Acids and Bases MORE WORK TO BE ADDED SOON.

Acid base titration [Please note some equations didn't come up here, hopefully get them later but if you requested a CD with this work then it can be included. The CD are made as you required, I can include any model question and answer from any module or topic, so what do you want in your CD?

Question : Suppose 25.0 cm 3 of NaOH was neutralized by 23.5 cm 3 of Hydrochloric acid of concentration 0.1 mol dm -3 . Find the concentration of the NaOH sloution.

Method: First write down the equation for the reaction.

NaOH(aq) + HCl(aq) ---- >NaCl(aq) + H 2 O

The above equation suggests that the reaction is a 1:1 ratio hence by working out the number of moles of HCl, the number of moles of NaOH can be found out. The concentration of the acid means how many moles of it is present in 1000 cm 3 of the solution, ie in 1000 cm 3 of the solution there are 0.1 moles hence 23.5 cm 3 of the solution contains,

0.1 x 23.5/1000 = 2.35 x 10 -3 mol

The number of moles of NaOH = The number of moles of HCl

Thus the number of moles of NaOH = 2.35 x 10 -3 mol

This number of moles of NaOH is present in 25.0 cm 3 of the solution hence the number of moles in 1000 cm 3 is, 2.35 x 10 -3 x 1000/25

2.35/25 = 0.094 mol dm -3

Mass/Relative atomic mass = Number of moles [The diagram is missing]

Mass = Relative atomic mass x Number of moles.

Let the mass of sodium present in the solution be x.

x/23 = 2.35 x 10 -3 , x = 23 x 2.35 x 10 -3

,, ,, x = 54.05 x 10 -3 ie = 0.05405

Let the mass of OH be y

y/17 = 2.35 x 10 -3, y = 17 x 2.35 x 10 -3

,, ,, y = 39.95x 10 -3 ie = 0.03995

Now add X+Y = 0.05405 + 0.03995 = 0.094 moles, this is the total number of moles present in NaOH.

More worked examples to be added by reference to GCSE past papers.

Useful hints to remember in volumetric calculations

No: of moles =(Volume/1000) x conc:

No: of moles = Volume x 10 -3 x conc:

Conc: = No: of moles / Volume x 10 -3

Weight =Relative molar mass x No: moles

No: moles = Weight/Mr (Relative molar mass)

Weight of Na = 23 x No: of moles of sodium.

Weight of Cl (Chloride) = 35.5. x No: of moles

Weight of Cl 2 (Chlorine) = 71 x No: of moles

Weight of Iron (Fe) = 55 x No: of moles

Weight of Iodine = 2 x 127 x No: of moles

Weight of HCl = 36.5 x No: of moles

Weight of Chloride = 35.5 x No: of moles

Weight of Copper = 63.5 x No: of moles

If 25.0 cm 3 of solution contains 1.75 x 10 -3 moles of Fe, calculate concentration per liter. The number of moles per dm 3 means, 25.0 cm 3 contains 1.75 x10 -3 moles, then in 100cm 3 it should be 4 x 1.75 x 10 -3 and per 1000 cm 3 it should be 10 x 4 x 1.75 x 10 -3

To calculate the number of moles of water (Water of crystallisation)

284 + n18 = 508, then n18 = 508 - 284, 18n = 224, n = 12