Useful tips on IA's
Link to IB site for IA criteria
Monday, November 28, 2011
Wednesday, November 23, 2011
Revision note sites
There are lots of good links here. The first one seems a detailed great summary of the whole course. Take a look and tell me what you think.
Revision notes
Revision notes (whole syllabus)
IB site acid base
Student room notes
Revision notes
Revision notes (whole syllabus)
IB site acid base
Student room notes
Test and IA dates
Please remember that your IA is due on monday and you will have a test on acids and bases on Thursday(dec 1st)
Sunday, November 20, 2011
Acid/ Base Indicators
The choice of indicator depends on the type of acid / base titration that is being carried out. A strong base with a strong acid has a large pH range at the equivalence point (long vertical section of the graph). Therefore, many indicators would change colour at the equivalence point and would work. However, if there is a weak acid/ strong base or weak base/ strong acid, the pH change at the equivalence point is less and the pH at the equivalence point is not 7. Therefore, care needs to be taken with the choice of indicator, so that it changes colour at the equivalence point and not before or after it. Weak acid with weak bases, don't have clear distinct equivalence points and no indicator would work well. The following notes from chem guide, explain clearly how indicators work. They are simply weak acids where the acid and its conjugate base have different colours, and so the colour will be different in acidic or basic conditions. The indicator should be chosen so that the PkIn value is at or close to the pH at the equivalence point.
Thursday, November 17, 2011
Buffer Practical Data
Drops of Acid Water (pH) Buffer ( PH)
0 6.67 10.15
1 3.00 10.02
2. 2.57 9.85
3 2.39 9.74
4 2.17 9.55
5 2.08 9.30
Note: It would have been better to have chosen a buffer with pH 7 so that it started at the same pH as water. However, this does clearly show the effectiveness of a buffer at keeping the pH relatively constant, compared to the massive change as seen when a drop of acid is added to water
Drops of Alkali Water (pH) Buffer (pH)
0 6.67 10.15
1 11.23 10.31
2 11.74 10.46
3 11.98 10.66
4 12.15 10.76
5 12.35 11.06
Homework:
1. Put the data into an appropriate table
2. Plot appropraite graphs
3. Discuss the results. What does it illustrate?
4. The buffer (pH 10.01) conntained sodium hydrogen carbonate and sodium carbonate. Explain how it worked and write equations to illustrate your answer.
Sunday, November 13, 2011
Notes on Acid / Base titrations and Buffer solutions
Acid/Base notes:
The equivalence point is when the acid base reaction is complete. For example if 25cm3 of 0.1M of sodium hydroxide is added to 25cm3 of 0.1M of ethanoic acid the reaction will be complete. Adding more base will cause the pH to shoot up to the pH of the base.
The pH at the equivalence point is not necessarily 7. For strong bases with strong acids it is. This is because the salts formed are neutral and have no reaction with water. For example sodium chloride (NaCl) is a neutral salt and is formed from the reaction of sodium hydroxide with hydrochloric acid.
For weak acids and a strong base the pH at the equivalence point will be above 7, in other words it will be an alkaline solution. This is because the salt that is formed is a conjugate base of the weak acid. For example the salt of the reaction with ethanoic acid and sodium hydroxide is sodium ethanoate (CH3COONa). In solution the ethanoate ions (CH3COO-) act as a base, and accept a proton from water, therefore, releasing OH- ions.
CH3COO- + H2O ↔ CH3COOH + OH-
For weak bases and strong acids the pH at the equivalence point will be acidic. This is because the salt formed is a conjugate acid of the weak base. For example the salt formed when a solution of ammonia reacts with hydrochloric acid is ammonium chloride (NH4Cl). In solution the NH4+ ions will donate a proton and therefore, the pH of the solution will be below 7.
NH4+ ↔ NH3 + H+
Or more truly: NH4+ + H20 ↔ NH3 + H3O+
The buffer region in titrations:
When a strong base is added to a weak acid or a strong acid is added to a weak base, around the half equivalence point (if the equivalence point is when 25cm3 has been added, the half point is when 12.5cm3 has been added). At this point and around this point there is the unreacted weak acid and the salt formed which is the conjugate base. Or the unreacted base and the salt formed which is the conjugate acid. Let’s take the weak acid strong base example first:
At the half equivalence point, half of the weak acid remains, E.g. CH3COOH, in addition there is the same amount of salt in the solution. This means that there is the acid to react with small amounts of base that is added. Therefore, pH remains relatively constant around the half equivalence zone. There is also the conjugate base which can react with small amounts of acid, if this was added. Buffer solutions can resist changes in pH upon small amounts of acid or alkali added.
For a weak base and a strong acid which is added, at the half equivalence point, there are equal amounts of the weak base and its salt (a conjugate acid). This means that the pH stays relatively constant around this area as there is some weak base to react with the added acid. It would also remain relatively constant if a small amount of base was added as there is the conjugate acid to react with it.
Buffer solutions
You have seen that a buffer solution is formed when a weak acid is partially reacted with a strong base or a weak base is partially reacted with a strong acid. A buffer solution could also be formed by adding the salt of a weak acid to the weak acid or adding the salt of a weak base to the weak base. Effectively you end up with the same thing.
Weak acid buffer: E.g. ethanoic acid and sodium ethanoate
The acid can react with small amounts of alkali added: CH3COOH + OH- → CH3COO- + H2O
The conjugate base can react with small amounts of acid added: CH3COO- + H+ → CH3COOH
Or CH3COO- + H3O+ → CH3COOH + H2O
Note that the weak acid is even less ionised due to the presence of the conjugate base and the salt is therefore, pretty much totally from the salt. So we could put the following equation:
CH3COOH ↔ CH3COO- + H+
As: Acid ↔ Salt + H+
Therfore Ka can be expressed as: Ka = [salt] [H+]/[acid]
[H+]= Ka [acid]/[salt]
Or pH = pKa + log [salt]/[acid]
And with this equation you can see clearly that at the half equivalence point the pH = pKa. You can therefore, use this to obtain the pka value of a weak acid.
Now follow the same logic for a weak base and its salt.
The relationship between Ka and Kb for an acid base conjugate pair
Pick and example or just use HA and A- to represent a weak acid and its conjugate base, and then write the appropriate equations and see if you can work out the relationship between them. This can be useful if you are given one value and you need to use the other.
Wednesday, November 2, 2011
Acid /Base equations
Acid and Base Equations
[H+] to pH = log (1/[H+])
pH to [H+] = 10-pH
kw = 1 x10-14 at 250C
[OH-] to pOH = log (1/[OH-])
pOH to [OH-] = 10-pOH
At 250C pH + pOH = 14
Ka = [H+] [A-]/[HA]
Kb = [BH+] [OH-]/[B]
Ka to pKa = log (1/Ka)
pKa to Ka = 10-pKa
Kb to pKb = log (1/Kb)
pKb to Kb = 10-kb
Ka x Kb = 1 x 10-14 at 25OC
PKa + pKb = 14 at 25OC
pH of an acidic buffer = PKa + log [salt]/[acid]
pOH of a basic buffer = pKb + log [salt]/[base]
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