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