Mass of empty beaker: 157.599g Temperature: 24°C Ba(NO3)2 mass added to beaker: 5.091g Mass of Beaker and contents after decanting barium nitrate solution and heating to dryness: 157.96g The molar solubility of Ba(NO3)2 is equal to the moles of Ba(NO3)2 dissolved in the saturated solution divided by the volume of the solution, 0.050 L. Once you have calculated S, using this equation Ksp = (S)(2S)2 = 4S 3
Mass of empty beaker: 157.599g Temperature: 24°C Ba(NO3)2 mass added to beaker: 5.091g Mass of Beaker and contents after decanting barium nitrate solution and heating to dryness: 157.96g The molar solubility of Ba(NO3)2 is equal to the moles of Ba(NO3)2 dissolved in the saturated solution divided by the volume of the solution, 0.050 L. Once you have calculated S, using this equation Ksp = (S)(2S)2 = 4S 3
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter11: Solutions
Section: Chapter Questions
Problem 66AP: Relative solubilities of salts in liquid ammonia can differsignificantly from those in water. Thus,...
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Given:
Mass of empty beaker: 157.599g
Temperature: 24°C
Ba(NO3)2 mass added to beaker: 5.091g
Mass of Beaker and contents after decanting barium nitrate solution and heating to dryness: 157.96g
The molar solubility of Ba(NO3)2 is equal to the moles of Ba(NO3)2 dissolved in the saturated solution divided by the volume of the solution, 0.050 L. Once you have calculated S, using this equation Ksp = (S)(2S)2 = 4S 3
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