Electrolytes

• Some solutes can dissociate into ions.

• Electric charge can be carried.

Types of solutes

Na+

Cl-

Strong Electrolyte -100% dissociation,all ions in solution

high conductivity

Types of solutes

CH3COOH

CH3COO-

H+

Weak Electrolyte -partial dissociation,molecules and ions in solution

slight conductivity

Types of solutes

sugar

Non-electrolyte -No dissociation,all molecules in solution

no conductivity

Types of Electrolytes

• Weak electrolyte partially dissociates.– Fair conductor of electricity.

• Non-electrolyte does not dissociate. – Poor conductor of electricity.

• Strong electrolyte dissociates completely.– Good electrical conduction.

Representation of Electrolytes using Chemical Equations

MgCl2(s) → Mg2+(aq) + 2 Cl- (aq)

A strong electrolyte:

A weak electrolyte:

CH3COOH(aq) ← CH3COO -(aq) +H+(aq)→

CH3OH(aq)

A non-electrolyte:

Strong ElectrolytesStrong acids: HNO3, H2SO4, HCl, HClO4

Strong bases: MOH (M = Na, K, Cs, Rb etc)

Salts: All salts dissolving in water are completely ionized.

Stoichiometry & concentration relationship

NaCl (s) Na+ (aq) + Cl– (aq)

Ca(OH)2 (s) Ca2+(aq) + 2 OH– (aq)

AlCl3 (s) Al3+ (aq) + 3 Cl– (aq)

(NH4)2SO4 (s) 2 NH4 + (aq) + SO42– (aq)

Acid-base Reactions HCl (g) H+ (aq) + Cl– (aq)

NaOH (s) Na+ (aq) + OH– (aq)

neutralization reaction: H+ (aq) + OH– (aq) H2O (l)

Explain these reactions

Mg(OH)2 (s) + 2 H+ Mg2+ (aq) + 2 H2O (l)

CaCO3 (s) + 2 H+ Ca2+ (aq) + H2O (l) + CO2 (g)

Mg(OH)2 (s) + 2 HC2H3O2 Mg2+ (aq) + 2 H2O (l) + 2 C2H3O2 – (aq)

acetic acid

Precipitation Reactions

Ag+ (aq) + NO3– (aq) + Cs+ (aq) + I– (aq) AgI (s) + NO3

– (aq) + Cs+ (aq)

Ag+ (aq) + I– (aq) AgI (s) (net reaction)or

Ag+ + I– AgI (s)

Heterogeneous Reactions

Spectator ions or bystander ions

Soluble ions

Alkali metals, NH4+

nitrates, ClO4-,

acetate

Mostly soluble ions

Halides, sulfates

Mostly insoluble

Silver halidesMetal sulfides, hydroxidescarbonates, phosphates

Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) →

AgI(s) + Na+(aq) + NO3-(aq)

Spectator ionsAg+(aq) + NO3

-(aq) + Na+(aq) + I-(aq) →

AgI(s) + Na+(aq) + NO3-(aq)

Net Ionic Equation

AgNO3(aq) +NaI (aq) → AgI(s) + NaNO3(aq)

Overall Precipitation Reaction:

Complete ionic equation:

Ag+(aq) + I-(aq) → AgI(s)

Net ionic equation:

How to write chemical equations

Suppose copper (II) sulfate reacts with sodium sulfide. a) Write out the chemical reaction and name the

precipitate.

CuSO4 (aq) + Na2S (aq) CuS (s) + Na2SO4 (aq) a) Write out the net ionic equation.

Cu+2 (aq) SO4

-2 (aq) + 2Na+

(aq) + S-2 (aq) CuS (s) + 2Na+ + SO4

-2 (aq)

Cu+2 (aq) + S-2 (aq) CuS (s)

Suppose potassium hydroxide reacts with magnesium chloride.

a) Write out the reaction and name the precipitate. b) Write out the net ionic equation.

Units of Concentrations

amount of solute per amount of solvent or solution

Percent (by mass) =g solute

g solutionx 100

g solute

g solute + g solvent

x 100=

Molarity (M) =

moles of solute

volume in liters of solution

moles = M x VL

Examples

What is the percent of KCl if 15 g KCl are placed in 75 g water?

%KCl = 15g x 100/(15 g + 75 g) = 17%

What is the molarity of the KCl if 90 mL ofsolution are formed?

mole KCl = 15 g x (1 mole/74.5 g) = 0.20 mole

molarity = 0.20 mole/0.090L = 2.2 M KCl

Examples:

Example 1: What is the concentration when 5.2 moles of hydrosulfuric acid are dissolved in 500 mL of water?

Step one: Convert volume to liters, mass to moles.

500 mL = 0.500 L

Step two: Calculate concentration.

C = 5.2 mol/0.500 L = 10mol/L

• Example 2: What is the volume when 9.0 moles are present in

5.6 mol/L hydrochloric acid?

• Example 3: How many moles are present in 450 mL of 1.5

mol/L calcium hydroxide?

• Example 4: What is the concentration of 5.6 g of magnesium

hydroxide dissolved in 550 mL?

• Example 5: What is the volume of a 0.100 mol/L solution that

contains 5.0 g of sodium chloride?

How many Tums tablets, each 500 mg CaCO3, would it take to neutralize a quart of vinegar, 0.83 M acetic acid (CH3COOH)?

2CH3COOH(aq) + CaCO3(s) Ca(CH3COO)2(aq) + H2O + CO2(g)

moles acetic acid = 0.83 moles/L x 0.95 L = 0.79 moles AA

mole CaCO3 = 0.79 moles AA x (1 mole CaCO3/2 moles AA)= 0.39 moles CaCO3

mass CaCO3 = 0.39 moles x 100 g/mole = 39 g CaCO3

number of tablets = 39 g x (1 tablet/0.500g) = 79 tablets

a quart

the mole ratio

molar mass