CHEMISTRY CALCULATION INDEX for GCSE and Advanced LevelDoc Brown's GCE AS A2 Level Chemistry

Advanced Level Chemistry Revision on Titrations

GCE A Level AS-A2 IB Chemistry Volumetric Analysis: Acid-base and other non-redox volumetric titration quantitative calculation answers to PART 1 Questions 1 to 20

PART 1 Questions * PART 2 Questions

PART 2 Question Answers *  Redox Titration Q's

Qualitative Analysis

If you find these useful or spot a silly error please EMAIL query? comment

ILLUSTRATIONS OF ACID-ALKALI TITRATIONS and SIMPLE STARTER CALCULATIONS

I DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do,  BUT I AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE EMAIL ME ASAP TO SORT IT OUT!

Most of the answers have been rounded up or rounded down to three significant figures (3sf)

Q1 ANSWERS (a) NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l)

(b) (i) pipette (ii) burette

(c) everything with distilled water, then pipette with a little of the NaOH(aq) and the burette with a little of the HCl(aq)

(d) pink to colourless, the first drop of excess acid removes the pink alkaline colour of phenolphthalein

(e) moles sodium hydroxide neutralised: 0.250 x 25.0/1000 = 0.00625 mol NaOH

(remember: moles = molarity x volume in dm3 and its two rearrangements and 1 dm3 = 1000 cm3)

(f) moles HCl = moles NaOH (equation) = 0.00625 mol HCl (in 22.5 cm3)

(g) concentration hydrochloric acid = 0.00625 x 1000 22.5 = 0.278 mol dm-3 (3sf)

(scaling up to 1 dm3 = 1000 cm3 to get the molarity)

Another way to work it out is 22.5 cm3 = 22.5 1000 = 0.0225 dm3

Therefore molarity = 0.00625 0.02250.278 mol dm-3

 

Q2 ANSWERS (a) Ba(OH)2(aq) + 2HCl(aq) ==> BaCl2(aq) + 2H2O(l)

(b) formula mass of Ba(OH)2 = 171, moles = 2.74 171 = 0.016 mol in 100 cm3, (scaling up x 10)

therefore 0.16 mol in 1000 cm3, so molarity of Ba(OH)2 = 0.16 mol dm-3

(c) moles Ba(OH)2 used in titration = 0.16 x 20/1000 = 0.0032 mol

(d) moles HCl titrated = 2 x moles of Ba(OH)2 used (2 : 1 in equation) 2 x .0032

= 2 x 0.0032 = 0.0064 mol HCl in 18.7 cm3 of the acid solution, 18.7 cm3 = 0.0187 dm3

(e) therefore molarity of HCl(aq) = 0.0064/0.0187 = 0.342 mol dm-3

 

Q3 ANSWERS (a) 2NaOH(aq) + H2SO4(aq) ==> Na2SO4(aq) + 2H2O(l)

(b) moles H2SO4 = 4.90 98 = 0.050 mol in 200cm3

scaling up to get molarity of the sulphuric acid solution, 0.050 x 1000 200 = 0.25 mol dm-3

(c) moles of sulphuric acid neutralised = 0.250 x 20.7/1000 = 0.005175 mol

(d) moles of sodium hydroxide neutralised = 2 x 0.005175 = 0.01035 mol (2 : 1 in equation)

(e) concentration of the sodium hydroxide = 0.01035 x 1000 10 = 1.035 mol dm-3 (molarity 1.04, 3sf)

 

Q4 ANSWERS (a) Mg(OH)2(aq) + H2SO4(aq) ==> MgSO4(aq) + 2H2O(l)

(b) moles of sulphuric acid neutralised = 0.100 x 4.5/1000 = 0.00045 mol

(c) moles of magnesium hydroxide neutralised also = 0.00045 (1:1 in equation) in 100 cm3

(d) concentration of the magnesium hydroxide = 0.00045 x 1000 100 = 0.0045 mol dm-3

(scaling up to 1000cm3 = 1dm3,  to get molarity)

(e) molar mass of Mg(OH)2 = 58.3

so concentration of the magnesium hydroxide = 0.0045 x 58.3 = 0.26 g dm-3 (= g per 1000 cm3),

so concentration = 0.26 1000 = 0.00026 g cm-3

 

Q5 ANSWERS (a)(i) The magnesium oxide is insoluble in water and so cannot be titrated directly by a volumetric analysis method. By dissolving in excess acid you are sure to get all the MgO into solution, and then determine the excess acid, from which you can derive the amount of MgO that dissolved.

(a)(ii) MgO(s) + 2HCl(aq) ==> MgCl2(aq) + H2O(l)

NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l)

(b) moles of hydrochloric acid added to the magnesium oxide = 2 x 100/1000 = 0.20 mol HCl

(c) moles of excess hydrochloric acid titrated = 19.7 1000 x 0.200 = 0.00394 mol HCl

{mole ratio NaOH : HCl is 1 : 1 from equation (ii)}

(d) moles of hydrochloric acid reacting with the magnesium oxide = 0.20 - 0.00394 = 0.196 mol HCl

(e) mole MgO reacted = 0.196 2 = 0.098 {1: 2 in equation (i)}

the formula mass of MgO = 40.3

therefore mass of MgO reacting with acid = 0.098 x 40.3 = 3.95 g

(f) % purity = 3.95 4.06 x 100 = 97.3% MgO

(g) Mg(OH)2 from MgO + H2O, MgCO3 from the original mineral source, both of these compounds react with acid and would lead to a false titration value.

 

Q6 ANSWERS (a) CaCO3(s) + 2HCl(aq) ==> CaCl2(aq) + H2O(l) + CO2(g)

(b) moles of hydrochloric acid was spilt = 2.00 x 10.0 = 20 mol HCl

(c) moles of calcium carbonate to neutralise the acid = 20 2 = 10.0 mol CaCO3 (1:2 in equation)

(d) formula mass of CaCO3 = 100,

so mass of limestone powder needed to neutralise the acid = 100 x 10 = 1000g CaCO3

(e) the neutralisation reaction is MgO + H2SO4 ==> MgSO4 + H2O,

moles H2SO4 = 1000 x 2 = 2000 mol acid, 2000 mol MgO needed (1:1 in equation),

mass MgO needed = 2000 x 40.3 = 80600 g or 80.6 kg

 

Q7 ANSWERS (a) 2NaOH(aq) + H2SO4(aq) ==> Na2SO4(aq) + 2H2O(l)

(b) moles of sodium hydroxide used in the titration = 25.0 x 1/1000 = 0.025 mol NaOH

(c) mol H2SO4 = mol NaOH 2 = 0.0125 mol in 20.0 cm3,

so scaling up to 1000 cm3 to get molarity of diluted acid = 0.0125 x 1000 20 = 0.625 mol dm-3

(or molarity = 0.0125 mol/0.02 dm3 = 0.625 mol dm-3)

(d) scaling up from 50 to 1000 cm3, gives the concentration of the original concentrated sulphuric acid solution,

 = 0.625 x 1000 50 = 12.5 mol dm-3

 

Q8 ANSWERS (a) NaHCO3(s) + HCl(aq) ==> NaCl(aq) + H2O(l) + CO2(g)

(b) mol = molarity x volume in dm3, mol acid = 0.200 x 23.75/1000 = 4.75 x 10-3 mol HCl

from equation HCl:NaHCO3 is 1:1 by ratio

so mol HCl = mol NaHCO3 = 4.75 x 10-3

(c) mass = mol x formula mass, f. mass NaHCO3 = 23 + 1 + 12 + (3 x 16) = 84

mass NaHCO3 = 4.75 x 10-3 x 840.399 g

% purity of NaHCO3 = 0.399 x 100/0.4099.75% (99.8%, 3sf)

(d)(i) mol = molarity x volume in dm3, mol acid = 0.200 x 20.00/1000 = 4.00 x 10-3 mol HCl

from equation above HCl : MHCO3 is 1 : 1 by ratio

so mol HCl = mol MHCO3 = 4.00 x 10-3

(ii) to get to the relative formula mass

moles = mass / MrMr = mass / mol = 0.400 / 4.00 x 10-3 = 100

(iii) for MHCO3, Mr(HCO3) = 1 + 12 + 48 = 61, Ar (M) = 100 - 61 = 39,

which, from the periodic table relative atomic mass values, corresponds to potassium

So the formula of the group 1 hydrogencarbonate titrated was KHCO3

 

Q9 ANSWERS (a)(i)  Na2CO3(aq) + 2HCl(aq) ==> 2NaCl(aq) + H2O(l) + CO2(g)

9a(ii) 20.0 cm3 of 1.0 mol dm-3 hydrochloric acid contains 1.0 x 20.0/1000 = 0.02 mol HCl

From the equation, 0.020 mol HCl reacts with 0.010 mol Na2CO3, Mr(Na2CO3) = 106

therefore mass Na2CO3 titrated = 0.01 x 106 = 1.06 g per aliquot,

since 250 cm3 is 1/10th of the aliquot, 10 x 1.06 = 10.6 g of Na2CO3 would be used to make up the solution.

Molarity of Na2CO3(aq) = (10.6 g/106 g mol-1)/0.25 dm3 = 0.40 mol dm-3

9(b)(i) CH3COOC6H4COOH + NaOH ==> CH3COOC6H4COO-Na+ + H2O

9b(ii) 23.0 cm3 of 0.100 mol dm-3 NaOH contains 0.100 x 23.0/1000 = 0.0023 mol NaOH

From the equation, mol Aspirin = mol NaOH, Mr(CH3COOC6H4COOH) = 180

so need Aspirin mass of 0.0023 x 180 = 0.414 g

9b(iii) The last stage in the synthesis of 2-ethanoylhydroxybenzoic acid ('Aspirin') is made by esterifying 2-hydroxybenzoic acid with ethanoic anhydride.

Mr(HOC6H4COOH) = 138, is 42 units less than aspirin. In terms of this particular impurity the % aspirin will be overestimated for the following reason. The 2-hydroxybenzoic acid will also be titrated with the aspirin, and, with its smaller molecular mass, it will need more alkali to neutralise than aspirin per equivalent mass of material. This can result in >100% purity!!!!

9(c)(i) It is a ligand substitution/replacement reaction.

[Ca(H2O)6]2+(aq) + EDTA4-(aq) ==> [CaEDTA]2-(aq) + 6H2O(l)

more simply Ca2+(aq) + EDTA4-(aq) ==> [CaEDTA]2-(aq)

or [Ca(H2O)6]2+(aq) + H2EDTA2-(aq) ==> [CaEDTA]2-(aq) + 2H+(aq) + 6H2O(l)

more simply Ca2+(aq) + H2EDTA2-(aq) ==> [CaEDTA]2-(aq) + 2H+(aq)

9(c)(ii) Mr(CaCO3) = 100, mol CaCO3 = mol Ca2+ in solution = 0.250/100 = 0.00250 mol

since 250 cm3 = 0.25 dm3, molarity Ca2+ = 0.0025/0.25 = 0.010 mol dm-3

9(c)(iii) mole CaCO3 = mol Ca2+ = mol EDTA used in titration.

Therefore from c(ii) mol Ca2+ = mol EDTA = 0.01 x 25.0/1000 = 0.00025 mol in 25.70 cm3 (0.0257 dm3) EDTA solution,

so molarity EDTA = 0.00025/0.0257 = 0.00973 mol dm-3 (3 sf)

9(c)(iv) Mr(apatite) = (5 x 40) + 3 x (31 + 4 x 16) + (16 + 1) = 502

% Ca in apatite = 200 x 100/502 = 39.8%

9(c)(v) In the titration mol Ca2+ = mol EDTA,

therefore mol Ca2+ = 22.5 x 0.0200/1000 = 0.00045,

since 10/250 of the original solution was used in the titration,

the total mol of calcium in the tooth solution = 0.00045 x 250/10 = 0.01125 mol Ca

total mass of Ca in tooth = 0.01125 x 40 =  0.45 g

% by mass Ca in the tooth = 0.45 x 100/1.40 = 32.1 %

 

Q10 ANSWERS (a) Ag+(aq) + Cl-(aq) ==> AgCl(s) (sodium ions and nitrate ions etc. are spectator ions)

(b) from equation: moles silver nitrate (AgNO3) = moles chloride ion (Cl-)

moles = molarity AgNO3 x volume of AgNO3 in dm3

= 0.100 x 13.8/1000 = 1.38 x 10-3 mol Cl- (in 25.0 cm3 aliquot)

(c) moles in 1 dm3 of diluted seawater = 1.38 x 10-3 x 1000/25 = 0.0552 (scaling up to 1000 cm3)

So molarity of chloride in diluted seawater is 0.0552 mol dm-3

(d) Now in the titration 25.0 cm3 of the 250 cm3 was used,

so the molarity of chloride ion in the original seawater must be scaled up accordingly.

molarity of chloride ion in seawater = 0.0552 x 250/25.0 = 0.552 mol dm-3

(e) Mr(NaCl) = 23 + 35.5 = 58.5

concentration of NaCl in g dm-3 = molarity x formula mass = 0.552 x 58.532.3 g dm-3

 

Q11 ANSWERS (a) moles = molarity AgNO3 x volume in dm3 = 0.100 x 19.7/1000 = 1.97 x 10-3 mol Cl- ion

[AgNO3:NaCl or Ag+ : Cl- is 1 : 1, see Q10(a)/(b)] f. mass NaCl = 23 + 35.5 = 58.5

(b) mass = mol x formula mass = 1.97 x 10-3 x 58.5 = 0.1152 g NaCl

(c) % purity = 0.1152 x 100/0.12 = 96.0 % in terms of NaCl (3sf)

 

Q12 ANSWERS (a) mole Cl- = moles Ag+ [=AgNO3, see Q10(a)/(b)]

mole Cl- = molarity AgNO3 x vol AgNO3 = 0.100 x 21.2/1000 = 2.12 x 10-3 mol Cl-

(b) Since calcium chloride is CaCl2, mol CaCl2 = mole Cl-/2  = 2.12 x 10-3/2 = 1.06 x 10-3 mol CaCl2

(c) Mr(CaCl2) = 40 + (2 x 35.5) = 111

mass = mol x f. mass = 1.06 x 10-3 x 111 = 0.1177 g CaCl2

(d) Since 1/10th of original solution titrated, original mass of CaCl2 in mixture

= 10 x 0.1177 g = 1.177g CaCl2 (1.78g 3sf)

(e) Therefore % = 1.177 x 100/5.0 =  23.5% CaCl2 (3 sf)

and % NaNO3 = 100 - 23.5  = 76.5% (3 sf)

 

Q13 ANSWERS (a) moles = mass/f. mass, f. mass Na2CO3 = 106, mol Na2CO3 = 13.25/106 = 0.125

molarity = mol/vol. in dm3, 250 cm3 = 0.250 dm3,

molarity Na2CO3 = 0.125/0.250 = 0.50 mol dm-3

(b) Na2CO3(aq) + 2HCl(aq) ==> 2NaCl(aq) + H2O(l) + CO2(g)

(c) mol = molarity x volume

mol Na2CO3 titrated = 0.5 x 25.0/1000 = 0.0125 mol Na2CO3 (in the 25 cm3 aliquot pipetted)

(d) from equation, mole ratio Na2CO3:HCl is 1:2,

so mol HCl = 2 x mol Na2CO3  = 2 x 0.0125 = 0.025 mol HCl (in the 24.65 cm3 titre)

(e) molarity = mol/vol. in dm3, dm3 = cm3/1000, 24.65/1000 = 0.02465 dm3

therefore molarity HCl = 0.025/0.02465 = 1.014 mol dm-3 (1.01 3sf)

 

Q14 ANSWERS (a) mol HCl = 1.00 x 25.3/1000 = 0.0253 mol

(b) reacting mole ratio, Na2CO3:HCl is 1:2

so mol Na2CO3 titrated = mol HCl/2 = 0.0253/2 = 0.01265 mol

(c) mass = mol x f. mass, so mass Na2CO3 = 0.01275 x 106 = 1.34 g

therefore % purity of Na2CO3 =  1.34 x 100/1.3599.3% (3sf)

 

Q15 ANSWERS (a) An appropriate quantity of the acid is weighed out, preferably on a 4 sf electronic balance. It can be weighed into a conical flask by difference i.e. weight acid added to flask = (weight of boat + acid) - (weight of boat).  The acid is dissolved in water, or aqueous-ethanol if not very soluble in water. The solution is titrated with standard sodium hydroxide solution using phenolphthalein indicator until the first permanent pink.

The burette should be rinsed with the sodium hydroxide solution first. During the titration the flask should be rinsed around the inside to ensure all the acid and alkali react, and drop-wise addition close to the end-point to get it to the nearest drop - the first permanent pink colour.

The pKind for phenolphthalein is 9.3, and its effective pH range is 8.3 to 10.0. The pH of a solution of the sodium salt of the acid (from strong base + weak acid) is in this region and so the equivalence point can be detected with this indicator.

(b) moles = molarity x volume in dm3 (dm3 = cm3/1000)

mol NaOH = 0.100 x 20.5/1000 = 0.00205 mol

(c) mol NaOH = mol RCOOH = 0.00205

because 1:1 mole ratio for a monobasic acid: RCOOH + Na+OH- ==> RCOO-Na+ + H2O

(d) moles = mass (g)/Mr, so Mr = mass/mol = 0.279/0.00205 = 136.1

(e) The simplest aromatic carboxylic acid is benzoic acid C6H5COOH, Mr = 122

136-122 = 14, which suggests an 'extra' CH2 (i.e. -CH3 attached to the benzene ring instead of a H), so, since the COOH is attached to the ring, there are three possible positional/chain isomers of CH3C6H4COOH (Mr = 136)

(c) doc b(c) doc b(c) doc b2-, 3- or 4-methylbenzoic acid.

 

Q16 ANSWERS (a) moles = molarity x volume in dm3 (dm3 = cm3/1000)

mol NaOH = 0.0995 x 19.85/1000 = 0.001975 mol

(b) The titration reaction for complete neutralisation is:

R(COOH)2 + 2Na+OH- ==> R(COO-Na+)2 + 2H2O

this 1 : 2 reaction mole ratio means that mol dibasic acid = mol NaOH/2 = 0.001975/2 = 0.0009875

(c) moles = mass (g)/Mr, so Mr = mass/mol = 0.103/0.0009875 = 104.3 (approx. 104 3sf)

(d) Since a dibasic acid, and 2 x COOH = 2 x 45 = 90 mass units, the remaining 14 units could be CH2,

and so the structure is likely to be HOOC-CH2-COOH,

propanedioic acid (malonic acid), Mr = 144

 

Q17 ANSWERS (a) mol NaOH = 0.100 x 19.25/1000 = 0.001925

(b) mol NaOH = mol acid = 0.001925, mol ratio 1:1, C6H5COOH + Na+OH- ==> C6H5COO-Na+ + H2O

Mr (C6H5COOH) =122, mol = mass (g)/Mr or mass = mol x Mr

so mass acid = 0.001925 x 122 = 0.2349 g

(c) % purity = actual mass of acid titrated x 100 / mass of original sample

% purity = 0.2349 x 100/0.236 = 99.5% (3 sf)

 

Q18 ANSWERS (a) Mr (C6H5COOH) =122, mol acid = mass (g)/Mr = 0.250/122 = 0.002049 mol

(b) mol alkali = 0.002049 mol, since mol acid, 1:1 mole ratio in reaction (see Q17(b)).

(c) Since 0.002049 mol NaOH in 22.5 cm3, so scaling up

molarity NaOH = 0.002049 x 1000/22.5 =  0.0911 mol dm-3 (3 sf)

 

Q19 ANSWERS (a) The ratio of 2M/0.1M is 20, so need to do 20 fold dilution.

So 25.0 cm3 of 2M diluted to 500 cm3 gives an approximately 0.1 mol dm-3 solution. You can do this with a measuring cylinder and beaker.

In practice you could make up 12.5 cm3 of the 2M acid up to 250 cm3. You could do this with a burette and a 250 cm3 standard volumetric flask, but standardisation of the acid is still required.

(b) The solubility of calcium hydroxide is low, so it would give a very inaccurate tiny titration value with relatively concentrated acid. For any accurate work e.g. to 3sf, standardisation of reagents is required.

(c) 1 x 50.0 cm3 pipette or 2 x 25.0 cm3 pipette would be the most convenient (or accurate burette?). Phenolphthalein is used for strong base-strong acid titrations.

(d) Ca(OH)2(aq) + 2HCl(aq) ==> CaCl2(aq) + 2H2O(l)

(e) from equation, mole ratio Ca(OH)2 : HCl is 1:2

and since moles solute = molarity x volume in dm3

mol HCl used in titration = 0.1005 x 15.22/1000 = 0.001530 mol HCl

therefore mol Ca(OH)2 = 0.001558/2 = 0.000765 mol Ca(OH)2

(f) Scaling up from mol Ca(OH)2 in 50 cm3 to 1 dm3 (1000 cm3)

molarity Ca(OH)2 = 0.00765 x 1000/50 = 0.1558 = 0.153 mol dm-3 (to 3sf)

(g) Mr[Ca(OH)2] =74, so solubility in g dm-3 = 0.153 x 74 = 1.13 g dm-3 (3 sf)

Since density of water is ~1.0 g cm-3, the solubility of Ca(OH)2 is about 0.113 g/100 cm3 H2O

 

Q20 ANSWERS (a) (c) doc b + CO2 === heat/pressure/NaOH ==> (c) doc b

(sodium salts status ignored)

(b) (c) doc b + (c) doc b == reflux ==> + (c) doc b

(c) 2-hydroxybenzoic acid Mr(C7H6O3) = 138.1, aspirin Mr(C9H8O4) = 180.3

(d) Aspirin is not very soluble in pure water and is much more soluble in ethanol solvent.

(e) The titre/mass data: Note: If you use average mass/average titre you get 55.73 cm3 0.1000M NaOH/g

Note that Q9b explains why 0.35-0.40g is a convenient mass to weigh out for an individual titration.

mass of aspirin (g) titration/cm3 of 0.1M NaOH titre/mass
0.3591 20.05 55.83
0.3532 19.65 55.63
0.3686 20.60 55.89
0.3583 19.90 55.54
0.3635 20.25 55.71
av titre/mass =  55.72 cm3/g

(f) (i) 55.72 cm3

(ii) + NaOH ==> + H2O

(iii) From the titration equation, 1 mole of aspirin = 1 mole of NaOH

therefore mol aspirin = mol NaOH = 55.72 x 0.1000 / 1000 = 0.005572 mol (mol = molarity x volume in dm3)

(iv) mass = Mr(aspirin) x 0.005572 = 180.3 x 0.005572 = 1.00463g

(v) % purity = 100 x g mass titrated/ 1g = 100.5% (1dp, 4sf)

(g) You need to re-read the opening paragraph of the question, particularly the last line. The problem with this titration is that you also titrate the 2-hydroxybenzoic acid impurity. This has a smaller molecular mass than aspirin, therefore for equivalent masses, the 2-hydroxybenzoic acid will give more moles to react with the NaOH and so give a false, and over estimated value of moles of aspirin, hence a % purity of >100.

(h) (i) The moles of acid = 0.005572 for 1g, since Mr = mol/mass, Mr(av) = 1/0.005572 = 179.5

(ii) Mr(av) = 179.5 = {x X 138.1(C7H6O3) + (100-x) X 180.3(C9H8O4)} / 100

where x = the % of 2-hydroxybenzoic acid, and multiplying through by 100 gives ...

17950 = 138.1x + 18030 - 180.3x

which on rearranging gives ...

42.2x = 80, x = 80/42.2 = 1.9

therefore the aspirin contained 1.9% impurity of 2-hydroxybenzoic acid.

(i) (i) % error = ~ 0.1 x 100 /20 = 0.5% (this would be good for any student!)

(Other sources of error are ignored for the purpose of this calculation)

(ii) The error range would be 179.5 0.5% of 179.5, 179.50.9, giving a range of 178.6 to 180.4.

(iii) Using an Mr(av) of 178.6 gives 4.0% of 2-hydroxybenzoic acid, 180.4 gives -0.24% !

(iv) This means, for a not unreasonable error range based on the titration, that a wide range of values of the % impurity in the aspirin, including the possibility of a -ve percentage. One must conclude that this is not a very accurate method for small percentages of this particular acidic impurity. It seems at first that the 0.5% error seems ok, but it isn't in reality, the error in the method is comparable to the actual % of the impurity!

(v) For low % impurity you need, if possible, a method that directly, as well as accurately measures the impurity. In this case you can use colorimetry, e.g. measure the absorbance of the colour produced by reacting the 2-hydroxybenzoic acid with iron(III) chloride solution. This gives a purple colour and is a test for the phenol group. (I haven't written this up yet!, but I have described the use of colorimetry in determining the formula of a transition metal complex which outlines the essential principles.)


PART 1 Questions * PART 2 Questions * PART 2 Question Answers *  Redox Titration Q's

I DO MY BEST TO CHECK MY CALCULATIONS, as you yourself should do,  BUT I AM HUMAN! AND IF YOU THINK THERE IS A 'TYPO' or CALCULATION ERROR PLEASE EMAIL ME ASAP TO SORT IT OUT!

ILLUSTRATIONS OF ACID-ALKALI TITRATIONS and SIMPLE STARTER CALCULATIONS


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