25cm3 of a sample of vinegar (ethanoic acid) was pipetted into a volumetric flask and the volume was made up to 250cm3. this solution was placed in a burette and 13.9cm3 were required to neutralise 25cm3 of 0.1 moldm-3 NaOH. calculate the molarity of the original vinegar solution and its concentration in gdm-3?

So the ratio of ethanoic acid and NaOH is 1:1. I have done a solution myself but I am not sure if this solution is correct.

Firstly, I did n=c x v
= 0.1 x 0.025
=0.0025
Then I divided it by 10 to get it to the original solution but I am not sure about this.
0.0025 / 10
= 0.025 mol

Then for concentration, I did
0.025 x60
= 1.5 g/dm^3
Original post by itsyahg
25cm3 of a sample of vinegar (ethanoic acid) was pipetted into a volumetric flask and the volume was made up to 250cm3. this solution was placed in a burette and 13.9cm3 were required to neutralise 25cm3 of 0.1 moldm-3 NaOH. calculate the molarity of the original vinegar solution and its concentration in gdm-3?

So the ratio of ethanoic acid and NaOH is 1:1. I have done a solution myself but I am not sure if this solution is correct.

Firstly, I did n=c x v
= 0.1 x 0.025
=0.0025
Then I divided it by 10 to get it to the original solution but I am not sure about this.
0.0025 / 10
= 0.025 mol

Then for concentration, I did
0.025 x60
= 1.5 g/dm^3

Hi, I've moved this to Chemistry for you.
Original post by rayquaza17
Hi, I've moved this to Chemistry for you.

thank you i'm not sure how to do that
Original post by itsyahg
thank you i'm not sure how to do that

I've sorted it out for you, so don't worry! I don't know enough about chemistry to help you though, so you'll have to wait until someone who does comes along!
You've worked out the amount (mol) in 25 cm3 of NaOH.

There will be that many mol of acid in 13.9 cm3 of diluted acid.

If there were that many mol of acid in 13.9 cm3, how many must there have been in 250 cm3 of the acid?

Which happens to be how many mol of acid there were in the original (undiluted) acid.

Now you've got an amount and a volume, shazam.

You won't use the Mr of ethanoic - there is no mention of a mass anywhere.
Original post by itsyahg
25cm3 of a sample of vinegar (ethanoic acid) was pipetted into a volumetric flask and the volume was made up to 250cm3. this solution was placed in a burette and 13.9cm3 were required to neutralise 25cm3 of 0.1 moldm-3 NaOH. calculate the molarity of the original vinegar solution and its concentration in gdm-3?

So the ratio of ethanoic acid and NaOH is 1:1. I have done a solution myself but I am not sure if this solution is correct.

Firstly, I did n=c x v
= 0.1 x 0.025
=0.0025
Then I divided it by 10 to get it to the original solution but I am not sure about this.
0.0025 / 10
= 0.025 mol

Then for concentration, I did
0.025 x60
= 1.5 g/dm^3

Okay so you need to write down what you've been told:

Ethanoic acid:
Volume - 13.9
Concentration - (need to find out)
Moles - (need to work out)
Ratio - 1

NaOH:
Volume - 25
Concentration - 0.1
Moles - (need to work out)
Ratio - 1

Okay so we've been told the volume and concentration of NaOH therefore we can work out the moles of that

moles = concentration x volume / 1000
moles = 25 * 0.1 /1000
moles = 0.0025

so we can add that into the list of things we know:
we can also write the moles of ethanoic acid as its a 1:1 ratio

Ethanoic acid:
Volume - 13.9
Concentration - (need to find out)
Moles - 0.0025
Ratio - 1

NaOH:
Volume - 25
Concentration - 0.1
Moles - 0.0025
Ratio - 1

now we know most of the things needed, now we need to work out the concentration by rearranging the equation to equal concentration

concentration = moles x 1000 / volume
concentration = 0.0025 x 1000 / 13.9
concentration = 0.18 (2dp)

its really simple if you write out what you already know and what you need to find out

im not quite sure how to work out the second part on gdm-3 im sorry but i think you multiply the answer by the mr of ethanoic acid
(edited 8 years ago)
Ooops, I didn't read the OP.

25.0 / 13.0 = 1.80 mol dm-3

to convert to g dm-3, just multiply 1.80 by the Mr of ethanoic, i.e. 1.80 x 60
Original post by itsyahg
25cm3 of a sample of vinegar (ethanoic acid) was pipetted into a volumetric flask and the volume was made up to 250cm3. this solution was placed in a burette and 13.9cm3 were required to neutralise 25cm3 of 0.1 moldm-3 NaOH. calculate the molarity of the original vinegar solution and its concentration in gdm-3? So the ratio of ethanoic acid and NaOH is 1:1. I have done a solution myself but I am not sure if this solution is correct. Firstly, I did n=c x v = 0.1 x 0.025=0.0025Then I divided it by 10 to get it to the original solution but I am not sure about this.0.0025 / 10 = 0.025 mol Then for concentration, I did 0.025 x60 = 1.5 g/dm^3
why is the concentration times 60?
Original post by lexiee29
why is the concentration times 60?

As I mentioned seven years ago, it is the Mr of the ethanoic acid. It is how you convert from mol dm^-3 to g dm^-3

Sheesh.