Question

A 21.524 g paint sample was analyzed for barium (Ba²⁺, MW = 137.327 g/mole) by an EDTA back titration: Ba²⁺(aq) + Y^4–(aq) à BaY^2–(aq). The sample was dissolved in acid and sufficient water was added to produce a volume of 100 mL. 10.00 mL of this concentrated solution was diluted to a volume of 50.00 mL. 25.00 mL of the diluted solution was treated with 33.95 mL of excess 0.09456 M EDTA. The excess EDTA was titrated to the endpoint with 11.04 mL of 0.07238 M Zn²⁺ by Zn²⁺(aq) + Y^4–(aq) à ZnY^2–(aq).

a. Use the results of the back-titration to calculate [Ba²⁺] in the diluted solution.

b. Calculate [Ba²⁺] in the concentrated solution.

c. Calculate the total moles of Ba in the sample.

d. Calculate the total mass of Ba in the sample.

e. If the barium in the sample is present as barium oxide (BaO, MW = 153.33 g/mole), calculate the mass of BaO in the sample.

f. Calculate the percentage of BaO in the sample.

Answer 1

@ Let the concentration of 25 ml of diluted Ba2+ solution bex'm 25 mL of De M Baltsoh + 11.04 ml of 0.07238 M. Zn2+ som = 33.95 mb of 0.09456 M EDTA sola op 25 *x + 11.04x0.07238 = 33.95 x 0.09456 of 25% = 3.210312 - 0:7990 752, Z 2.4112368 or x = ( 2:4112368 ) = 0.096449 (M) (Ans) 25 6 Now, V, 4 = V2 C2 17.50 mix 0.096449 = iomL x C2 of C2 = 5040:096449 (M) = 0.482247 (M) - The concentration of [Ba2+1 in the concentrated som = 0.482247M © Total moles of Bazt in the sample -100% 0.482247 ) . 1000 = 0.0482247 (ans) d mass of Ba en the sample = (137.327 x 0.048 2247) a = 6.6226 g. Cans) mass of Bab = (153:33. * 6.6226) g = 6.9515g. Cans) "137.327 Dy Bad in the samiple = 6952*100) = 32:30 % (ans)