1.
Solution Equilibrium and Factors affecting Solubility – Expressing Solution Concentration

2. Solution Equilibrium and Factors Affecting Solubility
When a solid solute is placed into a liquid solvent, the solid begins to dissolve into the solvent
But, as it dissolves, the solution becomes gradually more and more concentrated
Once the solution becomes more concentrated, some of the solute can recrystallize
When the rate of dissolution is equal to the rate of recrystallization, dynamic equilibrium has
been reached (like the dynamic equilibrium between phase changes talked about in the last chapter)
When a solution has reached this dynamic equilibrium, it is called a saturated solution
If you add solute it will not dissolve
If the solution does not have enough solute to reach the dynamic equilibrium,
it is called an unsaturated solution
If you add solute it will dissolve (until the dynamic equilibrium is reached).

3. Supersaturated Solutions
This occurs when the solution contains more than the amount of solute necessary for dynamic equilibrium
These solutions are unstable, excess solute will often precipitate out
But, sometimes they can exist for an extended time period, but when a piece of the solute is added, all of the excess solute will precipitate.

4. The Temperature Dependence of the Solubility of Solids
The solubility of solids in water often increases with increased temperature
To purify a solid, an excess of the solid can be dissolved in water at a higher temperature, and then the temperature can be lowered, leading to the solution becoming supersaturated, the excess solid precipitates out of the solution
If it cools slowly, crystals form, their structure tends to reject impurities, leaving a ‘purer’ solid

5. Factors Affecting the Solubility of Gases in Water
As temperature rises, the solubility of gases in liquids decreases, (that is why bubbles of gases are released before water boils)
This is why warm soda bubbles more than cold soda, and why fish are more lethargic when the water is warm (lower oxygen)
Practice
A solution is saturated in both nitrogen gas and potassium bromide at 75 degrees Celsius. When the solution is cooled to room temperature, what is most likely to happen?
1. Some nitrogen gas will bubble out of the solution
2. Some potassium bromide will precipitate out of the solution
3. Some nitrogen gas will bubble out of the solution and some potassium bromide will precipitate out of the solution
4. Nothing

6. The Effect of Pressure on Solubility of Gases in Water
The higher the pressure of a gas above the liquid is, the more soluble it is
In soda, the high pressure of gaseous CO2 results in much of the CO2 gas remaining in the solution. When the can is opened, the pressure decreases, allowing CO2 to bubble out
When the solution and the gaseous CO2 are in a sealed container, dynamic equilibrium occurs because the CO2 entering the solution equals the CO2 that is leaving (into the space above)
But, when the volume for the gaseous CO2 is decreased, the pressure of the CO2 increases, making more of the gaseous CO2 enter the solution, until that equilibrium is attained again.

7. Sgas is the solubility of the gas
Henry’s Law
Sgas = kHPgas
With this equation the solubility of gases with increasing pressure can be quantified
Sgas is the solubility of the gas
kH is a constant of proportionality, depends on specific solute and solvent, as well as on temperature
Pgas is the partial pressure of the gas
What pressure of Carbon dioxide is required to keep the carbon dioxide concentration in a bottle of club soda at 0.12 M at 25 degrees Celsius?
SCO2 = kH,CO2 PCO2
    PCO2 = SCO2/kH,CO2
    PCO2 = 0.12 M/ 3.4 x 10-2 M/atm
    PCO2 = 3.5 atm

8. Expressing Solution Concentration
Dilute Solute – contains small quantities of solute relative to the amount of solvent
Concentrated Solute – contains large quantities of solute relative to the amount of solvent
Solution concentration can be reported through
Molarity
Molality
Parts by Mass
Parts by Volume
Mole Fraction
Mole Percent

9. Molarity Moles of solute per liters of solution
Molarity is volume based, volume varies with temperature, and so, molarity does as well
Volume is greater at higher temperatures, a solution at room temperature may be slightly less than if the solution was at a higher temperature

10. Molality Moles of solute per kilogram of solvent, not solution
Independent of temperature (not volume based)

11. Parts by Mass Parts by Volume
It is a ratio of mass of solute to mass of solution, which is all multiplied by the same factor
Ex. Percent by mass, the multiplication factor is 100
For more dilute solutions, parts per million, with a multiplication factor of 10^6, or parts per billion, with a factor of 10^9
Parts by Volume
It is a ratio of the volume of the solute to the volume of solution, which is multiplied by a multiplication factor
These factors are the same as the above ones

12. Practice
What mass of sucrose (C12H22O11), in g, is contained in 355 mL of a soft drink that is 11.5% sucrose by mass? (Density of 1.04g/mL)
355 𝑚𝐿 𝑠𝑜𝑙𝑛 × 1.04𝑔 1𝑚𝐿 × 11.5 𝑔 C12H22O 𝑔 = g C12H22O11
A water sample is found to contain the pollutant chlorobenzene with a concentration of 15 ppb. What volume of this water contains 5.00 x 102 mg of chlorobenzene? (Density of 1.00 g/mL).
Ppb = (mass of solute/mass of solution) x 109
15 x 10-9 = g/mass of solution
= 3.33 x 107 g
3.33 × 𝑔 × 1 𝑚𝐿 1.00 𝑔 =3.33 × 𝑚𝐿

13. Mole Fraction and Mole Percent
The mole fraction is the moles of solute divided by the moles of solute and the moles of solvent
Multiplying the mole fraction (Xsolute) by 100% will give you the mole percent

14. Practice
A solution is prepared by dissolving 17.2 g of ethylene glycol (C2H6O2) in kg of water. The final volume of the solution is 515 mL. For this solution, calculate the concentration in each unit. A. Molarity B. Mole Fraction C. Mole percent

15. Practice (cont.) A.
Given; 17.2 g of C2H6O2 , 515 mL (0.515 L) of solution, the solution contains kg of water
Convert from grams to moles (molar mass of C2H6O2 is grams)
17.2 𝑔 𝐶 2 𝐻 6 𝑂 2 × 1 𝑚𝑜𝑙 𝐶 2 𝐻 6 𝑂 𝑔 𝐶 2 𝐻 6 𝑂 2 = 𝑚𝑜𝑙 𝐶 2 𝐻 6 𝑂 2
Use Molarity formula
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑀 = 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝑜𝑙) 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝐿)
Plug in moles of solute and L of solution
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦= 𝑚𝑜𝑙 𝐶 2 𝐻 6 𝑂 𝐿 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 =0.538 𝑀

16. Practice (cont.)
B. Given; 17.2 g of C2H6O2 , 515 mL (0.515 L) of solution, the solution contains kg of water
First solve for moles of H2O using the given of kg (500g) of water; (molar mass of water is g H2O) and for moles of C2H6O2 (solved for in part A)
500 𝑔 𝐻 2 𝑂 × 1 𝑚𝑜𝑙 𝐻 2 𝑂 𝑔 𝐻 2 𝑂 =27.75 𝑚𝑜𝑙 𝐻 2 𝑂
17.2 𝑔 𝐶 2 𝐻 6 𝑂 2 × 1 𝑚𝑜𝑙 𝐶 2 𝐻 6 𝑂 𝑔 𝐶 2 𝐻 6 𝑂 2 = 𝑚𝑜𝑙 𝐶 2 𝐻 6 𝑂 2
Use formula for mole fraction
𝑋 𝑠𝑜𝑙𝑢𝑡𝑒 = 𝑛 𝑠𝑜𝑙𝑢𝑡𝑒 𝑛 𝑠𝑜𝑙𝑢𝑡𝑒 𝑛 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑚𝑜𝑙 𝑚𝑜𝑙 𝑚𝑜𝑙 =9.89 × 10 −3 C.
𝑚𝑜𝑙 %= 𝑋 𝑠𝑜𝑙𝑢𝑡𝑒 ×100%
9.89 × 10 −3 ×100% =0.989%

17. More Practice
What is the molarity of 10.5% by mass glucose (C6H12O6) solution? (The density of the solution is 1.03 g/mL)
For this equation, make the solution 100g for simple math, that would make 10.5 grams of it glucose. Then convert from grams to moles (molar mass of 𝐶 6 𝐻 12 𝑂 6 is g)
10.5 𝑔 𝐶 6 𝐻 12 𝑂 6 × 1 𝑚𝑜𝑙 𝐶 6 𝐻 12 𝑂 𝑔 𝐶 6 𝐻 12 𝑂 6 = 𝑚𝑜𝑙 𝐶 6 𝐻 12 𝑂 6
You also must convert the grams of solution (which we earlier stated was 100g) to liters. For this, use density as a conversion factor into mL, then convert to Liters.
100 𝑔 𝑠𝑜𝑙𝑛 × 1 𝑚𝐿 1.03 𝑔 × 10 −3 𝐿 𝑚𝐿 = 𝐿 𝑠𝑜𝑙𝑛
Use the molarity equation
𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑀 = 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 (𝑚𝑜𝑙) 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝐿)
𝑚𝑜𝑙 𝐶 6 𝐻 12 𝑂 𝐿 𝑠𝑜𝑙𝑛 =0.600 𝑀 𝐶 6 𝐻 12 𝑂 6