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Topic: DESPERATE! Conc/n of HCl in titration?  (Read 3709 times)

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Offline cremeegg

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DESPERATE! Conc/n of HCl in titration?
« on: December 04, 2006, 11:59:08 AM »
Hi all,

This is my first question on this site and I am really confused and struggling.

Last week I did a titration with 0.1 M HCl in the burette and Na2CO3 diluted in water (1,4473g) in the beaker.

I did 3 titrations and discovered that it took on average 28.5ml of HCl to neutralise the Na2CO3.

After the lab we had to do some calculations and this is where I am stuckā€¦. The question says:

If the average titre is v mL, and the (still unknown) conc/n of HCl is X mol L-1, then the amount of HCl added is;
v/1000 x X mol

The amount of Na2CO3 in each titration is mass Na2CO3 in each flask/RMM Na2CO3

So

X =2000 x mass/v x RMM

Complete the cal of the conc of HCl expressing the measured conc as the average conc plus or minus the range.

------------------

I inputted my figures in and got 0.92 mol L-1 but now I have NO idea what to do regarding this range average and measured concentration.  Does anyone have any idea what this means??

I would REALLY appreciate some help, thankyou

 :-\


Offline FeLiXe

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Re: DESPERATE! Conc/n of HCl in titration?
« Reply #1 on: December 04, 2006, 05:54:32 PM »
you did three titrations so you have three values

giving the range means giving the largest and smallest value (or the difference between them)
Math and alcohol don't mix, so... please, don't drink and derive!

Offline natrium

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Re: DESPERATE! Conc/n of HCl in titration?
« Reply #2 on: December 14, 2006, 05:10:40 PM »
the most logical thing you can do... calculate the average conc with the 3 values and later compare it with the max and min values of that concentration

I hope this can be useful    ;D

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