I am doing a post lab for the conductometric titration of Carbonate Ion (CO3^-2)
ID: 1046177 • Letter: I
Question
I am doing a post lab for the conductometric titration of Carbonate Ion (CO3^-2). And I am super confused. Please show all steps, Thank you. Endpoint was reached at 30.00 ml of HCL added (titrant), HCL .098 M, used in experiment was .00294 mol HCL. And total Na2CO3 added to original beaker was 0.164g. Balanced equation is 2HCl + Na2CO3 --> H2O + CO2 + 2NaCl. A. Determine the moles of analyte that were added to the beaker. (Hint: you will need a balanced chemical equation). B. Calculate the percent mass (wt/wt) of analyte in the sample. C. Suppose you had attempted to titrate a salt that was only slightly soluble in water. How would conductivity be affected as you added more and more titrant? What sort of slope would that yield and why? (Hint: consider how LeChatelier's principle applies.)
Explanation / Answer
Ans. Moles of Na2CO3 consumed = Mass / MW = 0.263 g / (105.99 g/ mol)
= 0.002481 mol
#a. Following stoichiometry, 1 mol Na2CO3 is neutralized by 2 mol HCl.
So,
Required moles of Na2CO3 = 1/2 x Moles of HCl consumed =
= ½ x 0.00294 mol = 0.00147 mol
#B. Mass of Na2CO3 in sample = Moles x MW = 0.00147 mol x (105.988736 g/mol)
= 0.1558 g
Now, % Na2CO3 = (mass of Na2CO3 / Mass of sample) x 100
= (0.1558 g / 0.164 g) x 100
= 95.00 %
#C. Following Le Chatelier’s principle, addition of HCL shifts the equilibrium to the right, i.e. forms more NaCl into the solution. Increase in [NaCl] increases the conductivity of the solution.
The slope obtained for conductivity vs [NaCl] is linear because conductivity of the solution is proportional to [NaCl] – which gradually increases with addition of HCl till all the Na2CO3 is consumed up.
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