|Matters and Properties of Matters|
|Atomic Structure with Examples|
|The Mole Concept with Examples|
|Gases with Examples|
|Chemical Reactions with Examples|
|Nuclear Chemistry (Radioactivity)|
|--Solubility and Factors Affecting Solubility|
|--Concentration with Examples|
|--Dilution and Density of Solutions|
|--Concentration of Ions with Examples|
|--Properties of Solutions and Solution Calculations|
|--Solutionns Cheat Sheet|
|Acids and Bases|
|Rates of Reactions (Chemical Kinetics)|
|Exams and Problem Solutions|
Solubility and Factors Affecting Solubility
Solubility is the amount of solute in 100 cm3 (100 mL) solvent.
Example: In 100 g water at 20 0C, 36 g salt can be dissolved. Thus solubility of salt at 20 0C 100 g water is 36g/100g
Solubility is characteristic property of matters, we can distinguish matters by knowing their solubility values at same temperature. Table given below shows solubility of some matters at 20 0C;
Example: 25 g X salt is put into 40 cm3 water at 20 0C. After dissolution process, 15 g X stays undissolved at the bottom of the tank. Find solubility of X at 20 0C 100 g water.
25-15=10 g X dissolves in 40 cm3 water.
40 cm3 water dissolves 10 g X
100 cm3 water dissolves ? g X
?=25 g X
Solubility of X in 100 g water at 20 0C is 25g/100 cm3
Example: If solubility of KCl in water at room temperature is 25g/100cm3, which ones of the following solutions are saturated.
I. 50 g water - 15 g KCl
II. 30 g water - 10 g KCl
III. 20 g water - 3 g KCl.
I. 100 g water dissolves 25 g KCl
50 g water dissolves X g KCl
X=12,5 g KCl dissolves. Thus, 15-12,5=2,5 g KCl stays undissolved at the bottom of tank.
II. 100 g water dissolves 25 g KCl
30 g water dissolves X g KCl
X=7,5 g KCl dissolves. Thus, 10-7,5=2,5 g KCl stays undissolved at the bottom of tank.
III. 100 g water dissolves 25 g KCl
20 g water dissolves X g KCl
X=5 g KCl dissolves. Thus, if we add 5-3=2 g KCl it can also be dissolved in 20 g water.
Thus, I and II are saturated solutions and III is unsaturated solution.
Factors Affecting Solubility
Solvent and types of solute, temperature, pressure and common ion effect are factors affecting solubility.
1) Solvent and Types of Solute:
Some of the matters dissolve better by increasing temperature, on the contrary some of them dissolve better by decreasing temperature. Solutions taking heat are called endothermic solutions and solutions giving heat are called exothermic solutions.
a) Endothermic Solutions: Most of the solids need heat to dissolve like;
X(s) + Heat → X(aq)
In this type of solutions, solubility increases with increasing temperature.
a) Exothermic Solutions: Most of the gases give heat to dissolve like;
Y(g) → Y(aq) + Heat
In this type of solutions, solubility decreases with increasing temperature.
Example: Look at following reactions and find which ones of them have solubility increasing with temperature.
I. XY(s) + Heat → X+2(aq) + Y-2
II. XY2(s) → X+2(aq) + 2Y-1(aq) + Heat
III. XY3(s) → X+3(aq) + 3Y-1(aq) + Heat
Solution: In endothermic solutions, solubility increases with increasing temperature. Thus, since I. is endothermic reaction solubility of it increases with increasing temperature. II and III are exothermic reactions, so solubility of them decreases with temperature.
Pressure changes only solubility of gases in liquids. Solubility of gases in liquids increases with increasing partial pressure and decreases with decreasing partial pressure.
4) Common Ion:
Solubility of any solid matter having common ions with solvent is lower than solubility in pure solvents. For example, solubility of AgNO3 in pure water is larger than solubility of AgNO3 in NaNO3 since they have common ion NO3-.
Example: Compare solubility of NaCl in following solvents;
I. Pure water
Solubility of NaCl in pure water is larger than others since they have no common ions. NaCl has one common ion with NaNO3 and 2 common ion with Na2SO4. Increasing in the number of common ion decreases solubility. Thus;
Factors Effecting Speed of Solvation