Identifying Substance Z Using Titration Method: Results and Analysis"

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SUMMARY

The discussion focuses on the titration method used to identify Substance Z, represented by the reaction Z2CO3(aq) + 2HCl(aq) -> 2ZCl(aq) + CO2(g) + H2O(l). The experiment involved titrating a 250 cm³ solution of Z2CO3 with 0.1 M HCl, yielding a mean titre of 26.8 cm³. Calculations determined that there were approximately 1.134 x 10^-2 moles of Substance Z, leading to a molar weight of 36.1375 g, suggesting that Substance Z is likely potassium. The participant questioned the significance of a 3 g discrepancy from the expected potassium weight, which was deemed too large by another user.

PREREQUISITES
  • Understanding of acid-base titration principles
  • Familiarity with molarity calculations
  • Knowledge of stoichiometry in chemical reactions
  • Ability to use indicators like methyl orange in titrations
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  • Study the principles of acid-base titration in detail
  • Learn about the use of indicators in titration, specifically methyl orange
  • Explore stoichiometric calculations in chemical reactions
  • Investigate the properties and identification methods of alkali metals, particularly potassium
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Chemistry students, laboratory technicians, and educators involved in analytical chemistry and titration methods.

paulhunn
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Hi

I have been set the following problem. Z2CO3(aq)+2HCl(aq) -> 2ZCl(aq) + CO2(g) + H2o(l)

We did the experiment in class to find the amount of HCl it would take to titrate the soloution. First by measuring an amount of Z2CO3 and dissolving it into 250cm^3 of water and using methl orange indicator to know when the reaction had taken place. We did this three times using 25cm^3 samples and i got a mean titre of 26.8cm^3
We are told that it is a group one element and the concentration of the HCl is 0.1
So using the formula M=number of moles/volume in dm^3 i rearranged and got 0.1*0.0268=2.68*10^-3 meaning that it would be 2.68*10^-2 for the entire soloution.
Knowing it will take 2 moles of HCl to react one mole of substance Z it would mean there was 1.134*10^-2 moles of substance Z in the 250ml soloution and knowing the original mass that was dissolved (1.5g) i can use the formula weight(g)/number of moles=molar weight which gave me 132.275g. Then knowing that the rest of substance Z is CO3 i can subtract their weights (12+16+16+16) which gives 72.275 and then halfing to find the molar weight of Z (because in the formula we have Z2) we get 36.1375. which is relatively close to potassium.

Does my method follow correctly?
And is the 3g that i am off from potassium enough to say there was a significant error in working or is it ok?

Thanks in advance

Paul
 
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Approach is correct, error is too large for my liking (but it was very likely potassium as you think).

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Ok thanks Borek
 

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