Calculate the molarity of a NaOH solution

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SUMMARY

The discussion centers on calculating the molarity of a NaOH solution through a back-titration method involving Na2CO3 and HClO4. Participants utilized two key equations: the first relates the total amount of perchloric acid to the moles of Na2CO3 and NaOH, while the second equates the moles of NaOH to those of HClO4. Despite calculations yielding a molarity of 2.239 M for NaOH, the correct value is established as 1.974 M, indicating a potential error in the application of the equations or stoichiometry.

PREREQUISITES
  • Understanding of stoichiometry in acid-base reactions
  • Familiarity with back-titration techniques
  • Knowledge of molarity calculations
  • Basic grasp of chemical equations and mole conversions
NEXT STEPS
  • Review the principles of back-titration in analytical chemistry
  • Study the stoichiometry of acid-base reactions
  • Learn how to accurately calculate molarity from titration data
  • Explore common pitfalls in titration calculations and how to avoid them
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Chemistry students, laboratory technicians, and educators involved in analytical chemistry and titration methodologies will benefit from this discussion.

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Homework Statement



A 3.125 g sample of primary standard of Na2CO3 was treated with 40.00 mL of dilute perchloric acid. The solution was boiled to remove CO2, following which the excess HClO4 was back-titrated with 10.12 mL of dilute NaOH. In a separate experiment, it was established that 27.43 mL of the HClO4 neutralized the NaOH in a 25.00 mL portion. Calculate the molarity of the NaOH solution. Hint: Solve this problem by considering the two equations below


Homework Equations


For the back-titration (1): total amount of perchloric acid: n1(HClO4) = 2n(Na2CO3) + n1(NaOH)

For the acid-base titration (2): n2(NaOH) =n2(HClO4)



The Attempt at a Solution



I first calculate the moles for equation (1)
(40/1000)[HClO4]=2*(3.125g/ (105.99g/mol) +(10.12/1000)[NaOH]

then eq. (2)

(27.43/1000)[HClO4]=(25/1000)[NaOH]

I then substituted equation 2 into equation 1. My final answer for [NaOH] was 2.239. However the actual answer was 1.974. Can somebody please point out what I did wrong. Thanks
 
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I don't see anything wrong with the solution, I got 2.239 M as well.
 

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