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

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Presentation on theme: "Electrolytes Some solutes can dissociate into ions. Electric charge can be carried."— Presentation transcript:

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

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

3 Types of solutes CH 3 COOH CH 3 COO - H+H+ Weak Electrolyte - partial dissociation, molecules and ions in solution slight conductivity

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

5 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.

6 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:

7 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)

8 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

9 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

10 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:

11 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.

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13 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

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15 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

16 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 = L Step two:Calculate concentration. C = 5.2 mol/0.500 L = 10mol/L

17 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 mol/L solution that contains 5.0 g of sodium chloride?

18 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

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