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### Topic: Concentration results  (Read 1590 times)

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#### sciencenoob85

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##### Concentration results
« on: January 27, 2016, 12:40:01 AM »
So my understand when titrating and acid and base I can use the data from the known substance to calculate the concentration of the unknown

My calculation feel a little short

Have I missed anything? or is it just wrong?

The number of mols present in the titre of the known substance can be used to determine the concentration of the unknown acid.

NaOH:   n=vm
n= 25.00 cm3 x 1.0 mol
n= 25.00 moldm3/1000
n= 0.025 moldm3

CH3COOH(aq) + NaOH(aq)         CH3COONa(aq) + H2O(l)

CH3COOH:
n= 0.025 moldm3

Using mols and volume it is possible to work out the concentration;

c=m/v
c= 24.10 cm3/0.025mol
c= 964moldm3/1000
c= 0.964moldm3

#### mjc123

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##### Re: Concentration results
« Reply #1 on: January 27, 2016, 04:38:17 AM »
Quote
NaOH:   n=vm
n= 25.00 cm3 x 1.0 mol
n= 25.00 moldm3/1000
n= 0.025 moldm3
Your units are all over the place
n = 25 cm3 x 1.0 mol/dm3 (I presume)
n = 0.025 dm3 x 1.0 mol/dm3
n = 0.025 mol

NaOH reacts with AcOH in a 1:1 molar ratio, so at the end point you have 0.025 mol AcOH. I presume 24.10 cm3 is the volume of AcOH solution at the end point.

Quote
c=m/v
You have already used m for molar concentration, and n for moles. Be consistent. c = m = n/v
Quote
c= 24.10 cm3/0.025mol
Why are you dividing volume by moles? c = n/v = 0.025 mol/24.10 cm3
Proceed from here, and get the units right.