Author Topic: Titration Question (Read 609 times) Tweet Share

andy456 · « **on:** April 18, 2010, 10:18:22 am »

Ok, so I'm having a little trouble with this question. Any help would be great.

A 5.0ml solution of HCl is diluted to 100.0ml in a volumetric flask. 20.0ml aliquots are then titrated against a base and the concentration of the aliquots is found to be 0.12M. The concentration of the original acid would be?

I found the number of mole in each aliquot first.
$n=cv$
$n=0.12M * 0.02L$
$n=0.0024mole$

Then I did this:
$n * (20/100)$
$n(in 100ml)= 0.0024 * 0.2$
$n(in 100ml)= 0.00048$

Then I'm stuck. I don't even know if what I have done is right.

Edmund · « **Reply #1 on:** April 18, 2010, 11:02:16 am »

You found that the number of mole in each aliquot (20mL) is 0.0024 mol. How much in 100mL volumetric flask? So you have to multiply this by (100/20) (As a rough guide, there should be more moles in the volumetric flask than in the 20mL aliquot).

$n(in 100mL)=n*\frac{100}{20}=0.0024mol*5=0.012mol$

There will also be 0.012mol in the 5mL solution so you can work on the concentration from here

$C=\frac{0.012}{0.005}$

andy456 · « **Reply #2 on:** April 18, 2010, 11:04:10 am »

Thanks, I had the dilution factor the wrong way around. :buck2:

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