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Topic: Titration  (Read 4502 times)

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Offline enantiomorph

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Titration
« on: October 13, 2006, 06:04:27 AM »
Hi there, I have a question involving a titration problem that I'm not sure on how to solve I was wondering if anyone could please give me a hand with this.  Thank you.

Question: A 1.00 g solid sample containing a mixture of table salt NaCl and citric acid H3C6H5O7 is dissolved in 15 mL of water.  Titration of the acid solution requires 16.30 mL of 0.5020 M NaOH solution to reach the endpoint.  Calculate the mass percent of H3C6H5O7 in the solid mixture.

I think the first one I should do is to calculate the number of moles of NaOH required to reach the endpoint: n=(0.0163 L)(0.5020 M)

Now, how can I relate this information to the (volume of the mixture = 15 mL) and the (mass of the sample = 1.00 g)?

Offline DrCMS

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Re: Titration
« Reply #1 on: October 13, 2006, 06:31:12 AM »
You've got the first bit right.

Caclulate the no. of moles of NaOH.
Next look up the structure of citric acid (make a note of the molecular weight for the next step) and work out how many moles of citric acid would react with the no. of moles of NaOH you've already calulated.
From the no. of moles of citric acid work out the weight of cirtic acid in the sample.
Once you have the weight of citric acid you already have the total sample weight, 1g, divided one by the other to get the weight % of citric acid.  You don't need the volume.

Offline enantiomorph

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Re: Titration
« Reply #2 on: October 14, 2006, 12:25:52 AM »
Hi DrCMS.  I tried your approach to the problem and I think I got it.  I was wondering if you or anyone else here at chemical forum can please check over this.  Thanks!

So I've done this:
# of moles of NaOH = (0.0163 L)(0.5020 M) = 0.0081826 mol
Because citric acid is a triprotic acid, the number of moles needed to reach the end point is (1/3)(# of moles of NaOH)
=(1/3)(0.0081826 mol)
=0.0027 mol
The mass of citric acid = (# of moles) x (Molar mass)
mass = (0.0027 mol) x (192.14 g/mol)
mass = .524 g
The mass % of citric acid = (.524 g/1 g) x 100% = 52.4 %

Is this right?
« Last Edit: October 14, 2006, 01:43:54 AM by enantiomorph »

Offline DrCMS

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Re: Titration
« Reply #3 on: October 27, 2006, 06:46:17 AM »
Yes that's how to do it.

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