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Old Feb 21st 2017, 08:48 PM   #1
starrysky
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Titration question. help

A 0.2638 g impure sodium carbonate sample is analyzied by titrating the sodium carbonate with the 0.1288 M hydrochloride solution requiring 38.27 mL.

The reaction is CO3^2- + 2H^+ --> H2O + CO2
Calculate the percent sodium carbonate in the sample.

It will be great if you could help me solve this question. thank you!
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Old Mar 15th 2017, 07:00 AM   #2
MatthijsM
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First you calculate how much mole Hydrochloride there is.

n = c * V = 0.1288 * 0.03827 = 0.004929 mole.
You divide this by 2 so you get the amount of mole of the sodiumcarbonate, this is 0.00246 mole.

If you multiply this with the molecular mass of sodiumcarbonate you have the mass of your pure sodium carbonate. I think you can solve the rest
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