Presentation on theme: "Electrolytes Some solutes can dissociate into ions. Electric charge can be carried."— Presentation transcript:
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 CH 3 COOH CH 3 COO - H+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 MgCl 2 (s) → Mg 2+ (aq) + 2 Cl - (aq) A strong electrolyte: A weak electrolyte: CH 3 COOH(aq) ← CH 3 COO - (aq) +H + (aq) → CH 3 OH(aq) A non-electrolyte:
Strong Electrolytes Strong acids: HNO 3, H 2 SO 4, HCl, HClO 4 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) Ca 2+ (aq) + 2 OH – (aq) AlCl 3 (s) Al 3+ (aq) + 3 Cl – (aq) (NH 4 ) 2 SO 4 (s) 2 NH 4 + (aq) + SO 4 2– (aq)
Acid-base Reactions HCl (g) H + (aq) + Cl – (aq) NaOH (s) Na + (aq) + OH – (aq) neutralization reaction: H + (aq) + OH – (aq) H 2 O (l) Explain these reactions Mg(OH) 2 (s) + 2 H + Mg 2+ (aq) + 2 H 2 O (l) CaCO 3 (s) + 2 H + Ca 2+ (aq) + H 2 O (l) + CO 2 (g) Mg(OH) 2 (s) + 2 HC 2 H 3 O 2 Mg 2+ (aq) + 2 H 2 O (l) + 2 C 2 H 3 O 2 – (aq) acetic acid
Precipitation Reactions Ag + (aq) + NO 3 – (aq) + Cs + (aq) + I – (aq) AgI (s) + NO 3 – (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, NH 4 + nitrates, ClO 4 -, acetate Mostly soluble ions Halides, sulfates Mostly insoluble Silver halides Metal sulfides, hydroxides carbonates, phosphates
Ag + (aq) + NO 3 - (aq) + Na + (aq) + I - (aq) → AgI(s) + Na + (aq) + NO 3 - (aq) Spectator ions Ag + (aq) + NO 3 - (aq) + Na + (aq) + I - (aq) → AgI(s) + Na + (aq) + NO 3 - (aq) Net Ionic Equation AgNO 3 (aq) +NaI (aq) → AgI(s) + NaNO 3 (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. CuSO 4 (aq) + Na 2 S (aq) CuS (s) + Na 2 SO 4 (aq) a)Write out the net ionic equation. Cu +2 (aq) SO 4 -2 (aq) + 2Na + (aq) + S -2 (aq) CuS (s) + 2Na + + SO 4 -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 solution x 100 g solute g solute + g solvent x 100 = Molarity (M) = moles of solute volume in liters of solution moles = M x V L
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 of solution 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 CaCO 3, would it take to neutralize a quart of vinegar, 0.83 M acetic acid (CH 3 COOH)? 2CH 3 COOH(aq) + CaCO 3 (s) Ca(CH 3 COO) 2 (aq) + H 2 O + CO 2 (g) moles acetic acid = 0.83 moles/L x 0.95 L = 0.79 moles AA mole CaCO 3 = 0.79 moles AA x (1 mole CaCO 3 /2 moles AA) = 0.39 moles CaCO 3 mass CaCO 3 = 0.39 moles x 100 g/mole = 39 g CaCO 3 number of tablets = 39 g x (1 tablet/0.500g) = 79 tablets a quart the mole ratio molar mass