Solutions Notes

Words to Know• Solution – homogenous mixture• Solvent – substance present in the largest amount• Solutes – substance present in the smallest amount• Aqueous solution – solutions with water as the

solvent• Concentration – the amount of solute in a given

volume of solution• Concentrated – large amount of solute dissolved in

solvent• Dilute – small amount of solute dissolved in solvent

• Saturated – a solution that contains as much solute as will dissolve at that temperature

• Unsaturated – a solution that hasn’t reached that limit of solute that will dissolve

• Supersaturated - a solution that contains more solute than should dissolve at that temperature

Effect of Temperature on Solubility

• Increasing the temperature of a solution, increases the amount of solute that can be dissolved

• Decreasing the temperature of a solution, causes the solute to recrystallize

Effect of Pressure on Solubility

• Pressure has a major effect on the solubility of gas-liquid systems

• An increase in pressure increases the solubility of a gas in the liquid

Generally, “like dissolves like.” Polar molecules dissolve other polar molecules and ionic compounds. Nonpolar molecules dissolve other nonpolar molecules. Alcohols, which have characteristics of both, tend to dissolve in both types of solvents, but will not dissolve ionic solids.

“Like dissolves like” – a solvent usually dissolves solutes that have polarities similar to itself

 SOLUTES

SOLVENTSWater CCl4 Alcohol

 NaCl

 

     

 I2 

     

 C3H7OH

 

     

 benzene

(nonpolar)

     

 Br2

 

     

 KNO3

 

     

 toluene (polar)

     

 Ca(OH)2

 

     

 methanol

 

     

 NH3

 

     

 CO2

 

     

Colligative propertiesColligative properties - the physical changes that result from adding solute to a solvent. Colligative Properties depend on how many solute particles are present as well as the solvent amount, but they do NOT depend on the type of solute particles.

• Boiling Point Elevation

• Freezing Point Depression

• Osmotic Pressure

• Vapor Pressure Lowering

Solution Composition - Mass Percent

Mass percent – describes a solution’s composition expresses the mass of solute present in a given mass of solution

Mass Percent = mass of solute x 100% mass of solution*

* mass of solution = mass of solute + mass of solvent

Example – A solution is prepared by mixing 1.00g of C2H5OH, with 100.0g of H2O. Calculate the mass percent of ethanol.

Given

mass of solute = 1.00 g

mass of solution = 100.0 g + 1.00 g = 101.0 g

Mass Percent = mass of solute x 100% mass of solution

Mass % = 1.00 g x 100 %101.0 g

Mass % = 0.990 %

Solution Composition – MolarityMolarity – measure of concentration - number of moles of solute per volume of solution in liters

Molarity = moles of solute = mol = M L of solution L

Example – Calculate the molarity of a solution prepared by dissolving 11.5 g NaOH in enough water to make 1.50L solution.

=x __________1

1.50 L NaOH

x ___________mol NaOH

40.0

1

g NaOH

11.5 g NaOH 0.192 M

Ex: Calculate the mass of solid AgCl formed when 1.50L of a 0.100M AgNO3 solution is reacted with excess NaCl.

NaCl + AgNO3 AgCl + NaNO3

1.50 L0.100 M

? g

1.50 L AgNO3 x ______________ L AgNO3

mol AgNO30.100

1 x ____________

mol AgNO3

mol AgCl

1

1

mol AgCl

g AgCl

1

x ____________143.2 = 21.5 g

Example – How many moles of Ag+ ions are present in 25mL of a 0.75M Ag2SO4 solution?

Ag2SO4 2 Ag+1 + SO4-2

25 mL Ag2SO4 L Ag2SO4

mL Ag2SO41000

1 x _____________ x ______________ L Ag2SO4

mol Ag2SO4

1

0.75 x ______________ mol Ag2SO4

mol Ag+12

1=

0.038 mol Ag+1

Standard Solution• Standard Solution – a solution whose

concentration is accurately known

Example – A chemist needs 1.0 L of a 0.200M K2Cr2O7 solution. How much solid K2Cr2O7 must be weighed out to make this solution?

x ______________g K2Cr2O7

mol K2Cr2O7

294.2

1

x ________________mol K2Cr2O7

1

0.200

L K2Cr2O7

1.0 L K2Cr2O7 =

59 g K2Cr2O7

DilutionDilution – process of adding more solvent to a solutionMoles of solute before dilution = Moles of solute after dilution

M1V1 = M2V2

Example: What volume of 16M H2SO4 must be used to prepare 1.5L of a 0.10M H2SO4 solution?

Given

V1 = ?

M1 = 16 M

V2 = 1.5 L

M2 = 0.10 M

V1 = M2V2

M1

_____ V1 = (0.10 M)(1.5 L)____________16 M

V1 = 0.0094 L

Given

V1 = 500.0 mL

M1 = 1.00 M

M2 = 17.5 M

V2 = ?

V2 = M1V1

M2

_____ V2 = (1.00 M)(500.0 mL)_______________17.5 M

V2 = 28.6 mL

Example: Prepare 500.0mL of 1.00 M HC2H3O2 from a 17.5 M stock solution. What volume of the stock solution is required?