Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Moss on September 15, 2009, 06:13:17 AM
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Hey
Additional information:a)1.461g of dry sodium carbonate,Na2CO3, is dissolved in water in a 250mL volumetric flask
b) 20mL aliquots of this solution were titrated with nitric acid, HNO3,three times. The average concordant titre was found to be 22.17mL
i) find the concentration of the nitric acid
i know you have to use the formula c=n/v
how do you find the mole of nitric acid when they havent given you the mass?
thanks
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Start with the reaction equation.
How many moles of sodium carbonate in titrated sample?
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Start with the reaction equation.
How many moles of sodium carbonate in titrated sample?
i got 0.01378 mol
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No, you have not titrated everything at once. You diluted and took part of the solution.
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No, you have not titrated everything at once. You diluted and took part of the solution.
im not sure what you mean :S. could you explain it step by step what i need to do
thanks alot
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Your sample was dissolved in the 250 mL of water. That should give you concentration.
You took 20 mL of the solution - that, plus concentration, should give you number of moles of sodium carbonate used in each titration.
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the concentration i got is
0.01378/0.25
=0.05502 mol L-1
use this plus 0.020
=.075
the answer is 0.99mol L-1 though
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I told you to start with the reaction equation, instead of writing it you are trying to juggle numbers till they will miraculously give you the right answer. It won't work.
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the answer is 0.99mol L-1 though
No it is not.