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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: _cheers on October 01, 2006, 06:04:42 PM

Title: back titration
Post by: _cheers on October 01, 2006, 06:04:42 PM

 
if you have a .6g sample of K2CO3 (mm:138.2) and dissolve it in enough water to make a 200 ml sol'n (A). 20 ml of this is take out and has added to it 20 ml of .1700M HCl which is then titrated with .1048 M NaOH. How many ml of NaOH is used?

K2CO3 + 2HCl -> 2KCl + H20 + CO2

I tried working it backwards, but arrghh!!! I think my brain fried.

Title: Re: back titration
Post by: Borek on October 01, 2006, 06:36:47 PM

Strange, K₂CO₃ amount is very small compared to the amount of HCl used. No need for going backwards here - start with known amount of carbonate, calculate how much HCl was left and you are almost ready.

Title: Re: back titration
Post by: _cheers on October 01, 2006, 10:13:29 PM

so, if I find that HCL added is .0014 mole, can I subtract that from the moles of carbonate to find the moles of excess HCL? then use that to find the moles of NaOH used and divide that by the molarity? its come out weird...not the best idea to jump into first year after a 15 year break..sigh

Title: Re: back titration
Post by: Borek on October 02, 2006, 02:33:04 AM

Quote from: _cheers on October 01, 2006, 10:13:29 PM

so, if I find that HCL added is .0014 mole, can I subtract that from the moles of carbonate to find the moles of excess HCL? then use that to find the moles of NaOH used and divide that by the molarity?

Sounds like you will get negative volume this way, although I think you are on the right track. Show the figures.