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Topic: Titration of the Unknown  (Read 5284 times)

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Offline ItalianChick0188

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Titration of the Unknown
« on: April 02, 2009, 01:26:28 PM »
Here I am again. We did a experiment in lab: Analysis of Soda Ash Sample

Na2CO3 + 2 HCl+ 2 NaCl + H20 + C02

I figured out that my average molarity for the Standardization of HCl was 0.199 M.

I dno if this is relevant but the weight of my first flask was: 71. 17 g and with the Na2CO3 was 72.126 g.

That part was pretty easy but now I'm having a little trouble with the Titration of our unknown.

We used 5-6 drops of bromocresol green indicator and titrated with 0.30 M HCl until we got a light blue color.

My first Na2CO3 tablet weighed 0.2851 g and my volume was 7.250 mL.


1) How many moles of HCl were added to the titration flask?

Do I divide 0.2851 g by the molar mass of Na2CO3 and then take my answer and times it by 2? Since its a 1:2 ratio?


2) How many moles of Na2CO3 are in the sample?


3) How many grams of Na2CO3 are in the sample?
Once I figure out moles I'll be able to convert it to grams to get this answer.

4) Calculate the % Na2CO3 in the sample?

Offline ItalianChick0188

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Re: Titration of the Unknown
« Reply #1 on: April 02, 2009, 01:59:24 PM »
I took 0.007250 * 0.199 mol/L = 0.00144 moles added to titration flask.

For moles of Na2CO3 in the sample

I did 2 X 0.00144= 0.00288 but on Excel it said this was wrong?

I multiplied by 2 because it takes 2 HCl for 1 Na2CO3

Offline macman104

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Re: Titration of the Unknown
« Reply #2 on: April 02, 2009, 02:08:06 PM »
Please don't post the same topic in multiple sections of the forum.  There are people helping you in your other thread already.  Continue your discussion there.

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