Properties of Solutions

Classification of Matter

Solutions are homogeneous mixtures

A. Concentration

• The amount of solute in a solution.

• Describing Concentration– % by mass - medicated creams– % by volume - rubbing alcohol– ppm, ppb - water contaminants– molarity - used by chemists– molality - used by chemists

SoluteA solute is the dissolved substance in a solution.

A solvent is the dissolving medium in a solution.

Solvent

Salt in salt water Sugar in soda drinks

Carbon dioxide in soda drinks

Water in salt water Water in soda

MOLARITY

MOLARITY

A unit on concentration that is the ratio between moles of DISSOLVED substance and liters of solution

Molarity (M) =

moles of solute (mol)volume of solution (L)

There are many ways to represent molarity:

• Molarity• M• Molar• mol / L

MAKE SURE YOUR UNITS ARE CORRECT!!!

ALWAYS mol/L

MOLARITYA measurement of the concentration of a solutionMolarity (M) is equal to the moles of solute (n) per liter of solution M = n / V = mol / L

Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in 500.0 mL of water.First calculate the moles of solute:1.5 g NaCl (1 mole NaCl) = 0.0257 moles of NaCl

58.45 g NaCl

Next convert mL to L: 0.500 L of solutionLast, plug the appropriate values into the correct variables in the equation:M = n / V = 0.0257 moles / 0.500 L = 0.051 mol/L

MOLARITYM = n / V = mol

/ L

How many grams of LiOH is needed to prepare 250.0 mL of a 1.25 M solution?First calculate the moles of solute needed: M = n / V , now rearrange to solve for n: n = MV n = (1.25 mol / L) (0.2500 L) = 0.3125 moles of solute neededNext calculate the molar mass of LiOH: 23.95 g/mol

Last, use deminsional analysis to solve for mass:0.3125 moles (23.95 g LiOH / 1 mol LiOH) = 7.48 g of LiOH

Problem

• 4:67 pg 153 a,b

2211 VMVM

C. Dilution

• Preparation of a desired solution by adding water to a concentrate.

• Moles of solute remain the same.

C. Dilution

• What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?

GIVEN:M1 = 15.8MV1 = ?M2 = 6.0MV2 = 250 mL

WORK:M1 V1 = M2 V2

(15.8M) V1 = (6.0M)(250mL)

V1 = 95 mL of 15.8M HNO3

Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar).

M1 = 0.05 mol/L M2 = ?

V1 = 25.0 mL V2 = 50.0 + 25.0 = 75.0 mL

M1V1 = M2V2

M1 V1 = M2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI V2 75.0 mL

MOLARITY& Dilution

Problem

• 4:75 pg 153 a,b

Chapter 4Aqueous Reactions and Solution Stoichiometry

Electrolytes

• Substances that dissociate into ions when dissolved in water.

• A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

Electrolytes and NonelectrolytesSoluble ionic compounds and strong acids tend to be electrolytes.

There are only seven strong acids:

• Hydrochloric (HCl)• Hydrobromic (HBr)• Hydroiodic (HI)• Nitric (HNO3)• Sulfuric (H2SO4)• Chloric (HClO3)• Perchloric (HClO4

Electrolytes and NonelectrolytesMolecular compounds tend to be nonelectrolytes, except for acids and bases.

Electrolytes

• A strong electrolyte dissociates completely when dissolved in water.

• A weak electrolyte only dissociates partially when dissolved in water.

Strong Electrolytes Are…• Strong acids• Strong bases• Soluble ionic

salts

1. Pure water2. Tap water3. Sugar solution4. Sodium chloride solution5. Hydrochloric acid solution6. Lactic acid solution7. Ethyl alcohol solution8. Pure sodium chloride

Electrolytes?

ELECTROLYTES: NONELECTROLYTES:

Tap water (weak)

NaCl solution

HCl solution

Lactate solution (weak)

Pure water

Sugar solution

Ethanol solution

Pure NaCl

Answers to Electrolytes

Problem

• 4:15 pg 150

Precipitation Reactions

When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.

Metathesis (Exchange) Reactions• Double replacement• Metathesis comes from a Greek word

that means “to transpose”• It appears the ions in the reactant

compounds exchange, or transpose, ions

AgNO3 (aq) + KCl (aq) AgCl (s) + KNO3 (aq)

Net Ionic EquationNeed to know which ionic bonds are soluble and

which form precipitatesNeed to use solubility chart

Writing Net Ionic Equations1. Write a balanced molecular equation.2. Dissociate all strong electrolytes.3. Cross out anything that remains

unchanged from the left side to the right side of the equation.

4. Write the net ionic equation with the species that remain.

Net Ionic Equation

• Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions.

Ag+(aq) + NO3-(aq) + K+

(aq) + Cl-(aq)

AgCl (s) + K+(aq) + NO3-(aq)

AgNO3 (aq) + KCl (aq) AgCl (s) + KNO3 (aq)

Not soluble

Net Ionic Equation

• Ag+(aq) + Cl-(aq) AgCl (s)

Problem

• 4.19pg 151• 4.39 b and C

Acids

There are only seven strong acids:• Hydrochloric (HCl)• Hydrobromic (HBr)• Hydroiodic (HI)• Nitric (HNO3)• Sulfuric (H2SO4)• Chloric (HClO3)• Perchloric (HClO4)

Bases

The strong bases are the soluble salts of hydroxide ion (OH-):• Alkali metals• Calcium• Strontium• Barium

Neutralization ReactionsWhen a strong acid reacts with a strong base, the

net ionic equation is…

HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)

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

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

Neutralization ReactionsWhen a strong acid reacts with a strong base, the

net ionic equation is…

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

Neutralization Reactions

Observe the reaction between Milk of Magnesia, Mg(OH)2, and HCl.

Problem

• 4.81g 153