Titration calculations help PLEASE

[Ocean Breeze]

Can any of you very clever people please help me with these titration calcs. been so long since I have done it i am confused.

I have just done titration in order to standardise Na OH.

I used oxalic acid as the standard, at 0.0500 M/L

My Na OH is approx 0.1 M/L. the tect book says that
2Na OH + H₂ C₂ O₄ ---> Na₂ C₂ O₄ +2H₂0.
These are my averaged results.

20 ml of 0.0500 M/L oxalic acid is neutralised by 10.17 ml of Na OH

From here, how do I get to the molarity of the Na OH from here? and if possible, can you lay out the steps? Cnt remeber what to do about the 2Na OH compared to the single oxalic acid.
Thanks in advance (hopefully!)

[rae]

I think you do
M1V1 = M2V2
So molarity of Oxalic acid x volume used ( in litres) = Molarity of NaOH x volume used

So 0.05x 0.02 = M2 x 0.01012

So the molarity = 0.099
Would that make sense??

[sunray18]

that looks good to me

[Ocean Breeze]

Ok. that gets me out of trouble
So where does the 2Na OH neutralising the 1 oxalic acid come into play at all?

[Jazz]

this is once a year calculation and I strugle to remember. Hope this SWM will help a bit

[DavidPeterson]

2 OH- is neutralising 2H+ (from H₂C₂O₄) therefore mole ratio is 1:1.

Or another way to look at it. 0.05M H₂C₂O₄ is equivalent to 2 x 0.05M = 0.10M H+

[rae]

Its taken in to account by only using approx. 1/2 the volume to neutralise the oxalic acid.

[Ocean Breeze]

many thanks everyone for your assistance.
so 0.99 M/L it is

and I will keep that useful work method sheet for future reference

[DavidPeterson]

Not to be technical, however it is Friday pm and I feel like stirring the pot (it's been a long week and I'm sure it has been for you too)......

I say this in the friendliest way....it's either 0.99M or 0.99mol/L......(not 0.99M/L)

[Ocean Breeze]

Thanks David. That too helps

[lada]

I hate to add the wrong idea to the calculation, but should you not use the 2OH- ?
so
0.05x0.02=2xM2x0.01017
which gives you only 0.049mole/L?
Lada

[lada]

Had a look again, and I think it should be
2x0.05x0.02=M2x0.01017
=0.197M
Anyone can confirm??
Lada

[Ocean Breeze]

eeeeeekkkkkk!
this is getting scary.
I have to have the correct result for the molarity of the Na OH, as after the studetns have done their individual titrations, the teacher will ask me for the precise molarity. (its a titration assessement)

is this becasue of the 2Na OH neutralising with the one of the Oxalic acid?

[lada]

I am so sorry!!! Yes, you have to take in account the 2:1 ratio. I am pretty certain with my LAST answer. Again, sorry about the confusion.
Lada
Where are all the brains when you need them>???

[nickykinz]

I got 0.197M but as it was different to everyone else I thought I had done it wrong! I'll check with our Chem teacher as she is pretty good with this stuff.

[DavidPeterson]

I'm embarrased to say I've had a mental blank.....probably because I'm in long Service Leave mode.....last week until term 3 yay!

[Ocean Breeze]

Thanks for this "brains trust"
I await a definitive answer ..pretty please

[lada]

I am with David. 0.197M NaOH is the right answer.
Lada

[nickykinz]

Ok. Have spoken with chem teacher and she agrees with me and Lada. O.197M.
I'm not very good with formulas as I can never remember them so I will go through how I worked it out. 0.02L of 0.05M oxalic acid means you have 0.001 moles of oxalic acid. Every one mole of oxalic acid reacts with 2 moles of NaOH therefore you used 0.002 moles of NaOH (2 x 0.001).
You know you have 0.002 moles in 0.01017L so you need to work out how many moles in 1L.
(0.002 x 1)/0.01017=0.19665M
Hope this makes sense. My brain hurts now.

[RosalieM]

eeeeeekkkkkk!
this is getting scary.
I have to have the correct result for the molarity of the Na OH, as after the studetns have done their individual titrations, the teacher will ask me for the precise molarity. (its a titration assessement)

is this becasue of the 2Na OH neutralising with the one of the Oxalic acid?

Give the teacher your results and let them do the calculation!