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Topic: double indicator titration  (Read 26956 times)

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Offline k329

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double indicator titration
« on: April 12, 2010, 12:04:08 AM »
i feel very confuse about the calculation...
i have been sitting in front of the lab report for more than one hour..
can someone help me :-[?

the question is to demination of the amount of NaOH in a NaOH/Na2Co3 mixture (method of double indicators)
i titre the mixture with a standardized HCl (standardized by Na2Co3)
here are the reaction involved:
NaOH + HCl -->NaCl + H2O
Na2CO3 + HCl --> NaHCO3 + NaCl   1st-end-pt
NaHCO3 + HCl --> NaCl + H2O + CO2    2nd-end-pt

here is the titration result:
Part a : standardized HCl by Na2Co3
(we prepare a standard Na2Co3 solution by dissolving 1.37G Na2Co3 into 250cm3 of 0.050M <~ i dun understand what that means...?)
->  26.05cm3 of HCl used

Part b : Double indicator method
1st end-pt :  16.875cm3 of HCl used
2nd end-pt : 6.2cm3 of HCl used
                  16.875 - 6.2 = 10.675cm3 of HCl used in 2nd end-pt



how can i find out the Molarity of HCl in part a?
i should use 0.050M X 26.05/1000 or use the 1.37g to find out the mole of Na2Co3 first?

and how can i calculate the mass of NaOH by the Part b result?
i should use 1st end pt result or 2nd end point result???
i know they are very simple   (actually i have learnt already... sorry /.\)
OMG i am so confused!!!!
Please help me
Thanks at all  :)
« Last Edit: April 12, 2010, 12:18:20 AM by k329 »

Offline Borek

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Re: double indicator titration
« Reply #1 on: April 12, 2010, 03:27:29 AM »
(we prepare a standard Na2Co3 solution by dissolving 1.37G Na2Co3 into 250cm3 of 0.050M <~ i dun understand what that means...?)

No idea. Using this amount of sodium carbonate I got 0.0517M. Could be this is some correction for the fact that solid is wet, it is not easy to prepare dry sodium carbonate.

Assume your solution was 0.050M and continue from there.

http://www.titrations.info/acid-base-titration-sodium-hydroxide-and-carbonate
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline k329

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Re: double indicator titration
« Reply #2 on: April 12, 2010, 03:38:19 AM »
(we prepare a standard Na2Co3 solution by dissolving 1.37G Na2Co3 into 250cm3 of 0.050M <~ i dun understand what that means...?)

No idea. Using this amount of sodium carbonate I got 0.0517M. Could be this is some correction for the fact that solid is wet, it is not easy to prepare dry sodium carbonate.

Assume your solution was 0.050M and continue from there.

http://www.titrations.info/acid-base-titration-sodium-hydroxide-and-carbonate

Thank you first :)

oops i forgot to tell you that my teacher said the Na2Co3 is anhydrous

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