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Electrolytes

Electrolytes. Some solutes can dissociate into ions. Electric charge can be carried. Types of solutes. high conductivity. Strong Electrolyte - 100% dissociation, all ions in solution. Na +. Cl -. Types of solutes. slight conductivity. Weak Electrolyte - partial dissociation,

kylee-bennett
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Electrolytes

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  1. Electrolytes • Some solutes can dissociate into ions. • Electric charge can be carried.

  2. Types of solutes high conductivity Strong Electrolyte - 100% dissociation, all ions in solution Na+ Cl-

  3. Types of solutes slight conductivity Weak Electrolyte - partial dissociation, molecules and ions in solution CH3COOH CH3COO- H+

  4. Types of solutes no conductivity Non-electrolyte - No dissociation, all molecules in solution sugar

  5. Types of Electrolytes • Strong electrolyte dissociates completely. • Good electrical conduction. • Weak electrolyte partially dissociates. • Fair conductor of electricity. • Non-electrolyte does not dissociate. • Poor conductor of electricity.

  6. A weak electrolyte: CH3COOH(aq)← CH3COO-(aq) +H+(aq) → A non-electrolyte: CH3OH(aq) Representation of Electrolytes using Chemical Equations MgCl2(s) → Mg2+(aq) + 2 Cl-(aq) A strong electrolyte:

  7. Strong Electrolytes Strong 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)

  8. 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

  9. Precipitation Reactions Heterogeneous Reactions Spectator ions or bystander ions 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) Mostly insoluble Silver halidesMetal sulfides, hydroxidescarbonates, phosphates Soluble ions Alkali metals, NH4+nitrates, ClO4-, acetate Mostly soluble ions Halides, sulfates

  10. Spectator ions Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) → AgI(s) + Na+(aq) + NO3-(aq) Net Ionic Equation Overall Precipitation Reaction: AgNO3(aq) +NaI (aq) → AgI(s) + NaNO3(aq) Complete ionic equation: Ag+(aq) + NO3-(aq) + Na+(aq) + I-(aq) → AgI(s) + Na+(aq) + NO3-(aq) Net ionic equation: Ag+(aq) + I-(aq) → AgI(s)

  11. Suppose copper (II) sulfate reacts with sodium sulfide. Write out the chemical reaction and name the precipitate. CuSO4 (aq) + Na2S (aq) CuS (s) + Na2SO4 (aq) 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. Write out the reaction and name the precipitate. Write out the net ionic equation. How to write chemical equations

  12. g solute g solute x 100 x 100 = g solution g solute + g solvent moles of solute volume in liters of solution Units of Concentrations amount of solute per amount of solvent or solution Percent (by mass) = Molarity (M) = moles = M x VL

  13. 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

  14. 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

  15. 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?

  16. 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) a quart moles acetic acid = 0.83 moles/L x 0.95 L = 0.79 moles AA the mole ratio mole CaCO3 = 0.79 moles AA x (1 mole CaCO3/2 moles AA) = 0.39 moles CaCO3 molar mass mass CaCO3 = 0.39 moles x 100 g/mole = 39 g CaCO3 number of tablets = 39 g x (1 tablet/0.500g) = 79 tablets

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