[James_Bond]

Ok, heres the problem mates.
I know it seems to be an easy experiment, but i am stuck.

We are doing a Titration lab, consisting of NaOH and an unknown acid (liquid)
Now the problem is, I know what data I need to collect during the experiment but have no idea on where to start with the data.

So we are given a solid form NaOH, now assume i got all the data during the lab. What must I do to find the Molarity of the Acid.
if You want, to show me how its done you can make up the data to show, thanks

Please Help this is due tomorrow.

Thanks
Mr.Bond in need of Fast help

[enahs]

You took that solid form of NaOH and made a solution, correct?
What is the molarity of that solution?

You were probably told the acid was monoprotic (or if not, what it was).

So, how many mols of NaOH was required to neutralize the acid?
If you know the number of mols in the original solution and its volume you know its molarity (mol/volume).

[James_Bond]

yeh, thanks you helped out a bit.

alright, well heres what im thinking.

NaOH dissolved in water, so molarity= mole of NaOH used while in solid/Volume of water Use= 2.13599 mol/L
(we used 5.15 grams of NaOH solid, in 60 ML of water, thus giving us mol of 0.1288 mol)

then if 6.7 ml of NaOH used and 7 ml Of acid used
i would say the molarity is pretty close and a little bit above, since i looked at the ratio

I got 2.054.

Im pretty sure thats how its done, not totally sure.

[Borek]

NaOH dissolved in water, so molarity= mole of NaOH used while in solid/Volume of water Use= 2.13599 mol/L
(we used 5.15 grams of NaOH solid, in 60 ML of water, thus giving us mol of 0.1288 mol)

Check your numbers.

then if 6.7 ml of NaOH used and 7 ml Of acid used
i would say the molarity is pretty close and a little bit above, since i looked at the ratio

Something like that, but there is no nedd to guesstimate - if you start with the balanced reaction equation calculations are exact. As is your final result.