Reading graph: at 38 °C the solubility of copper sulphate, CuSO4, is

28g of anhydrous salt per 100g of water.

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

Step 2: Use this constant to find the solubility (concentration) when P is 0.78 atm:

44 16.8 10

6.8 101.00

c x Mk x M atm

P atm

4 1 4(6.8 10 )(0.78 ) 5.3 10c kP x M atm atm x M

How do I get sugar to dissolve faster in my iced tea?

Stir, and stir, and stir

Add sugar to warm tea then add ice

Grind the sugar to a powder

Fresh solvent contact and interaction with solute

Greater surface area, more solute-solvent interaction

Faster rate of dissolution at higher temperature

http://academic.pgcc.edu/~ssinex/Solutions.ppt

Revision Units of Concentrationsamount of solute per amount of solvent or solution

Percent (by mass) =g solute

g solutionx 100

g solute

g solute + g solvent

x 100=

Molarity (M) =

moles of solute

volume in liters of solution

moles = M x VL

http://academic.pgcc.edu/~ssinex/Solutions.ppt

Exampleshttp://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 ofsolution 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 at10,000 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.

Calculate the Freezing Point of a 4.00 molal Calculate the Freezing Point of a 4.00 molal glycol/water solution.glycol/water solution.

KKff = 1.86 = 1.86 ooC/molal (See KC/molal (See Kff table) table)

SolutionSolution

∆∆TTFPFP = K = Kff • m • i • m • i

= (1.86 = (1.86 ooC/molal)(4.00 m)(1)C/molal)(4.00 m)(1)

∆∆TTFP FP = 7.44 = 7.44

FP = 0 – 7.44 = -7.44 FP = 0 – 7.44 = -7.44 ooCC(because water normally freezes at (because water normally freezes at 0)0)

Freezing Point DepressionFreezing Point Depression

At what temperature will a 5.4 molal At what temperature will a 5.4 molal solution of NaCl freeze?solution of NaCl freeze?

SolutionSolution

∆∆TTFPFP = K = Kff • m • i • m • i

∆ ∆TTFPFP = (1.86 = (1.86 ooC/molal) • 5.4 m • 2C/molal) • 5.4 m • 2

∆ ∆TTFP FP = 20.1= 20.1 ooCC

FP = 0 – 20.1 = -20.1 FP = 0 – 20.1 = -20.1 ooCC

Freezing Point DepressionFreezing Point Depression

Calculate the Freezing Point of a 4.00 molal Calculate the Freezing Point of a 4.00 molal glycol/water solution.glycol/water solution.

KKff = 1.86 = 1.86 ooC/molal (See KC/molal (See Kff table) table)

SolutionSolution

∆∆TTFPFP = K = Kff • m • i • m • i

= (1.86 = (1.86 ooC/molal)(4.00 m)(1)C/molal)(4.00 m)(1)

∆∆TTFP FP = 7.44 = 7.44

FP = 0 – 7.44 = -7.44 FP = 0 – 7.44 = -7.44 ooCC(because water normally freezes at (because water normally freezes at 0)0)

Freezing Point DepressionFreezing Point Depression