Calculate Moles of NaOH: Homework Help

[homevolend]

Homework Statement

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.

Homework 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.

The Attempt at a Solution

 

[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?

 

[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?

 

[Borek]

Well, we did not get the concentrations.

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.

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

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.

 

[DeeNos]

First, let's start by identifying the known and unknown variables in this problem. We know the volume of NaOH used (0.02 L) and the volume of acetic acid neutralized (3 ml). We also know that the reaction is a titration, meaning that the moles of NaOH used must be equal to the moles of acetic acid neutralized. Therefore, our unknown variable is the moles of NaOH.

To calculate the moles of NaOH, we can use the formula: moles = concentration (M) x volume (L). Since we do not know the concentration of NaOH, we will have to rearrange the formula to solve for concentration.

Concentration (M) = moles / volume (L)

We know that the moles of NaOH and acetic acid are equal, so we can set up the equation:

Moles of NaOH / 0.02 L = Moles of acetic acid / 0.003 L

Solving for moles of NaOH, we get:

Moles of NaOH = (Moles of acetic acid x 0.02 L) / 0.003 L

Now, we need to find the moles of acetic acid. To do this, we can use the formula: moles = concentration (M) x volume (L). We know the volume of acetic acid (3 ml), so we just need to find the concentration.

To find the concentration, we can use the formula: concentration (M) = molarity x volume (L). We know the volume (3 ml), but we do not know the molarity. However, we can use the fact that the moles of NaOH and acetic acid are equal to find the molarity.

Moles of NaOH = Moles of acetic acid

Concentration of NaOH (M) x 0.02 L = Concentration of acetic acid (M) x 0.003 L

Solving for concentration of acetic acid, we get:

Concentration of acetic acid (M) = (Concentration of NaOH (M) x 0.02 L) / 0.003 L

Now that we have the concentration of acetic acid, we can substitute it into our original equation to find the moles of NaOH:

Moles of NaOH = [(Concentration of NaOH (M) x