Monday
September 1, 2014

Homework Help: Analytical chemistry

Posted by Gloria on Thursday, August 1, 2013 at 7:49am.

A dilute perchloric acid solution was standardized by dissolving 0.2445 g of primary standard sodium carbonate in 50 mL of the acid, boiling to eliminate CO2, and back-titrating with 4.13mL of dilute NaOH. In a seperate titration a 25mL portion of the acid required 26.88 mL of the base. Calculate the molar concentrations of the acid and the base.

First i wrote the equations
Na2CO3 + 2HClO4 -> CO2 + 2NaClO4 + H2O
n(Na2CO3)=n(HClO4)/2

back titration
HClO4 + OH -> ClO4- + H2O
n(HClO4)=n(OH)=c(OH)*0.00413L

from the seperate titration:
n(HClO4)=n(NaOH)
c(HClO4)*0.025L=c(NaOH)*0.02688L

from first equations i calculated the mols:
n(Na2CO3)=0.2445/105.99=0.0023068 mol
n(HClO4)=2*0.0023068=0.0046134 mol

i know that the mols from back titration represent the excess acid, but i just get stuck with the calculations.

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