Answer to Question #165586 in Physical Chemistry for agent

Question

The reaction between barium chloride and sodium carbonate gave a white precipitate which was filtered and dried. A total of 25.00 mL of 0.300 M HCl was added to 0.250 g of the resulting white precipitate with resultant fizzing and dissolution of the sample. After the reaction, 24.80 mL of 0.200 M NaOH was required too back-titrate the excess HCl. Determine the molar mass of the white precipitate. How would you confirm its identity? 

Answer 1

BaCl₂(aq) + Na₂CO₃(aq) = BaCO₃(s) + 2NaCl(aq)

BaCO₃(s) + 2HCl(aq) = BaCl₂(aq) + H₂O(l) + CO₂(g)

HCl(aq) + NaOH(aq) = H₂O(l) + NaCl(aq)

n(BaCO₃) = (n(HCl) - n(HCl)_excess)/2 = (0.025*0.3-0.0248*0.2)/2 = 0.00127 mol.

m(BaCO₃) = n(BaCO₃)*Mr(BaCO₃) so we can find precipitate mass knowing molar mass and vice versa

If we want to determine molar mass then we should have weighed precipitate and knowing its mass:

Mr(BaCO₃) = m(BaCO₃)/n(BaCO₃) = (0.25 g)/(0.00127 mol) = 196.9 g/mol.

To identify the precipitate we should divide it into two parts. Then use one part for the flame test. Barium ions will cause the flame to turn green. The second part should be treated with a strong acid, for example, HCl. The reaction mixture will start fizzing because of the reaction CO₃^2-(aq) + 2H⁺(aq) = H₂O(l) + CO₂(g). This way we confirm that the precipitate is BaCO₃.