# Calculating moles

1. Jan 25, 2012

### homevolend

1. The problem statement, all variables and given/known data

Calculate the moles of NaOH in the reaction of NaOH + acetic acid. This is a titration where the NaOH neutralized the acetic acid. It took 0.02 L of NaOH to neutralize 3 ml of acetic acid. I just do not know what to do, have been trying for a few days and can't seem to get it.

2. Relevant equations
Can't use the moles = M x L because I don't know the Molarity.

I am thinking that I have to use some other law but not sure what one...

I did the lab at room temp, not sure if that makes a differnce or not.

thanks, help greatly appreciated.

3. The attempt at a solution

2. Jan 25, 2012

### spaghetti3451

My friend, unless you know the concentration of the sodium hydroxide or the acetic acid that you used in the laboratory titration, there is no way you will be able to calculate the number of moles of NaOH used up in the process.

The room temperature cannot hint at the number of moles as the two quantities are apparently unrelated.

You were right to quote the formula c = n / V, where c is the concentration, n is the number of moles and V is the volume. However, you need the concentration of either of the acids to work out the number of moles of NaOH.

You didn't record the concentrations in your lab book, did you?

3. Jan 25, 2012

### homevolend

Well, we did not get the concentrations. We put the NaOH in a burette then recorded the initial and final volume. Also recorded the amount of acetic acid used.

I think some are calculating moles by using the number 22.4 not sure what that means though?

4. Jan 28, 2012

### Staff: Mentor

Then whole procedure doesn't make sense. You need a concentration. It is usually either written on the bottle or told by TA or whoever takes care of the lab.

It means nothing. 22.4L is a volume of 1 mole of gas at STP. There are also several other numbers they can use - like 3.14 (pi), or 2.78 (e). They don't make more sense though, they are just a way of showing creativity.