Reading graph: at 38 °C the solubility of copper sulphate, CuSO4, is
28g of anhydrous salt per 100g of
Reading graph: at 84 °C the solubility of potassium sulphate, K2SO4,
is
22g per 100g of water.
Ex Q1: How much potassium nitrate will dissolve in 20g of water at 34
°C?
At 34 °C the solubility is 52g per 100g of water, so scaling down, 52 x 20 / 100
= 10.4g will dissolve in 20g of water.
Ex Q2: At 25 °C 6.9g of copper sulphate dissolved in 30g of water,
what is its solubility in g/100cm3 of water?
Scaling up, 6.9 x 100 / 30 = 23g/100g of water (check on graph, just
less than 23g/100g water).
http://www.wsd1.org/grantpark/staff/patenaude/powerpoint/Solutions_30SE.ppt
Henry’s Law Application
• The solubility of pure N2 (g) at 25oC and 1.00 atm pressure
is 6.8 x 10-4 mol/L. What is the solubility of N2 under
atmospheric conditions if the partial pressure of N2 is
0.78 atm?
Step 1: Use the first set of data to find “k” for N2 at 25°C
c
6.8 x10 4 M
k 

 6.8 x10 4 M atm 1
P
1.00 atm
Step 2: Use this constant to find the solubility (concentration)
when P is 0.78 atm:
c  kP  (6.8 x104 M atm 1 )(0.78 atm)  5.3 x104 M
http://academic.pgcc.edu/~ssinex/Solutions.ppt
How do I get
sugar to dissolve
faster in my
iced tea?
Stir, and stir, and stir
Fresh solvent contact and interaction with solute
Add sugar to warm tea then add ice
Faster rate of dissolution at higher temperature
Grind the sugar to a powder
Greater surface area, more solute-solvent interaction
http://academic.pgcc.edu/~ssinex/Solutions.ppt
Revision Units of Concentrations
amount of solute per amount of solvent or solution
Percent (by mass) =
Molarity (M) =
g solute
g solution
x 100 =
g solute x 100
g solute + g solvent
moles of solute
volume in liters of solution
moles = M x VL
Examples
http://academic.pgcc.edu/~ssinex/Solutions.ppt
What is the percent of KCl if 15 g KCl are
placed in 75 g water?
%KCl = 15g x 100/(15 g + 75 g) = 17%
What is the molarity of the KCl if 90 mL of
solution are formed?
mole KCl = 15 g x (1 mole/74.5 g) = 0.20 mole
molarity = 0.20 mole/0.090L = 2.2 M KCl
Gas Pressure and Solubility
• Quiz: The amount of dissolved oxygen in a
mountain lake at
10,000 ft and 50oF is __?_ than the
amount of dissolved oxygen in a lake near
sea level at 50oF.
• Answer: Less at higher altitude because
less pressure.
• A Coke at room temperature will have __?_
carbon dioxide in the gas space above the
liquid than an ice cold bottle.
• Answer: More gas, because the warm coke
can hold less of the gas in solution.
Gas Pressure and Solubility
• Hyperbaric therapy, which involves
exposure to oxygen at higher than
atmospheric pressure may be used to
treat hypoxia (low oxygen supply in
the tissues). Explain how the
treatment works.
• Answer: The increase in pressure in
the chamber will cause more gases to
enter into lungs.
Freezing Point Depression
Calculate the Freezing Point of a 4.00 molal
glycol/water solution.
Kf = 1.86 oC/molal (See Kf table)
Solution
∆TFP = Kf • m • i
= (1.86 oC/molal)(4.00 m)(1)
∆TFP = 7.44
FP = 0 – 7.44 = -7.44 oC
(because water normally freezes at 0)
Freezing Point Depression
At what temperature will a 5.4 molal solution
of NaCl freeze?
Solution
∆TFP = Kf • m • i
∆TFP = (1.86 oC/molal) • 5.4 m • 2
∆TFP = 20.1 oC
FP = 0 – 20.1 = -20.1 oC
Freezing Point Depression
Calculate the Freezing Point of a 4.00 molal
glycol/water solution.
Kf = 1.86 oC/molal (See Kf table)
Solution
∆TFP = Kf • m • i
= (1.86 oC/molal)(4.00 m)(1)
∆TFP = 7.44
FP = 0 – 7.44 = -7.44 oC
(because water normally freezes at 0)