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Topic: Titration sum  (Read 5389 times)

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Offline shanker

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Titration sum
« on: October 07, 2009, 04:59:17 PM »
Na 2Co3(aq) + 2HCl (aq) -> 2NaCl (aq) + CO2 ( g) + H2O  (l)


1.4 g of pure anhydrous sodium carbonate  were dissolved in water and made up to a volume of 250 cm3. A 25 cm3 aliquot of this solution required 24.5 cm3 of a solution of HCl in order to neutralise it fully. Calculate the molarity if the HCl and its concentration in grams of HCl per dm3
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Offline Arctic-Nation

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Re: Titration sum
« Reply #1 on: October 07, 2009, 05:15:46 PM »
Triple cross-post? No prior attempt to solve the problem? Borek will have your ass for breakfast.
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Offline shanker

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Re: Titration sum
« Reply #2 on: October 07, 2009, 05:17:33 PM »
cant you answer?


this is how far i have got moles = mass/rfm
conc. = moles/volume

1:2 ratio
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Offline sjb

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Re: Titration sum
« Reply #3 on: October 07, 2009, 05:20:10 PM »
Quote from: Arctic-Nation on October 07, 2009, 05:15:46 PM
Triple cross-post? No prior attempt to solve the problem? Borek will have your ass for breakfast.

Triple? More than that now! ;)

Quote from: shanker on October 07, 2009, 05:17:33 PM
cant you answer?


this is how far i have got moles = mass/rfm
conc. = moles/volume

1:2 ratio

OK, how many moles of Na2CO3 (not Na2Co3, which may be some weird alloy?) were there in your titrated mixture?
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Offline shanker

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Re: Titration sum
« Reply #4 on: October 07, 2009, 05:26:53 PM »
moles = mass/rfm


1.4/106 =0.0103 moles i dont know if that is right


but my teacher said to work from the end ofthe question to the front?
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Offline shanker

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Re: Titration sum
« Reply #5 on: October 07, 2009, 05:30:32 PM »
yes?
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Offline Arctic-Nation

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Re: Titration sum
« Reply #6 on: October 07, 2009, 05:37:43 PM »
Shanker, you should've been here long enough by now to know the rules. And unless your homework should've been finished half an hour ago, a few minutes won't make a difference.

Anyway, you're going in the right direction. 1.4 g / 106 g/mole = 13.21 mmoles. That's the amount of sodium carbonate in 250 ml. How many moles are there then in 25 ml, and how many moles of HCl will you need for neutralisation?

Edit: And Jesus f&#$, stop posting this on other forums already, will you.
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Offline shanker

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Re: Titration sum
« Reply #7 on: October 07, 2009, 05:45:07 PM »
is this right?


in 25ml there are 1.321 moles


because it is 1:2 you multiply 1.321 by 2 so there are 2.642 moles of Hcl??
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Offline Arctic-Nation

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Re: Titration sum
« Reply #8 on: October 07, 2009, 05:54:36 PM »
That's correct. Next step is to calculate the molarity of the HCl solution, and from that the concentration in g/dm^3.
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Offline sjb

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Re: Titration sum
« Reply #9 on: October 08, 2009, 02:16:43 AM »
Quote from: shanker on October 07, 2009, 05:45:07 PM
is this right?


in 25ml there are 1.321 moles


because it is 1:2 you multiply 1.321 by 2 so there are 2.642 moles of Hcl??

No, 13.21 mmol / 10 is not 1.321 mol. Careful with your notation. The application of the balanced equations is correct though, so far as it goes.
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