1. Fundamentals of Chemistry: Theory and Practice:DH2K 34
Calculations from experiments

2. Volumetric Titrations
NaOH 0.2moll-1
If the average titre of the NaOH is 15.5cm3 calculate the concentration of HCl
25cm3 HCl

3. Method 1: Write a balanced equation NaOH + HCl NaCl + H2O
Write down what you know
NaOH HCl
Conc moll unknown
Vol cm3 = l cm3= 0.025l
No of moles

4. CaVa = CbVb
na nb
Rearrange the equation to get your unknown
Ca = Cb x Vb x na
Va x nb
Complete the calculation
Ca = 0.2 x x 1 = moll-1
0.025 x 1

5. Method 2: Write down what you know NaOH HCl Conc 0.2moll-1 unknown
Vol cm3 = l cm3= 0.025l
Calculate the number of moles of NaOH using
n = C x V
n = 0.2 x
=0.0031moles

6. Write a balanced equation and establish the mole ratio
NaOH + HCl NaCl + H2O
1 mole mole
0.0031moles moles
Now calculate the concentration of the acid using C = n ÷ V = ÷ =0.124moll-1

7. Redox Titrations
Volumetric analysis can be applied to redox reactions. For example the concentration of a reducing agent may be determined by titrating with an oxidising agent of known concentration provided:-
The balanced redox equation is known or can be derived
The volumes of the reactants may be accurately measured.
The end point of the titration can be readily determined

8. Example
Determine the concentration of iron (II) sulphate by titration with a potassium permanganate solution of known concentration.
20cm3 of iron (II) sulphate is placed in a conical flask and is titrated with a 0.02moll-1 solution of acidified potassium permanganate until a permanent purple colour is obtained an average titre of 24.0cm3 was obtained.

9. Calculations MnO4- + 8H+ + 5Fe2+ Mn2+ + 5Fe3+ + 4H2O 1mole 5 moles
nMnO4- = C x V = 0.02 x = 4.8 x10-4
nFe2+ = 5 x 4.8x10-4 = 2.4 x10-2
Conc.Fe2+ = n ÷ V = 2.4 x 10-2 ÷ 0.02 =0.12moll-1

10. Complexometric titrations
In a complexometric titration a metal ion is surrounded by a ligand.
A ligand is a neutral or negatively charged (anionic) species that has unbound electrons that can bind to an ion.
Examples include CN- NH3 and OH-
When a ligand binds to an ion it forms a chelate.
The ion is said to be chelated and forms a complex

11. Complexometric titrations
This technique can be used to determine the concentration of metal ions.
In biochemical analysis the most widely used chelating agent is ethylenediaminetetracetic acid
EDTA

12. EDTA

13. EDTA chelates metal ions
Metal ions are surrounded by 4 oxygen atoms and 2 nitrogen atoms.
EDTA can also form 2- ions
Resulting complex is very soluble due to ‘H’ bonding between water molecules and the outer oxygen atoms within EDTA

14. EDTA chelates metal ions
Using indictor sensitive to the ion under investigation the concentration of the ion can be determined.
EDTA can be used to
Treat lead poisoning
Remove excess calcium from the blood
Determine the hardness of water

15. Example:If 20cm3 ofa 0.05moll-1 EDTA is required to fully chelate a 25cm3 solution of calcium ions . Calculate the concentration of the calcium ions.
Ca H2Y2- CaY H+
Conc:-unknown
Vol:25cm3
n=1
Conc:- 0.05
Vol:20cm3
n=1
CCa2+ x VCa2+ CEDTA X VEDTA
nCa2+ nEDTA =

16. CCa CEDTA X VEDTA
VCa2+ =
CCa X 0.02
0.025 =
= 0.04moll-1

17. = More commonly the equation is shown as C1 V1 C2 V2 n1 n2
General Equation
More commonly the equation is shown as
C1 V1 C2 V2
n1 n2 =

18. Method 2: Write a balanced equation Ca2+ + H2Y2- CaY2- + 2H+
Write down what you know
Ca H2Y2-
Vol 25cm3 = 0.025l cm3 = 0.020l
Conc unknown moll-1
Calculate the number of moles of EDTA using n = C x V =0.05 x 0.02
=0.001moles

19. Look at the molar ratio and complete the calculation
Ca H2Y2- CaY H+
1 mole mole
0.001moles moles
C = n ÷ V = ÷ = 0.04moll-1
In general at this stage you can assume an EDTA:metal ion ratio of 1:1

20. Precipitation Reactions
During a precipitation reaction a solid is formed.
The solid is formed from the solution when it exceeds its solubility product.
The original ions are still present but one or more have formed a solid.

21. Gravimetric Analysis
This is NOT in the assessment but you will carry out a lab like this next term
An analysis of substances by weight
Carried out by converting the chemical into a product that can be easily isolated and weighed
This is often just a matter of driving off water in a hydrated chemical to form the anhydrous salt
Look at the example on p 13 of your booklet

22. Write a balanced chemical equation Look at the molar ratio Add gfm
When excess sodium sulphate was added to 2 litres of water a precipitate of lead sulphate was formed.
Following filtering and drying the precipitate was found to weigh 231.5mg (1g =1000mg)
Calculate the concentration of lead in mgl-1
Write a balanced chemical equation
Look at the molar ratio
Add gfm
Use proportion calculations or triangles to complete the calculation