a) Amount of Ba(NO3)2 dissolved in grams. Molar mass of Ba(NO3)2 = 261.3 g/mol b)  Solubility product, Ksp c) Vapor pressure of the solution in kPa

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a) Amount of Ba(NO3)2 dissolved in grams. Molar mass of Ba(NO3)2 = 261.3
g/mol
b)  Solubility product, Ksp
c) Vapor pressure of the solution in kPa

1. Non-ionic and ionic equilibrium
In an experiment to calculate the solubility product (Ksp) of barium nitrate (Ba(NO3)2),
an excess amount of Ba(NO3)2 was added to 2 liters of water at 25 °C until the solution is
saturated. Because Ba(NO:)2 is only slightly soluble in water, the excess Ba(NO3): was
filtered out to get a solid-free mixture. The proponents of the experiment thought about
using the concept of colligative properties, specifically boiling point elevation, to determine
Kp. It was observed in their experiments that the solution boils at 100.15 K. Barium
nitrate dissociates via the process:
Ba(NO;)26) ² Ba'aq) + 2NO5 (aq)
Ksp = [Ba2*][N0;
,2+
Calculate the following in no particular order if Kb.water = 0.51 K-kg/mol:
Transcribed Image Text:1. Non-ionic and ionic equilibrium In an experiment to calculate the solubility product (Ksp) of barium nitrate (Ba(NO3)2), an excess amount of Ba(NO3)2 was added to 2 liters of water at 25 °C until the solution is saturated. Because Ba(NO:)2 is only slightly soluble in water, the excess Ba(NO3): was filtered out to get a solid-free mixture. The proponents of the experiment thought about using the concept of colligative properties, specifically boiling point elevation, to determine Kp. It was observed in their experiments that the solution boils at 100.15 K. Barium nitrate dissociates via the process: Ba(NO;)26) ² Ba'aq) + 2NO5 (aq) Ksp = [Ba2*][N0; ,2+ Calculate the following in no particular order if Kb.water = 0.51 K-kg/mol:
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