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

Learning Check

1. How many grams of NaCl will dissolve in 100 g of H2O at 90°C?

2. 50 g of KCl is dissolved in 100 g of water at 50°C. Is the solution saturated, unsaturated or supersaturated?

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

• Polar molecules dissolve other polar molecules and ionic compounds.

• Nonpolar molecules dissolve other nonpolar molecules.

• Alcohols, which have characteristics of both polar & nonpolar, 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

 

     

polar non polar alcohol

Alcohols are organic, covalent molecules with an –OH group. Alcohol names end with “-ol.”

ionic

polar

non polar

non polar

non polar

ionic

ionic

alcohol

alcohol

and alcohols

and alcohols

and other alcohols

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 Increasing• Vapor Pressure Lowering• Conductivity Increasing

More particles/ions = greater change

Learning Check1. Which substance will provide the greatest change in freezing point of water?

A.NaCl B. CaCl2 C. C6H12O6 D. H2O

2. Which of the following reflect colligative properties? (I) A 0.5 m NaBr solution has a higher vapor pressure than a 0.5 m BaCl2 solution.

(II) A 0.5 m NaOH solution freezes at a lower temperature than pure water

(III) Pure water freezes at a higher temperature than pure methanol.

A. only I B. only II C. only III D. I and II E. I and III

2 ions 3 ions 1 particle no change in H2O

2 ions 3 ions

= vapor pressure lowering (more ions lower pressure)

2 ions 0 ions

= freezing point depression

Freezing point is a physical property, not a colligative property

3. A student measured the conductivity in water, of unlabeled liquids, after each added drop. The following graph was produced...

a. Identify the line that represents:– aluminum chloride – water– magnesium chloride – sugar– sodium chloride

b.Which line could

also represent

potassium iodide?

# of Drops

Co

nd

uct

ivity

(µ

s/cm

)

AlCl3, 4 ions

H2O, no change in H2O

MgCl2, 3 ions

C6H12O6, 1 particle

NaCl, 2 ions

H2O

C6H12O6

NaCl

MgCl2

AlCl3

KI, 2 ions = same as NaCl

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

40.00

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 ____________143.32 = 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

Learning check

Calculate the molarity of a solution prepared by dissolving 25.6 g NaC2H3O2 in enough water to make 200.0 mL solution.

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

294.20

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?

Learning Check

Notes-Acids and Bases

Acids and Bases

Arrhenius ACIDS – produces hydrogen ions in aqueous solutions, sour taste, low pH, and the fact that they turn litmus paper red

HCl (aq) H+ (aq) + Cl- (aq)

Arrhenius BASES – produces hydroxide ions in aqueous solutions, bitter taste, slippery feel, high pH, and the fact that they turn litmus paper blue

NaOH (aq) Na+ (aq) + OH- (aq)

Arrhenius definition – limits the concept of a base

Bronsted – Lowry definition – gives a broader definition of a base Bronsted – Lowry ACID – a proton (H+) donor Bronsted – Lowry BASE – a proton (H+) acceptor

General Reaction –

HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Acid Base Conjugate Conjugate Acid Base

Conjugate Base – everything that remains of the acid molecule after a proton is lost

Conjugate Acid – the base with the transferred proton (H+)Conjugate Acid – Base Pair – two substances related to

each other by the donating and accepting of a single proton

proton donor

proton acceptor

Examples: Finish each equation and identify each member of the conjugate acid –base pair.

H2SO4 (aq) + H2O (l) HSO4-1

(aq) + H3O+ (aq)

Acid Base Conjugate Base

Conjugate Acid

CO32- (aq) + H2O (l) HCO3

-1(aq) + OH-

(aq)

Base Acid Conjugate Acid

Conjugate Base

The hydronium ion, H3O+, forms when water behaves as a base. This happens when the two unshared pairs of electrons on O bond covalently with the H+.

Learning checkWrite the conjugate ACID

a.NH3

b.HCO3-1

Write the conjugate BASE

a.H3PO4

b.HBr

Finish each equation and identify each member of the conjugate acid –base pair.

a. H2SO3 (aq) + H2O (l)

b. SO4-2

(aq) + H2O (l)

Water as an Acid and a Base

Amphoteric – a substance that can behave as either an acid or a base

- water is the most common amphoteric substance

Ionization of Water –

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

In the shorthand form:

H2O (l) H+ (aq) + OH- (aq)

Ion-product constant – Kw refers to the ionization of water

Kw = [H+][OH-]

At 25C, Kw = [H+][OH-] = [1.0 x 10-7] [1.0 x 10-7] = 1.0 x 10-14

If [H+] increases, the [OH-] decreases, so the products of the two is still 1.0 x 10-14.

There are three possible situations –1. A neutral solution, where [H+] = [OH-]2. An acidic solution, where [H+] [OH-]3. A basic solution, where [H+] [OH-]

[ ] = concentration[H+] = hydrogen ion concentration in M[OH-] = hydroxide ion concentration in M

Example: Calculate [H+] or [OH-] as required for each of the following solutions at 25C, for each solution state whether it is neutral, acidic, or basic.

a. 1.0 x 10-5 M OH- b. 10.0 M H+

Kw = [H+][OH-]

1 x 10-14 = [H+][1.0 x 10-5 M]

[H+] = 1.0 x 10-9 M

BASIC

Kw = [H+][OH-]

1 x 10-14 = [10.0 M][OH-]

[OH-] = 1.00 x 10-15 M

ACIDIC

pH scalepH scale – because the [H+] in an aqueous solution is

typically small, logarithms are used to express solution acidity

pH = -log [H+] pOH = -log [OH-]

Graphing calculator Non graphing calculator1. Press the +/- key 1. Enter the [H+]2. Press the log key 2. Press the log key3. Enter the [H+] 3. Press the +/- key

Significant Figure Rule – The number of places to the right of the decimal for a log must be equal to the number of significant figures in the original number.

pH pH ScalScal

ee

Example – Calculate the pH or pOH

a. [H+] = 5.9 x 10-9 M b. [OH-] = 2.4 x 10-6 M

pH = - log [H+]

pH = - log (5.9 x 10-9 M)

pH = 8.23

pOH = - log [OH-]

pOH = - log (2.4 x 10-6 M)

pOH = 5.62

Since Kw = [H+][OH-] = 1.0 x 10-14 ,

pH + pOH = 14.00

Example - The pH of blood is about 7.4. What is the pOH of blood?

pH + pOH =14.00

7.4 + pOH = 14.00

pOH = 6.6

In order to calculate the concentration from the pH or pOH,

[H+] = 10-pH [OH-] = 10-pOH

Graphing calculator Non-graphing calculator1. Press the 2nd 1. Enter the pH

function, then log 2. Press the +/- key2. Press the +/- key 3. Press the

inverse3. Enter the pH log key

Example - The pH of a human blood sample was measured to be 7.41. What is the [H+] in blood?

[H+] = 10-pH

[H+] = 10-7.41

[H+] = 3.9 x 10-8 M

Example – The pOH of the water in a fish tank is found to be 6.59. What is the [H+] for this water?

[OH-] = 10-pOH

[OH-] = 10-6.59

[OH-] = 2.6 x 10-7 M

Kw = [H+][OH-]

1 x 10-14 = [H+][2.6 x 10-7 M]

[H+] = 3.8 x 10-8 M

Learning check

• Determine the pH of a solution with a hydrogen ion concentration of 3.2 x10-12 M. 

• What is the [OH-] concentration of a solution with a hydrogen ion concentration of 8.9x10-4M?

• What is the pH of a solution with a hydroxide ion concentration of 5.7x10-10 M?

How Do We Measure pH?

• For less accurate measurements, one can use– Litmus paper

• “Red” paper turns blue above ~pH = 8

• “Blue” paper turns red below ~pH = 5

– An indicator

How Do We Measure pH?

For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

Strong Acids

• seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.

• These are, by definition, strong electrolytes and exist totally as ions in aqueous solution.

Strong Bases

• Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+).

• Again, these substances dissociate completely in aqueous solution, strong electrolytes

Strong, Weak, or Nonelectrolyte• Electrolytes are substances which, when dissolved in water, break up into

cations (plus-charged ions) and anions (minus-charged ions). We say they ionize. Strong electrolytes ionize completely (100%), while weak electrolytes ionize only partially (usually on the order of 1–10%). The ions in an electrolyte can be used to complete an electric circuit and power a bulb.

• Strong electrolytes fall into three categories: strong acids, strong bases, and soluble salts.

• The weak electrolytes include weak acids, weak bases and insoluble salts.

• Molecules are nonelectrolytes.Substance Classification - Strong acid, weak

acid, strong base, weak base, soluble salt, insoluble salt, molecule

Strong electrolyte, weak electrolyte, nonelectrolyte

sodium

hydroxide

   

acetic acid   

potassium

nitrate

   

hydrobromic

acid

   

silver chloride   

Carbon dioxide   

strong base strong electrolyte

weak acid weak electrolyte

soluble salt strong electrolyte

strong acid strong electrolyte

insoluble salt weak electrolyte

molecule nonelectrolyte

Learning checkSubstance Classification - Strong

acid, weak acid, strong base, weak base, soluble salt, insoluble salt, molecule

Strong electrolyte, weak electrolyte, nonelectrolyte

chloric acid    

barium carbonate    

nitric acid    

sulfurous acid    

strontium sulfate    

ethanol    

octane (gasoline)    

Titration

A known concentration of base (or acid) is slowly added to a solution of acid (or base).

Titration

A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

Titration of a Strong Acid with a Strong Base

From the start of the titration to near the equivalence point, the pH goes up slowly.

Titration of a Strong Acid with a Strong Base

Just before and after the equivalence point, the pH increases rapidly.

Titration of a Strong Acid with a Strong Base

At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

Titration of a Strong Acid with a Strong Base

As more base is added, the increase in pH again levels off.

Neutralization

Neutralization Reaction =

Acid + Base Salt + Water

Salt – ionic compound containing a positive ion other than H+ and a negative ion other than OH-

Buffered solutions – resists a change in its pH even when a strong acid or base is added to it

- A solution is buffered in the presence of a weak acid and its conjugate base

pH and pOH Calculations

H + O H -

pH pO H

[O H -] = 1 x 10 - 1 4

[H + ]

[H + ] = 1 x 10 - 1 4

[O H -]

p O H = 14 - p H

p H = 14 - p O H

pOH

= -l

og[O

H- ]

pH =

-log

[H+

]

[OH

- ] = 1

0-pO

H

[H+

] = 1

0-pH