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# AQA Chemistry Unit 4 with Anon_98 Watch

1. Acids + Bases – Weak acids + bases. Ka for weak acids. – Topic 3.

Okay so, the equation for the dissociation of a weak acid, using ethanoic acid as an example is:

CH3COOH → CH3COO + H+

And the expression for its acid dissociation constant, Ka is:

Ka =[CH3COO-] [H+] / [CH3COOH]

~

pKa = -log10Ka

Ka = 10-pKa

Overall,

Acids with a Ka value much greater than 1 (negative PKa) are classified as strong acids and..
Acids with a Ka value much smaller than 1 (Positivite PKa) are classified as weak acids.
2. Hi dude I am doing chem 4 as well I like the way you do your notes but I think your missing some stuff like definition of rate of reaction which is change in conc of products or reacts / time or with equilbrium when volume is increased the pressure is lowered. Then you apply le chatlier's principle. Both these questions come up a lot idk if you know it or not. Also I asked in thread too... How many hours are you planning to finish in because I have done the first 3 already and am planning to do the next 8 in 10 hours

edit: was going to send as PM but you don't accept them...
3. Acids + Bases – PH curves, titrations + indicators –Topic 3.

Okay, so there are basically 3 types of PH curves we need to know.

Strong acid, strong base.
Strong acid, weak base.
Weak acid, strong base.

~

So, the example that we're going to use for a Strong acid, Strong base titration curve is HCl + NaOH.

HCl + NaOH -----------------------> NaCl + H2O

At the equivalence point:

- Ratio of HCl to NaOH is 1:1

- PH changes rapidly at this point (vertical)

PH at the equivalence point:

- Midpoint of the vertical line - 7.0

- At this point only Na+ and Cl- ions are present in solution.

- Both ions do not react with water and that PH is that of water.

Choice of indicator:

- An indicator must change colour at the equivalence point.

- The PH range over which it changes colour should therefore straddle the vertical portion of the graph.

- Methyl orange is the most suitable indicator bc it just fits + will change colour at the equivalence point.

Alright, I'll draw the curve for this below.
4. (Original post by EnterNamehereplz)
Hi dude I am doing chem 4 as well I like the way you do your notes but I think your missing some stuff like definition of rate of reaction which is change in conc of products or reacts / time
Mhm, I've already got that on page 1 here:

r = rate equation, I.E. The change in concentration of a reactant (say A), per unit time.
Units: moldm-3 s-1

(Original post by EnterNamehereplz)
or with equilbrium when volume is increased the pressure is lowered. Then you apply le chatlier's principle. Both these questions come up a lot idk if you know it or not.
Yeah, I know but if I was to include that then I'd have to choose an example and I cba doing that bc you get practice with those kinda questions by doing the past papers so I'm not going to include every single possible question they might ask otherwise it'll take me forever.

(Original post by EnterNamehereplz)
Also I asked in thread too... How many hours are you planning to finish
Yeah, I know you did + I was going to come back to it after I've finished bc I don't really like having a conversation in between all the notes bc it messes up the layout + ..disrupts my mind.. + now this message I'm sending in response has ruined everything bc I wanted the picture of the titration curve to be directly below my last post. >.>

I've obviously absolutely no idea. I'm not planning, I'm just doing it as fast as I can and whenever I get it all done, then that's how long it takes me. I'm pretty slow sometimes, especially bc I'm having to digest the info first, then think about how I'm going to condense it etc etc. I think Chem 2 took me about 18hrs, this unit is a lot bigger so I don't know.

in because I have done the first 3 already and am planning to do the next 8 in 10 hours
I don't understand what you mean by this. First 3 topics? Cool - keep it up!

(Original post by EnterNamehereplz)
edit: was going to send as PM but you don't accept them...
Yep, I know .. Uh, I can open them up later for you if you like. Perhaps after I've finished these notes.
5. Alright, so.. Strong Acid, Strong Base.

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6. Now onto Strong acid, weak base.

So, the example I'm going to use for a strong acid, weak base titration curve is HCl + NH3.

HCl + NH3 -----------------------> NH4Cl

At the equivalence point:

- Ratio of HCl to NH3 is 1:1

- PH changes rapidly at this point (vertical)

PH at the equivalence point:

- Mid point of the vertical - about 5.

- At this point only NH4+ and Cl- ions are present in solution.

- NH4+ hydrolyses so that PH is acidic:

NH4+ + H2O <-------------------> NH3 + H3O+

Choice of indicator:

- An indicator must change colour at the equivalence point.

- The PH range over which it changes colour should therefore straddle the vertical portion of the graph.

- Methyl orange is the most suitable indicator bc it just fits + will change colour at the equivalence point.

Phenolphthalein is a v poor choice, if it changes colour it will be way after the equivalence point.

Alright, so now I'm going to draw the curve for this below:
7. Curve for a strong Acid, weak Base

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8. Finally a weak acid, strong base.

The example I'm going to use is CH3COOH + NaOH.

At the equivalence point:

- Ratio of CH3COOH to NaOH is 1:1

- PH changes rapidly at this point (vertical)

PH at the equivalence point:

- Mid point of the vertical - about 9.

- At this point only Na+ and CH3COO- ions are present in solution.

- CH3COO- hydrolyses:

CH3COO- + H2O <-------------------> CH3COOH + OH-

Choice of indicator:

- An indicator must change colour at the equivalence point.

- The PH range over which it changes colour should therefore straddle the vertical portion of the graph.

- Methyl orange is v rubbish indicator, phenolphthalein will work best.

Curve below:
9. Curve for Weak acid, strong base.

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10. What is meant by the term ‘half-equivalence point’ in a weak acid (HA) strong base titration and what is it's significance in finding the pKa of the weak acid?

At the half-equivalence point exactly half the equivalence volume of strong base has been added to the weak acid. For instance, if the equivalence volume is 25.0 cm3 then the half-equivalence volume will be 12.5 cm3.

At this point [HA] = [A-] and the expression:

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

REDUCES TO:

[H+]= Ka

THEN.. multiplying both sides by –log gives:

PH= pKa

Basically, this means that the pH at the half-equivalence point has the same value as the pKa of the weak acid.
11. Acids + Bases – Buffer action –Topic 3.

Buffer solution - A solution which resists changes in PH when small amount of acid or base are added. It also maintains PH when diluted with water.

~

Acidic buffer + how it works when a small amount of acid is added.

An acidic buffer consists of a mixture of a weak acid and one of its salts. For example, CH3COOH and CH3COONa.

So, when a small amount of acid is added..

The H+ reacts with the CH3COO-ions:

CH3COO- + H+ → CH3COOH and the excess acid is removed.

Therefore the [CH3COOH] slightly increases and the [CH3COO-] slightly decreases.

However, since [CH3COOH] and [CH3COO-] are high, the ratio of [CH3COOH] to [CH3COO-]remains roughly constant in the expression:

[H+] = Ka [CH3COOH] /[CH3COO-]

and therefore the [H+] remains approximately the same.
12. Basic buffer + how it works when a small amount of base is added.

A basic buffer consists of a mixture of a weak base and one of it's salts. For example: NH3 and NH4Cl.

So, when a small amount of base is added...

the -OHreacts with the NH4+:

NH4+ + -OH → NH3+ H2O and the excess base is removed.

Therefore the [NH3] slightly increases and the [NH4+] slightly decreases.

However, since [NH4+] and [NH3]are high, the ratio of [NH4+] to [NH3] remains roughly constant in the expression:

[H+] = Ka [NH4+]/ [NH3]

and therefore the [H+] remains approximately the same.
13. Applications of buffers.

- Human blood - Has to be kept constant at around a PH of 7.4 otherwise illness occurs.This is achieved through buffering action in the blood.

- Fizzy drinks. - They usually contain citric acid and are buffered with sodium citrate.
14. Example question 1:

In a buffer solution, the concentration of ethanoic acid is 0.150moldm-3 and the concentration of sodium ethanoate is 0.100moldm-3. The Ka for ethanoic acid is 1.74 x 10-5 .

A) Calculate the PH of this buffer solution.

So, we'll do it in steps:

1)CH3COOH → CH3COO + H+

2) Ka = [CH3COO-] [H+] / [CH3COOH]

THEREFORE

[H+] = Ka[CH3COOH] / [CH3COO-]

= 1.74 x 10-5 x 0.150/0.100

= 2.61 x10-5 mol dm-3

3) PH = -log [H+]

THEREFORE:

pH = -log 2.61 x 10-5 = 4.58
15. B) A 10.0 cm3 portion of 1.00 mol dm–3 hydrochloric acid is added to 1000 cm3 of this buffer solution. Calculate the number of moles of ethanoic acid and the number of moles of sodium ethanoate in the solution after addition of the hydrochloric acid. Hence, find the pH of this new solution.

1) n(H+)= cV = 1.00 x 10/1000 = 0.01 mol

The H+ reacts with ethanoate ions:

CH3COO- + H+ → CH3COOH

THEREFORE

n(CH3COO-)= 0.100 – 0.01 = 0.09 mol
n(CH3COOH) = 0.150 +0.01 = 0.16 mol

2) [H+] = Ka [CH3COOH] / [CH3COO-]

=1.74 x 10-5 x 0.16 / 0.09
= 3.09 x 10-5

3) PH = -log [H+]

THEREFORE

pH = -log 3.09 x 10-5 = 4.51
16. Nomenclature + Isomerism – Naming organic compounds –Topic 4.

Here is the priority table that you need to remember when naming compounds. The higher up it is in the table, the higher priority it has and therefore will take dominance over the name.

17. Nomenclature + Isomerism – Isomerism –Topic 4.

Structural isomerism. -Molecules with the same molecular formula but with different structural formula.(chain, position, functional group)

Stereoisomerism - Molecules with the same structural formula but their bonds are arranged differently in space. (E-Z + Optical)

Features of a molecule that give rise to optical isomerism are:

- Asymmetry.

- Chiral C atom. I.E. Carbon atom with 4 different atoms/groups attached.

Optical isomers exist as non super-imposable mirror images and differ only in their effect on plane polarised light.

I'll draw an example of Butan-2-ol below to illustrate it:
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19. Enantiomer - it is the term used for one of the optical isomers which is a mirror image of the other.

How can two enantiomers be distinguished?

They are optically active, but otherwise have identical physical properties.

However, one enantiomer rotates the plane of polarised light clockwise (+) and the other rotates the plane anticlockwise(-)by an equal amount.

~

Racemate - An equal amount of two enantiomers and is optically inactive.·(Sometimes called a racemic mixture)

Alright, so I'm going to do an example reaction mechanism between Hydrogen Bromide and But-1-ene below to show how a racemate is formed.
20. HBr + But-1-ene.

(the blue dashed line is trying to show a mirror plane)

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