Question on Titration and Back Titration

GoodBoil SG

Hi all, I'd like to seek some confirmation on the concept of both titr and back titr as mentioned above as I usually just go about doing them w/o really understanding the logic.. As Im taking H1 Chem, I cant visualise doing the titr procedures hence I'd need a better understanding pls..

Firstly, when we solve titr questions, usually we have A (with known vol) titrated against B (with known vol) and using A to titrate against C (with known vol and conc).. We will be asked to find conc of A and B.. Usually we will find no of moles of A from the reaction between A and C right? Then we proceed to find the conc of A using the amt of A we find from stoichiometric ratio of the bal eq bet A and C since we can find no of moles of C.. Using the conc of A from the reaction bet A and C, we will multiply this conc with the vol of A (the vol here is the vol of A titrated against B in the earlier titration) to find no of moles of A then we find conc of B from here..

However, what I dont understand is why the conc of A seems to be constant in both titrations against B and C respectively while the amount of A changes? Pls enlighten, thanks! And when does this 'constant vol/ conc/ no of moles' apply? I know no of moles is constant for dilution but not sure about the rest.. Hmm..

GoodBoil SG

Ill come back to back titration later.. Just wanna clear up on titration 1st.. :p

GoodBoil SG

Hi :D

Now I'd use a rather common question to illustrate more clearly the concept that I have not grasped.. Pls bear w me..

A student was given soln of HCl and NaOH with unknown concentration. He carried out titration and found out that 27.50ml of NaOH soln was required to completely react with 100ml of the HCl soln, and used 18.40ml of the NaOH soln to titrate 50.0ml of a 0.0782moldm-3 H2SO4. What are the concentrations of the given HCl and NaOH solns?

I guess the qn which is bugging me most is-

we find out the no of moles of H2SO4 which is 3.91 X 10^-3 moles and from the bal eqn bet H2SO4 and NaOH, we derive no of moles of NaOH is 2X 3.91 X 10^-3= 0.00782moles.. And here is my very troubling issue.... Why is it wrong to find out the concentration of NaOH by using this amt of NaOH- .00782mol divide by (27.50/1000)

dm-3?? And when the qn says 'used 18.40ml of NaOH to titrate H2SO4'.. Are they using 18.40dm-3 from the earlier 27.50ml? Or is that 18.40ml of NaOH a 'fresh' amount from a new source? And what are we really doing when we take 0.00782mol to divide by the volume reacted with HCl (27.50/1000)??

Hope it's not too messy.. ^^

GoodBoil SG

About the concentration and no of moles of NaOH.. Does it mean that the same concentration of NaOH is used in both titration calculation ie when titrating with HCl and titrating with H2SO4 because there is no dilution so no chg in concentration? Then under what circumstances do no of moles of NaOH change? Why do no of moles of NaOH change in both titration? Sorry if it sounds really silly.. :/

GoodBoil SG

Now for back titration.. I think I have just one question about back titration..

One example: 1.60g of a metallic oxide MO is dissolved in 100cm3 of 1moldm-3 HCl. Resulting liquid is made up to 500cm3 with distilled water. 25cm3 of this soln required 21.05cm3 of a 0.1020moldm-3 NaOH for neutralisation.. Calculate

mass of oxide that reacts with 1 mole of HCl

the relative formula mass of oxide an the relative atomic mass of the metal

so....

Here comes the question..

We know that the no of moles of solute does not change in dilution.. Here, it seems that the HCl was diluted with distilled water which made the HCl soln 500cm3.. Why is it that we take vol ratio of HCl to find no of moles of HCl in 500cm3? (500/25)X no of moles of HCl (in 25cm3). if no of moles of solute doesnt change in dilution, is it true to say HCl as a soln is NOT a solute- hence this rule doesnt apply on HCl??

And secondly, is it true to say no of moles of a soln can increase if we increase the vol of solution? Why?

Sorry again for asking so many qns.. :/

Chemfreak022 SG

Q1. 27.50 ml of NaOH reacted COMPLETELT with HCl as stated in your question. Then the 18.40 ml of NaOH is fresh of course.

Why is it wrong to find out the concentration of NaOH by using this amt of NaOH- .00782mol divide by (27.50/1000)

Because the amount of NaOH in 18.40 ml and amt of NaOH in 27.50 ml are different.

GoodBoil SG

Q1. 27.50 ml of NaOH reacted COMPLETELT with HCl as stated in your question. Then the 18.40 ml of NaOH is fresh of course.

Why is it wrong to find out the concentration of NaOH by using this amt of NaOH- .00782mol divide by (27.50/1000)

Because the amount of NaOH in 18.40 ml and amt of NaOH in 27.50 ml are different.

Oh yes, it'd seem logical that the 18.40cm3 is from another source.. Haha, thanks.. Regarding the amount of NaOH.. Is it safe to say the number of moles will only change when there is a change in volume of a solution like when an acid of a smaller volume is made up to a larger volume using distilled water? Chg in volume= chg in no of moles of solute/ soln??

Chemfreak022 SG

If you dilute a solution, no. of mol of solute does not change.

After dilution, if you take out a part of the dilute solution, amt of solute in the small part will change.

What do u mean by change? Increase or decrease?

No. of mol of solute/ soln = conc

Conc. is inversely proportional to dilution.

You add 1 dm3 to 1 dm3 of 1 mol/dm3 of HCl, you end up with 0.5 mol/dm3 of HCl. But amt of HCl remains unchanged = 1 mol.

You take out 10 ml out of 2 dm3. Conc of 10 ml sample = 0.5 mol/dm3.

But amt of HCl in 10 ml = 10/2000 x 1 = 0.005 mol

You take out part of a solution, amt of solute will depend on the volume of sample. Bu conc. of sample will not change from that of the original bulk

GoodBoil SG

Thanks for ur reply, chemfreak.. :)

Just to check with u pls.. Like u said, if u remove a part of the diluted sol, the amt of solute in the 'new solution' will definitely decrease? Is that due to having less number of moles in this smaller, 'new solution' compared to the TOTAL no of moles of the solute in the original solution? Smaller volume of solution from an original solution will always have less no of moles than the original as it contains less of that solute? Thanksss!

Chemfreak022 SG

Yes. Yes. Yes.