Solutions

Chapter 12

Types of Mixtures

• Solutions– Solvent dissolves a solute

• Suspensions– The solute particles are too large to be

dissolved. They settle out unless the mixture is constantly stirred.

• Colloids– Particles intermediate in size between

solutions and suspensions.

Solutions

• Solvent– The component of a solution that does the

dissolving.

• Solute– The component of a solution that is dissolved.

• Soluble– Capable of being dissolved.

• Insoluble– Incapable of being dissolved.

Solutions

Solute State Solvent State example

Gas Gas Air

Gas Liquid CO2 in water

Liquid Liquid Alcohol in water

Solutions

Solute State Solvent State Example

Liquid Solid Hg in Ag and Sn (fillings)

Solid Liquid Sugar in water

Solid Solid Cu in Ni

Examples of Solutions• Sugar Water

– Sugar in water (aqueous solution)

• Salt Water– Salt in water (aqueous solution)

• Alloys– Brass (zinc and copper)– Sterling Silver (silver and copper)– Stainless Steel (iron, chromium, and

nickel)

• Air– Oxygen in nitrogen

Suspensions

• Oil and vinegar– Nonpolar and polar

• Oil and water– Nonpolar and polar

• Muddy water– Large particles

Colloids

• Paint• Gelatin• Milk• Mayonnaise• Shaving and whipped cream• Fog• Cheese• butter

Tyndall Effect• Light can be scattered by colloidal particles.

• Used to distinguish a solution from a colloid.– That is why it is hard to

see in the fog, because the light is scattering and is not focused in one direction.

Electrolytes

• Substances that dissolve in water that can conduct current.-Ionic Compounds-

NaCl, CaCl2, MgCl2, KCl

We need electrolytes in our bodies to keep our nervous system and muscular systems functioning properly.

Video

The Assault on SaltSalt vs. Sea Salt

Solubility

• Unsaturated solution– A solution in which more solute can still dissolve.

• Saturated Solution– A solution in which no more solute can dissolve.– **A solution can be saturated when at a warmer

temperature and then when cooled, form a supersaturated solution in which crystals form (rock candy)

• Solubility– The amount of substance required to form a saturated

solution with a specific amount of solvent at a specific temperature.

Solubility curves

Solubility curves Tutorial

Y axis- grams of solute/100g water

X axis temperature

Positive slopes- the higher the temperature, the more solute that will dissolve.

Negative slopes- the higher the temperature, the less solute that will dissolve.

What influences strength of a solution?

Sample Volume of water Mass of sucrose Mass of salt

A 50 mL 5 g 0

B 50 mL 0 5g

C 50 mL 10 g 0

D 50 mL 0 10g

Concentrations Of Solutions

• Molarity– The number of moles

of solute in 1 L of solution.

– Mol– L– Represented by M

• Molality– The number of moles

of a solute per kg of solvent.

– Mol– Kg– Represented by m

What influences strength of a solution?

Sample Mass of sucrose

Mass of salt Moles of solute

Molarity of solution

A 5 g 0 0.015 .30 M

B 0 5g 0.086 1.72

C 10 g 0 0.029 .58 M

D 0 10g 0.171 3.42

Practice Molarity

• You have 3.50 L of solution that contains 90.0 g of sodium chloride, NaCl. What is the molarity of the solution?

M= mol

L

90.0 g NaCl x 1 mol NaCl = 1.54 mol

58.44 g

1.54 mol NaCl = 0.440 M NaCl

3.50 L solution

Practice Molarity

• You have 0.8 L of a 0.5 M HCl solution. How many moles of HCl does this solution contain? How many grams of HCl would that be?

M = mol

L

0.5 = mol

0.8

Mol = 0.4 mol NaCl x 36.46 g/mol = 14.58 g HCl

Practice Molarity• To produce 40.0 g of Ag2CrO4 you will

need at least 23.4 g of K2CrO4 in solution as a reactant. All you have on hand is 5 L of a 6.0 M K2CrO4 solution. What volume of the solution is needed to give you the 23.4 g K2CrO4 needed for the reaction?

1 mol K2CrO4 = 194.2 g You need at least 23.4 g x 1 mol = 0.120 mol K2CrO4

194.2 g

M= mol therefore, 6.0 M = 0.120 L = 0.020 L K2CrO4 solution

L L

Practice molality

• A solution was prepared by dissolving 17.1 g of C12H22O11, sucrose, in 125 g of water. Find the molal concentration of the solution.

M = mol solute

kg of solvent

17.1 g C12H22O11 = 1 mol = 0.0500 mol C12H22O11

342.34 g

0.0500 mol = 0.400m C12H22O11

0.125 kg H2O

Practice molality

• A solution of Iodine, I2, in CCl4 is used for chemical tests. How much iodine must be added to prepare a 0.480m solution of iodine in CCl4 if 100.0 g of CCl4 is used?

100.0 g = .100 kg CCl4

0.480 m = x mol = 0.0480 mol I2

0.1 kg H2O

0.0480 mol I2 x 253.8 g I2 = 12.2 g I2

1 mol