia with water and explain why a solution of 1M of ammonia has a pH less than 14 at 298K despite the fact that 1M solution of OH- ions would numerically have a pH of 14.
NH3 + H2O REVERSIBLE NH4+ + OH–
(ii) Ammonia is weak base (1)
Equilibrium to left or incomplete reaction (1).
Essentially not all ammonia will have dissociated to form the same amount of OH- . so [OH-] < 1M
118
Identify a reagent which could be added to a solution of ammonia in order to form a buffer solution.
Reagent: NH4Cl
| That would act as NH4+ and Cl -.
119
The concentration of HA in a buffer solution is 0.250 mol dm–3. Calculate the concentration of A– in this buffer solution when the pH is 3.59. State also why you shouldn't panic when you come across a question like this.
Ka, for the weak acid HA, at 298 K, is 1.45 × 10–4 mol dm–3.
```
[H+] = 10–3.59 = 2.57 × 10–4 or 2.6 × 10–4 1 [A–] = Ka[HA] 1
[H+]
= (1.45×10−4]×0.25 1
2.57×10−4
=0.141(moldm−3)
```
You can simply sub into the ka expression or Buffer equation if you prefer to call it that and instead of solving for H+ since you already have pH you can instead solve for { A-} after you get HA
120
A titration curve is plotted showing the change in pH as a 0.0450 mol dm–3 solution of sodium hydroxide is added to 25.0 cm3 of a solution of ethanedioic acid, H2C2O4
The titration curve obtained has two equivalence points (end points).
(i) Write an equation for the reaction which is completed at the first equivalence point.
Hint include only the relevant Ions. (ii)
When the second equivalence point is reached, a total of 41.6 cm3 of 0.0450 mol dm–3 sodium hydroxide has been added.
Calculate the concentration of the ethanedioic acid solution. State how this is similar to a method you gave used in other areas.
H2C2O4 + OH– → HC2O4– + H2O
For part two it helps if you draw out an ethandioic acid molecule to you can understand that two OH- ions are required to neutralise one ethandioic acid.
The method of finding the moles at the second neutralisation volume of substance been added and then finding the concentration of substance in flask by dividing by the volume of the solution originally in the flask is not new its basic neutralisation mole calculation.
However when you get the value for moles of NaOH you need to divide it by 2 due to the structure of the ethandioic acid as discussed above.
mol OH– = (41.6 × 10–3) × 0.0450 (1) = 1.87 × 10–3
∴ mol H2C2O4 = 9.36 × 10–4 (1)
[H2C2O4] = 9.36 × 10–4 × 103/25
= 0.0374 (1)
121
Calculate the pH of a mixture formed by adding 25 cm3 of a 0.117 mol dm–3 aqueous
solution of sodium propanoate to 25 cm3 of a 0.117 mol dm–3 aqueous solution of propanoic acid.
And easy give away that this will be the method is the fact the exact same quantities and concentration of each are used and also the fact it is neutralisation.
Must be able to recognise that here that half the acid has been neutralised with the particular quantities that have been used and hence there is an equal supply of HA and A- ions. HA + OH- reversible A- + H2O
So at half neutralisation Ka = [H+]
OR
pKa = pH; 1 pH = 4.87;
