[James_Bond]

Ok, sorry for the last post will be more better this time

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.

now say i used 40 grams of NaOH, and the solution formed with water was 100 ML, how can i find the moles of NaOH, If i have this I think I can do the rest of the lab on my own.

20 ML of NaOH was used to reach the equal point, where H+ ions = OH- ions
meaning 20 ML of NaOH was used to turn the 30 Ml of unknown acid pink under the indicator

If someone can also help me for other calculations, then it would be great

Thanks
Mr.Bond in need of help

[Arkcon]

Great, now you can't write a complete balanced equation without knowing the acid's formula, but can you write a "pretend" one, using some generic name to symbolize acid?  A balanced equation will give you a hint as to what calculations you'll have to do.

[James_Bond]

If you do a titration lab correctly, then u dont need a balanced chemical eqn

if u know the balanced chemical eqn, then u dont even need to do the lab, you have the answer theoretically. but the point of titration is to find a unknown acid or base from.. that is what titration is. i just dont understand the calculation part?

[Borek]

If you do a titration lab correctly, then u dont need a balanced chemical eqn

if u know the balanced chemical eqn, then u dont even need to do the lab, you have the answer theoretically. but the point of titration is to find a unknown acid or base from.. that is what titration is. i just dont understand the calculation part?

I am afraid you are completely missing the point of the whole experiment. Yes: you need a balanced reaction equation. No: just knowing the reaction doesn't mean you don't have to do the experiment.

Do as Arkcon suggests.

[James_Bond]

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

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]

Look at your other thread, I have answered there moment ago.

You understand now why asking the same question in different places makes it hard to follow discussion?