Back-titration for Calculating Reagent Qtys for 7g/dm³ Cl (aq)

[briton]

Im writing up a plan to find the concentration of saturated aq chlorine - which has been given to me as an approx. concentration of 7 g/dm³. I need to use this value to calculate what suitable quantities of reagents to use.

first stage: solution of chlorine is added to an excess of a sutiable reducing reagent, eg an iron (II) compound.

second stage: amount of reducing agent left over (as it was excess) is determined by a titration (the concentration of Cl ions formed in stage1 is too low to affect titration.)

so for stage one:
I'm thinking it would be something like iron sulphate (reducing agent)?
is the equation right? $$FeSO_4 \ + \ 2Cl^- \ --> \ FeCl_2 \ + SO_4^{2-}$$

then the Fe created goes into stage 2: -->although I'm not sure in what state?? is it just as Fe2+ ?
where I'll use manganate as the oxidising agent for the actual titration..
$$MnO_4^- \ + \ 8H^+ \ + \ 5Fe^{2+} \ --> \ Mn^{2+} \ + \ 4H_2O \ + \ Fe^{3+}$$

assuming that's correct, I need to use the [Cl] of 7g/dm^3:
I thought it would be something like this...
conc. [Cl] = mass/Mr
= 7/35.5 = 0.197 mol/dm³

molar ratio of $$Cl^-$$: $$FeCl_2$$ = 2:1
so mol of FeCl_2 = 0.1972/2 = 0.0986 mol. ... this is transferred to

stage2:
so mol of Fe2+ = 0.0985 mol.

ratio of Fe2+ : MnO4 = 5:1 so
mol of MnO4 = 0.0986/5
= 0.01972 mol.

I would ideally want a 25cm³ (0.025dm³) titre of MnO4..
so concentration of [MnO4] = mol/vol
= 0.01972/0.025 = 0.789 mol/dm³


now that final conc of manganate is apparently (according to my calculations) what I should ideally use (to get a decent titre of about 25cm³), assuming the conc of Cl (aq) is approx 7g/dm³. However 0.789 mol/dm³ sounds much too high.

where have I gone wrong??!

 

[Gokul43201]

Stage one is incorrect, and hence the rest of the plan fails too. Look at stage one again and tell us if FeSO4 is really serving as a reducing agent (as you claim) ? But, anyway, why would you want to reduce Cl- (if it were even possible) ?

Start by asking yourself what the nature of a saturated aqueous solution of chlorine is ? Is it molecular chlorine dissolved in water ? Or is there a chemical reaction leading to the formation of ions ?

 

[briton]

Thanks for your reply. I have to follow the stage 1 and stage 2 method (hence reduce the solution of chlorine).

I always thought saturation (in this context) was just dissolving? Ie here the chlorine ions are dissolved in water (ie at 7 g per litre)?

 

[Gokul43201]

I always thought saturation (in this context) was just dissolving? Ie here the chlorine ions are dissolved in water (ie at 7 g per litre)?

If chlorine gas is "just dissolving" in water, where do the ions come from ? In reality, there will be some Cl- and OCl- ions, but these will be small enough to neglect. Most of the chlorine in solution is in the form of Cl2.

Now rewrite your stage 1 reaction with Cl2 in the LHS getting reduced to Cl- in the RHS. As a result, Fe(II) will be oxidized to Fe(III). If I were you, I'd use a better reducing agent, like thiosulphate, but that's just my opinion. Besides, FAS is pretty commonly used, so you will probably be okay.

 

[Gokul43201]

Also, the idea with back titration is that you use an excess of Fe(II) and titrate what's left of this excess against KMnO4. You are not titrating against the product of the first stage, but the unreacted excess.

 

[briton]

Also, the idea with back titration is that you use an excess of Fe(II) and titrate what's left of this excess against KMnO4. You are not titrating against the product of the first stage, but the unreacted excess.

Ok, so using thiosulphate would it be:

$$2FeS_2O_3 \ + \ Cl_2 \ + \ -> \ 2Fe^{3+} \ + \ 2Cl^- \ + \ 2S_2O_3^{2-}$$

Using this, assuming that I want a titre of MnO4 (concentration say 0.020 mol/dm³) of say 0.015 dm³, then I calculate the mass of $$FeS_2O_3$$ to be about 33 grams... have I gone wrong again, or is it meant to be this high?
Or unless of course I compensate the titre value, and assume I want the MnO4 titre to be say ~ 2 cm³, then my FeS2O3 mass needed would decrease to something more suitable?

 

[Gokul43201]

I'm sorry I brought up thio. Perhaps I shouldn't have. It's only misled you. (In thiosulphate, the S-atom sees the change in oxidation state, but that's a different story)

Let's stick with FeSO4.

$$2Fe^{2+} + Cl_2 \longrightarrow 2Fe^{3+} + 2Cl^-$$

But since you used an excess of FeSO4, you will have some left over Fe(II) ions in solution. These can be titrated against KMnO4.

Give it another shot.

 

[briton]

I'm sorry I brought up thio. Perhaps I shouldn't have. It's only misled you. (In thiosulphate, the S-atom sees the change in oxidation state, but that's a different story)

Let's stick with FeSO4.

$$2Fe^{2+} + Cl_2 \longrightarrow 2Fe^{3+} + 2Cl^-$$

But since you used an excess of FeSO4, you will have some left over Fe(II) ions in solution. These can be titrated against KMnO4.

Give it another shot.

Thanks, I think I've got it.

here I'm trying to work out the mass of the FeSO4 required for a decent titration:

mol of Cl2 = 7/(35.5*2)
= 0.09859

mol ratio Fe2+ : Cl2 = 2:1
so mol of Fe{2+} which reacts = 0.09859*2
= 0.1972

let total mol of Fe{2+} added to Cl2 = 0.1972 + x

where x = excess amount.

x goes into stage2.
mol of MnO4- = x/5
assuming I want 0.015dm^3 titre (of 0.02M MnO4-) then
as mol = conc * vol
x/5 = 0.02 * 0.015
x = 0.0015 mol

so total mol of FeSO4 = 0.0015 + 0.1972 = 0.1987

but I think the FeSO4 supplied is something like FeSO4.7H2O
so mass I will need = mol * Mr(FeSO4.7H2O)

which comes out to about 55.22 g.
but this is for a litre, and I'll make a standard solution of FeSO4 in a 250cm^3 flask so mass = 55.22*0.25 = 13.81 g

 

[Gokul43201]

I think you've got it. A few comments :

1. You are given the strength (g/L) of the chlorine solution. From this you get the number of moles of Cl2. But for this, you must assume some volume of solution. In your calculation, this volume is 1.0 L.

2. If you really meant to write M (or mol/L or mol/dm3) instead of mol, then your calculation is perfect...just change the mol to M and you are good.