# Year 11- Titraiton Calculations

Discussion in 'Science' started by thattallteacher, Apr 29, 2012.

1. ### thattallteacherNew commenter

With my year 11s we have 2 hours of lesson time left on Chemistry before they go on study leave.
I have tried to do titration calculations with them for 2 lessons and all im seeing is glazed over eyes and faces.
Has anyone got any tips/hints for teaching Titration calculations?

2. ### l_lblack

make sure they can do n=m/Mr and n=CxV. they chould be ok if they can do those and remember the difference between cm3 and dm3 and conversions thereof.

3. ### leftieMNew commenter

When i taught it I got them to focus on the data in the question and let that lead them to the right equation to use eg if you see mass, use n=m/rmm, if you see a volume of gas use n=v/24 and if you see a volume of solution use n=cv.
Very few of them get the concept of the mole so i would just teach the maths and teach them to identify what formula to use and keep the conceptual teaching very simple.

I try to encourage them to draw a table of concentration and volume of the acid and alkali. There should be one unknown in the table which should be the answer to the question. I find it's a useful way of extracting the important data from the question which simplifies things a little.

I find the glazing over part hard to get past too. One of those lessons where I feel as much a maths teacher as a science teacher!

5. ### thattallteacherNew commenter

I really like the idea of putting it in a table. It has deffinately helped me as a biologist to understand what is going on myself. going to try this out tomorrow and I will let you know how it goes

6. ### Beta_1

Thought of trying the C1V1=C2V2 approach.

Often they can get the answer (even if they don't quite know how)

7. ### weary willy

I tend to steer clear of formulae .... better to have a (high level) understanding of what is happening:
1. Write the neutralisation formula
2. Balance it
3. How much stuff have you got? (strength x volume)
4. How much of the other stuff will you need to neutralise it?
5. So do a simple sum
6. Does it feel right? E.g. if it's a 1:1 neutralisation (eg NaOH + HCl) and does a slightly stronger conc require a lower volume?

Makes you long for the discipline of slide rules... does the number feel right? Work out yourself where the decimal place should be

8. ### ThejumpingjewNew commenter

I'm having a similar problem with some students - They can all do it with thought in lessons, but not in the exam.

I have decided that next week will be Calculation week. All pupils will see me every lunchtime to do two calculation problems - Titration and Empirical from combustion data - before going to lunch (They will do it as they want to improve) - i'm hoping small repitition over a long term will help - I will then run an after school on Friday to mop up, and present a little award for encouragement and achievement over the week.

It always amazes me the fact that pupils ignore the power or repetition.

9. ### meaghersclasses

Chemistry calculations are no mystery, as they always follow the same format. iI wholeheartedly agree that students need to understand what they are doing, rather than slavishly following some magical formula, However, if the following steps are always followed then the students understand why they are performing the calculation, and are given a scaffold as to what they need do at each step of the procedure. ie

STEP 1 FIND THE NUMBER OF MOLES OF MOLES OF WHATEVER YOU CAN from whatever information that you have (via n=mM or n=cv or PV=nRT or n(e-)=It/(9.65x10^4) etc)

STEP 2 FIND A RATIO between the number of moles that you have just calculated and the number of moles of the substance that you want to know about. (in this case the ratio is the equation's stoichiometric ratio for the reaction between the acid and base or if a redox titration then between the oxidising agent and reducing agent etc)

STEP 3 Use this ratio to DETERMINE THE NUMBER OF MOLES OF THE UNKNOWN substance. (and yes you could use c1v1=c2v2 in a titration calculation - if the stoichiometric ratio is 1:1, since c(acid)v(acid)=c(base)v(base) is really another expression of n(acid)=n(base) since n=cv)

STEP 4 Use the number of moles determined in the previous step to SOLVE THE PROBLEM. (In the case of a titration this would be to determine the concentration of the solution using c=n/v)

Application of these four steps apply to any chemistry calculation that I throw up to my students, and having a set procedure means that they always know at least where to begin ie FIND THE NUMBER OF MOLES OF WHATEVER YOU CAN.

I must be a bit picky with terminology, weary *****, but above where you state: 3. How much stuff have you got? (strength x volume) is not strictly correct it should read 3. How much stuff have you got? (concentration x volume) since concentration and strength are two completely different concepts. Strength refers to the degree to which an acid or base will ionise or dissociate in solution whereas concentration refers to the number of moles per litre or some other measurement such as g/L etc. Eg hydrochloric acid is a strong acid, no matter its concentration, since it completely ionises in solution, whereas ethanoic (acetic) acid is always a weak acid, no matter its concentration, since it only partially ionises in solution. ie 0.00001mol/L HCl is a solution of a dilute, strong acid and 5mol/L CH3COOH is a solution of a concentrated, weak acid. I would therefore not say a stronger concentration (as in point 6 above), I'd rather use the term higher concentration in order to avoid confusing the students.

10. ### s0910796

I teach Titration Calculations as a "Murder Mystery" theme. It may sound crazy but it works. All the pupils love it and start doing titration calculations and forget about the fact that it also involves some maths.
The whole resource is on TES - Take my user id and check my resources - yoy should find it.
Let me know if you don't and I'll email you the full resource.
Regards
Phil Wootton - Chemistry Teacher Kirckcaldy High School