Characteristics of solutions

• Solution – homogeneous mixture

a) parts of a solutioni) solute – substance being

dissolvedii) solvent – substance doing

dissolving

both can be either solid, liquid, or gas

Solubility

• Soluble – substance can dissolve in a solvent

ex: salt in water

• Insoluble – substance cannot dissolve in a solvent

ex: sand in water

Solvation In Aqueous Solutions• Solvation – process of surrounding

solute particles with solvent particles

Why are some substances soluble in a solvent and some others are not?

must be compatibility between solute and solvent

Dissolution of sodium Chloride

“like dissolves like”

• Defn – rule used to determine if substance will dissolve in another

- based on attractive forces between solute and solvent

“like dissolves like”

• polar solvents – dissolve polar molecular compounds and ionic compoundsex: salt and water, alcohol and vinegar

• nonpolar solvents – dissolve nonpolar compounds onlyex: oil and gasoline

Factors Affecting Rate of Solvation• How can you dissolve something

faster???

a) increase temp of solvent

this accelerates particles creating more particle collisions

Factors Affecting Rate of Solvation

b) agitate the solution more particle collisions between

solute and solvent

c) Increase surface area of solutebreaking into smaller pieces allows more solute to be in contact w/ solvent

Solubility

• Defn – max amt of solute that can dissolve in a solvent at a specific temp

how much solute can be put into solvent?

Unsaturated Solution

• Defn – less than max amt of solute dissolved

if I put sugar into water and all sugar is dissolved, solution is unsaturated

Saturated Solution

• Defn – contains max amt of solute dissolved

if I put sugar into water and not dissolves (you can see the sugar), the solution is saturated

Supersaturated Solution

• Defn – contains more solute than saturated solution at the same conditions

a saturated solution made at high temp cools slowly. Slow cooling allows excess solute to remain dissolved in solution at lower temperature

very unstable

Solubility Curve (generic)Saturated-Line represents max amount solute that will dissolve at a given temperature

Temperature

Solu

bili

ty(g

solu

te/

10

0 g

H2O

)

Unsaturated(below line)

Supersaturated(above line)

How does temp affect solubility?• The higher the temp, higher the

solubility

(for most cases)

Solution Concentration

• Concentration – how much solute dissolved in amount of solvent

what is difference between concentrated and diluted?

Concentrated vs. Dilute

Concentration

• 3 different units of concentration

a) percent by mass

b) molarity (M)

c) molality (m)

Percent by mass

• Formula

OR

100 xsolution of mass

solute of mass

100 x solvent mass solute mass

solute mass

Percent by mass

• Ex prob: If 3.6 g NaCl is dissolved in 100 g H2O, what is the percent by mass?

What is the solute? NaCl

What is the solvent? H2O

Percent by mass

• Mass of solute (NaCl) =3.6 g

• Mass solvent (H2O) =

100 g• Mass solution =

3.6 + 100 = 103.6 g

Percent by mass

Percent by mass = 3.6 g x 100

103. 6 g

= 3.5 % NaCl

Molarity

• Defn - # of moles per liter of solution

• Formula mol solute L solution

• unit mol = M (capital M) L

Molarity Ex prob #1

• A solution has a volume of 250 mL and has 0.70 mol NaCl. What is the molarity?

2.8 mol/Lor

2.8 M0.250 L

=0.70 mol

Molarity ex prob #2

• What is the molarity of a solution made of 47.3 g NaOH in 500 mL water?

step 1: convert grams to moles47.3 g NaOH

40 g NaOH

1 mol NaOH=1.1825 mol NaOH

Molarity ex prob #2

Step 2: divide moles by volume (L)

2.37 mol/L NaOHor

2.37 M NaOH0.500 L

=1.1825 mol

Molarity ex prob #3

• How many moles of solute are present in 1.5 L of 2.4 M NaCl? How many grams?

# moles = volume x molarity

1.5 LL

2.4 mol NaClx = 3.6 mol NaCl

Molarity ex prob #3

• moles to grams

3.6 mol NaCl

1 mol NaCl

58.5 g NaCl= 210.6 g NaCl

Diluting Solutions

• Defn – add more solvent to original solution

• FormulaM1V1 = M2V2

M1 is more concentrated than M2

Diluting Solutions

• What volume of a 2.0 M stock solution is needed to make 0.50 L of a 0.300 M solution?

M1=

V1=

M2=

V2=

2.0 M 0.300 M

? 0.50 L

(2.0 M) V1 = (0.300 M)(0.50 L)

V1 = 0.075 L

Diluting Solutions

• If you dilute 20.0 mL of a 3.5 M solution to make 100.0 mL of solution, what is the molarity of the dilute solution?M1=

V1=

M2=

V2=

3.5 M ?

20.0 mL 100 mL

(3.5 M)(20.0 mL) = M2 (100.0 mL)

M2 = 0.7 M

Molality

• Defn - # moles of solute in one kg solvent

• Formula mol solutekg solvent

• Units mol = m (lower case m)

kg

Molality ex problem

• What is the molality of a solution with 8.4 g NaCl in 255 g of water?

Step 1: convert grams to moles

8.4 g NaCl

58.5 g NaCl

1 mol NaCl= 0.14 mol NaCl

Molality ex problem

Step 2: divide by mass (kg)

0.55 mol/kg NaClor

0.55 m NaCl0.255 kg

=0.14 mol NaCl

Colligative Properties of Solutions

• Solutes affect the physical properties of their solvents

• Colligative properties (defn) – properties that depend only on the number of solute particles present, not their identity

• Ex: boiling point, freezing point

Electrolytes

• Defn – substances that break up (ionize) in water to produce ions; can conduct electricity- consist of acids, bases, ionic compounds

Ex: NaCl Na1+ + Cl1-

H2SO4 2 H+ + SO42-

Nonelectrolytes

• Defn – do not break up (ionize) in water, they stay the same; doesn’t conduct electricity- usually molecular/covalent compounds

Ex: sugar C6H12O6 C6H12O6

ethanol C2H5OH C2H5OH

Determining # of solute particles

• For ionic cmpds/acids sum # moles of ions

ex: NaCl 1 Na+ + 1 Cl1- = 2 particles

CaBr2 1 Ca2+ + 2 Br1- = 3 particles

Determining # of solute particles

• For covalent compounds # is always 1

ex: sugar C6H12O6 1 particle

ethanol C2H5OH 1 particle

Freezing Point Depression

• Defn (ΔTf) – difference in temp between solution’s fp and pure solvent’s fp

• Formula ΔTf = Kf x m x i

# particlesmolalitymolal fp constant

Ex problem

• What is the freezing pt of water if 12.3 g NaCl is added to 200 g water?(Kf = 1.86 °C/m, fp = 0°C)

1 mol NaCl

58.5 g NaCl

12.3 g NaCl = 0.21 mol NaCl

m = 0.21 mol NaCl

0.200 kg water= 1.05 m

Ex problem

• ΔTf = Kf x m x i

ΔTf = (1.86 °C/m)(1.05 m) (2)

= 3.91 °C

New f.p. = 0°- 3.91° = -3.91°C

Boiling Point Elevation

• Defn (ΔTb) – difference in temp between solution’s bp and pure solvent’s bp

• Formula ΔTb = Kb x m x i

# particlesmolalitymolal bp constant

Ex problem

• What is the new boiling point of acetone if 13.7 g C6H12O6 is dissolved in 200 g acetone? (Kb = 1.71°C/m, bp = 56°C)

1 mol C6H12O6

180 g C6H12O6

13.7 g C6H12O6 = 0.076 mol

m = 0.076 mol C6H12O6

0.200 kg acetone= 0.381 m

Ex problem

• ΔTb = Kb x m x i

ΔTb = (1.71 °C/m)(0.381 m)(1)

= 0.65 °C

New bp = 56°+ 0.65° = 56.65°C