1. In this experiment you need to add enough of a 0.20 M BaCl2 solution to preci
ID: 711958 • Letter: 1
Question
1. In this experiment you need to add enough of a 0.20 M BaCl2 solution to precipitate all the sulfate from the solution of your unknown. For this calculation assume that the sample contains a maximum of 70% sulfate (mass percent) and that you weigh out 0.400 gram of sample. a. Wh at is the maximum mass of sulfate and number of moles of sulfate that could be in this sample? b. What volume of 0.20 M BaCl2 solution is needed to deliver as many moles of Ba2 (a) as there are of SO, (a)? 2. a. How many moles of sulfur are there in 0.244 g of BaSO,? b. A 0.188-g sample of an unknown solid containing sulfate ions was dissolved to make a solution. Upon addition of excess Ba2,0.224 g of BaSO4 were obtained. What was the percent by mass of sulfur in the unknown solid? 3. Why is a small amount of acid added to the solution containing the sulfate and the barium ions in this analysis?Explanation / Answer
0.2 M of BaCl2
0.4 g of 70 % sulfate
Maximum Mass of sulfate = 0.4 g * 0.7 = 0.028 g
Molar mass of sulfate = 96 g/mol
no of moles of sulfate = mass / molar mass = 0.028 g / 96 g/mol = 2.92 * 10-4 moles
Ba2+ + SO42- ---> BaSO4
so 2.92 * 10-4 moles of Sulfate will require 2.92 * 10-4 moles of barium ions
Molarity = 0.2 M ( or mol/L)
Volume = 2.92 * 10-4 moles / 0.2 mol/L = 0.0014 L = 1.4 mL
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