Question

1. Glucose has a solubility of 125 g /100 mL. What is the approximate concentration of glucose in a
saturated solution? How much glucose can be dissolved in 745 mL of water?
[glucose] _________ M
Glucose _________ g
2. Determine whether the following salts or molecules will have higher or lower solubility in water
at higher temperatures.
KCl(s) + H2O(l) KCl(aq), ?Hrxn > 0 __________
NH3(g) + H2O(l) NH3(aq), ?Hrxn< 0 __________
3. BaF2 has a Ksp of 2.45 x 10-5. What is the solubility of BaF2? What is the F-
concentration in a
saturated barium fluoride solution?
Solubility of barium fluoride ___________
[F-]= _______

Answer 1

Molar Mass of Glucose = 180.2

In 100 mL water, glucose dissolved = 125 g

In 1 mL water, glucose dissolved = 1.25 g

In 745 mL water, glucode dissolved = 1.25 x 745 = 931.25 g

Moles = 931.25 / 180.2 = 5.17 moles

Assumption: Since the density of glucose solution is not provided, I am assuming that the addition of glucose does not affect the volume of the solution.

Mass of glucose in 1000 mL water = 1250 g

Moles = 1250 / 180.2 = 6.94

Molarity = Moles / Volume (L) = 6.94 / 1 = 6.94 M

2)

KCl(s) + H2O(l) ---- KCl(aq), ?Hrxn > 0

Since delta H > 0(endothermic reaction) , according to le chateleirs principle if we increase the temp. of this reaction, it will move in the forward direction. Hence for the above reaction as temp. increases KCl salt will have higher solubility

NH3(g) + H2O(l) ----- NH3(aq), ?Hrxn< 0

Since delta H < 0 (exothermic reaction) , according to le chateleirs principle if we increase the temp. of this reaction, it will move in the backward direction. Hence for the above reaction as temp. increases NH3 will have lower solubility

3)

BaF2 ------> Ba2+ + 2F-

..................S.........2S

S = Solubility of BaF2 in moles/liter

Ksp = [Ba2+] [F-]^2

=> 2.45 x 10^-5 = S x (2S)^2 = 4S^3

=> S = 0.0183 M

Solubility of barium fluoride = 0.0183 moles/liter = 0.0183 x 175.34 = 3.21 g/L

[F-] = 2S = 2 x 0.0183 = 0.0366 M