x Turn on thread page Beta
 You are Here: Home

# Titration equations watch

1. I did a vinegar titration & I'm struggling with the calculations for mass, density & percentage by mass of ethanoic acid in vinegar. I used 25cm-3 0.1M NaOH, phenolphthalein indicator & 25 cm-3 diluted with distilled water to 250 cm-3.

The mean titre was 21.85 cm-3

The calculations I have so far are:

The chemical equation for this reaction is: CH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l)
This equation shows that 1 mol of NaOH reacts with 1 mol of CH3COOH so the molar ratio of CH3COOH to NaOH is 1:1

Moles NaOH = (0.1 x 25)/1000 = 2.5x10-3
Moles CH3COOH diluted =(0.1 x 250)/1000 = 2.5x10-4
Moles CH3COOH undiluted = (0.1 x (250/10))/1000 = 2.5x10-3
Concentration undiluted CH3COOH = 2.5x10-3 / (21.28/1000) = 0.117 moldm-3

Mr CH3COOH = 60 g mol-1

This is where I get stuck!

I think I should use mass = mol x Mr but I don't know if I should use the figures from the diluted or undiluted figures, then which figures to use to calculate density. No matter which figures I use I can't come up with a figure that's even close to what google says is the density of vinegar. All this is leading to me not being to even begin to calculate percentage by mass. Help!!
2. *resolved*

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 25, 2015
Today on TSR

### Loughborough better than Cambridge

Loughborough at number one

### Can I date a girl with no boobs?

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE