Chemistry help

[BlakBlastoise]

Hello Smogon, I'm not sure of how to work out a problem for my chemistry course and since there's plenty of smart people here, I thought I'd ask you.

One of the "active" ingediants in Alka-Seltzer is sodium bicarbonate, NaHC(O3). A student analyzes an Aka-Seltzer tablet to determine the % by mass of NaHC(O3) present in the sample. A 0.500 gram tablet was powdered and mixed with 50.00 ml of .200 M HCL (aq). This reaction mixture was allowed to proceed until all the available sodium bicarbonate had reacted.

Reaction 1) NaHCO3(aq) + HCL(aq) --> NaCl(aq) + H2O(l) + CO2(g)

Not all of the hydrochloric acid reacted and the amount of excess HCL(aq) was determined by titration with sodium hydroxide solution to a phenolphthalein end point. This "back-titration" required 47.5 mls of .125 M NaOH (aq)

Titration reaction: HCL(aq) + NaOH(aq) --> NaCl(aq) + H2O(l)

Based on this data, determine the mass percent of sodium Bicarbonate in the original Alka-Seltzer sample.
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This is what I did, but i'm not sure if i'm right since i've never seen a titration problem like this.

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.125M = [x/.0475 l]
x = .005937 mol NaOH
.005937 mol NaOH(1 mol/84 g. per mol.) = .49875 g NaHCO3
(The above step is where i'm pretty sure i'm wrong. My thought process was that since the mole to mole ratio of HCL to NaHCO3 is the same in both equations, that it would be right?)
.49875g/.5g = 99.75% ???

Thanks for your time/help.

 

[Jimbo]

I don't have the time to do the entire thing, but I'm pretty sure you'd go like this:

Get the moles of HCl that reacted with NaOH (using the info about NaOH), that is how many HCl ml was left over from the first reaction. Using that number (and the 50 original ml of HCl) you can see how many moles of HCl reacted with the tablet, allowing you to get the mass of the reacted tablet...

again I could be wrong someone less tired might be able to help you more (stellar is in chem now!)

 

[BlakBlastoise]

Thanks Jimbo. Can anyone else confirm what he said?

 

[lati0s]

jimbo's method is correct

 

[BlakBlastoise]

Thanks guys, I think i got it now :D
.2M(.o5L) = .01 Moles
.125(.04750L) = .0059375 Moles
.01 - .0059375 = .0040625 Moles NaHClO3
.0040625 x 84 g per mol = .34125 g NaHClO3
(.34125/.5)x100 = 68.25%

That took me all day and i would not have figured it out without you. Thank you very much. I hope there's none of this on the final.
Mods may lock now.

 

[ivar]

My mod powers are a bit tired but dont worry Ill lock this thread soon.

 

[TAY]

Mods may lock now.

lol, thanks for the permission.


Guys, this is not a "post your homework" forum. There was an old homework help thread a couple months back (in congregation); if anyone has questions then they can ask there. The last thing we need is people posting a new thread every time they need help on a homework problem.

Thanks.