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

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

Method 1: Write a balanced equation NaOH + HCl NaCl + H2O Write down what you know NaOH HCl Conc 0.2moll-1 unknown Vol 15.5cm3 = 0.0155l 25cm3= 0.025l No of moles 1 1

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 0.0155 x 1 = 0.124moll-1 0.025 x 1

Method 2: Write down what you know NaOH HCl Conc 0.2moll-1 unknown Vol 15.5cm3 = 0.0155l 25cm3= 0.025l Calculate the number of moles of NaOH using n = C x V n = 0.2 x 0.0155 =0.0031moles

Write a balanced equation and establish the mole ratio NaOH + HCl NaCl + H2O 1 mole 1 mole 0.0031moles 0.0031moles Now calculate the concentration of the acid using C = n ÷ V = 0.0031 ÷ 0.025 =0.124moll-1

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

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.

Calculations MnO4- + 8H+ + 5Fe2+ Mn2+ + 5Fe3+ + 4H2O 1mole 5 moles nMnO4- = C x V = 0.02 x 0.024 = 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

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

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

EDTA

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

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

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. Ca2+ + H2Y2- CaY2- + 2H+ Conc:-unknown Vol:25cm3 n=1 Conc:- 0.05 Vol:20cm3 n=1 CCa2+ x VCa2+ CEDTA X VEDTA nCa2+ nEDTA =

CCa2+ CEDTA X VEDTA VCa2+ = CCa2+ 0.05 X 0.02 0.025 = = 0.04moll-1

= 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 =

Method 2: Write a balanced equation Ca2+ + H2Y2- CaY2- + 2H+ Write down what you know Ca2+ H2Y2- Vol 25cm3 = 0.025l 20cm3 = 0.020l Conc unknown 0.05moll-1 Calculate the number of moles of EDTA using n = C x V =0.05 x 0.02 =0.001moles

Look at the molar ratio and complete the calculation Ca2+ + H2Y2- CaY2- + 2H+ 1 mole 1 mole 0.001moles 0.001moles C = n ÷ V = 0.001 ÷ 0.025 = 0.04moll-1 In general at this stage you can assume an EDTA:metal ion ratio of 1:1

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.

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

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