Electrolytes
-
Upload
kylee-bennett -
Category
Documents
-
view
44 -
download
0
description
Transcript of Electrolytes
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 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