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SOLUTIONS

SOLUTIONS. A homogeneous mixture in which the components are uniformly intermingled. Vocabulary Terms to Know. ELECTROLYTES. Substances that break up in water to produce ions. These ions can conduct electric current Examples: Acids, Bases and Salts (ionic compounds). ELECTROLYTES.

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SOLUTIONS

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  1. SOLUTIONS A homogeneous mixture in which the components are uniformly intermingled

  2. Vocabulary Terms to Know

  3. ELECTROLYTES • Substances that break up in water to produce ions. • These ions can conduct electric current • Examples: Acids, Bases and Salts (ionic compounds)

  4. ELECTROLYTES • When substances break up into their ions in water, this process is called ionization or dissociation. • For these compounds, we can write their dissociation equation: BaCl2(s)  Ba2+(aq) + 2Cl– (aq) Pb(NO3)2(s)  Pb2+(aq) + 2NO3–(aq)

  5. Terms Solvent – The substance present in the largest amount in a solution. The substance that does the dissolving. Solute – The other substance or substances in a solution. The substance that is dissolved.

  6. SOLUBILITY • Is the amount of a substance that dissolves in a given quantity of solvent (usually water) at a specific temperature to produce a saturated solution

  7. SOLUBILITY of Polar vs NonPolar • “Like dissolves Like” • Polar molecules dissolve polar molecules (Ionic compounds dissolve ionic compounds) • Nonpolar molecules dissolve nonpolar molecules (Molecular compounds dissolve molecular compounds) Lab/Demo for Electrolytes

  8. SOLUBILITY RULES • All common salts of Group I elements and ammonium are soluble • All common acetates and nitrates are soluble • All binary compounds of Group 7 (other than F) with metals are soluble except those of silver, mercury I and lead • All sulfates are soluble except those of barium, strontium, calcium, silver, mercury I and lead • Except for those in Rule 1, carbonates, hydroxides, oxides, sulfides and phosphates are insoluble

  9. Super Cooled Water Terms • Saturated • When a solution contains the maximum amount of solute • Unsaturated • When a solvent can dissolve more solute • Concentrated • When a relatively large amount of solute is dissolved • Dilute • When a relatively small amount of solute is dissolved • Supersaturated • When the solution contains more solute than a saturated solution will hold at that temperature

  10. Factors that affect solubility • Temperature: • Solids • Direct relationship; as temperature goes up solubility goes up • Gases • Indirect relationship; as temperature goes up solubility goes down

  11. Temperature vs Solubility

  12. Temperature vs Solubility

  13. Gas Solubility (Temp)

  14. Factors Affecting the Rate of Dissolution (dissolving) • Stirring • Temperature • Surface Area

  15. Gas Solubility (Pressure) S1 S2 P1 P2 Henry’s Law (for gases!) = Solubility and pressure are directly related (you will NOT need to solve this equation) Visualization

  16. MOLARITY • Molarity is the number of moles of solute per liters of solution • A way to quantify concentration • (a way to put a number to concentration) • M = molarity = moles of solute liter of solution

  17. moles liter M = Practice Problem #1 • Calculate the molarity of a solution prepared by dissolving 11.5 g of NaOH in enough water to make a 1.50 L solution. 1 mol NaOH 11.5g NaOH = .288 mol 40.0g NaOH .288 mol 1.50 L = .192M

  18. Practice Problem #2 • Calculate the molarity of a solution prepared by dissolving 1.56 g of HCl into enough water to make 26.8 ml of solution. .0427 moles of HCl 26.8 mL = .0268L M = 1.59M solution

  19. moles liter M = Practice Problem #3 • Calculate the amount of liters needed to make a 3.00 molar solution of NaNO3 with 44.0 grams of NaNO3? .518 moles of NaNO3 .518 mol ? Liters 3.00M = ? Liters = .173 liters

  20. DILUTIONS (uses molarity) • M1 x V1 = M2 x V2 • What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4? • 16mL of a 12M acid is diluted to 750 mL, what is the new molarity? 0.0094L or 9.4mL 0.25 M

  21. M1 x V1 = M2 x V2 DILUTIONS (Real solutions) • To how much water should 25.0mL of 6.00M HCl be added to produce a 4.00M solution? V2= 37.5 mL V2– V1 = amount of water 37.5 mL – 25.0 mL = 12.5 mL 12.5 mL of water

  22. Freezing Point Depression in Solutions • The particles hinder the solvent from freezing by slowing the formation of solid crystals • This is used in winter with icy roads & sidewalks! • Making Popsicles • Cold Ice

  23. Colligative Properties • A property that depends on only the number of solute particles, NOT their identity • There are 3 important colligative properties: • Boiling Point Elevation • Freezing Point Depression • Vapor-pressure Lowering

  24. WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES Molality (NOT molarity) • Similar to molarity in that it shows concentration… • Molality (m) = moles of solute Kilogram of solvent

  25. WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES Molality Practice (mol/kg) • 2.4 moles of sucrose are dissolved in 320 mL of water. What is the molality? 2.4 moles sucrose .320Kg water = 7.5m

  26. WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES Molality Practice (mol/kg) • 56.8g of NaCl is dissolve in 560 mL of water. What is the molality? .97mol/.560kg = 1.73m

  27. WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES Colligative Properties Freezing Point Depression/ Boiling Point Elevation • Tf = m x kf x i OR Tb = m x kb x i m = molality i = number of ions dissolved per molecule kf = freezing point constant (given) kb = boiling point constant (given) kf = 1.86 CKg/mol (for water) kb = 0.51 CKg/mol (for water)

  28. WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES Colligative Practice Problem • 56.8g of NaCl is dissolved in 560mL of distilled water. What is the freezing point of this solution? Tf = m x kf x i • Step 1 -> determine m (molality)=moles/Kg • Step 2 -> Plug into equation with kf(given) and i (number of ions, in this case i= 2) • Step 3 -> solve for Tf • Step 4 -> adjust original freezing point accordingly

  29. WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES Colligative Practice Problem m = 1.73 • Tf = m x kf x i Tf = (1.73m) x 1.86CKg/mol x 2 Tf = 6.4 What is new freezing point? -6.4 C

  30. WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES Colligative Problem: finding molar mass • A solution of a nonelectrolyte (does not dissociate in water) contains 30.0 g of solute dissolved in 250.0 g of water. The boiling point of the water is observed to be 101.04°C. What is the molar mass of this substance?

  31. A solution of a nonelectrolyte (does not dissociate in water) contains 30.00 g of solute dissolved in 250.0 g of water. The boiling point of the water is observed to be 101.04°C. What is the molar mass of this substance? • Game plan: • Molar mass (of solute) units grams/mole • We are given grams of solute; must solve for moles. • We are given BP of water; we must be using Tb = m x kb x i • Are there moles anywhere in the equation? • m = moles of solute kg of solvent

  32. WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES A solution of a nonelectrolyte (does not dissociate in water) contains 30.00 g of solute dissolved in 250.0 g of water. The boiling point of the water is observed to be 101.04°C. What is the molar mass of this substance? Tb = m x kb x i • Tb = 101.04°C - 100°C = 1.04 °C • m = ?moles of solute/.2500Kg of water • kb(always given)= 0.51 CKg/mol • i = 1 (for all nonelectrolytes 1.04 °C = ?moles x .51CKg/molx 1 .2500 kg

  33. WE WILL NOT BE CALCULATING MOLALITY OR COLLIGATIVE PROPERTIES A solution of a nonelectrolyte (does not dissociate in water) contains 30.00 g of solute dissolved in 250.0 g of water. The boiling point of the water is observed to be 101.04°C. What is the molar mass of this substance? • ? Moles = 0.5098 moles • Molar mass = grams/moles • 30.00g of solute 0.5098 moles = 58.8 g/mol

  34. Solution Stoichiometry • How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate? 2 AgNO3(aq) + K2CrO4(aq) Ag2CrO4(s) + 2 KNO3(aq)

  35. How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate? 2 AgNO3(aq) + K2CrO4(aq) Ag2CrO4(s) + 2 KNO3(aq) • Given: 150. mL of 0.500 M AgNO3 100. mL of 0.400 M K2CrO4 • Want: Grams of Ag2CrO4 • Game-Plan: • Convert from mL to Liters • Use molarity to convert to moles of reactants • Use mole ratio to convert to moles of Ag2CrO4 • Use molar mass to convert to grams of Ag2CrO4

  36. How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate? 2 AgNO3(aq) + K2CrO4(aq) Ag2CrO4(s) + 2 KNO3(aq) 0.150 L AgNO3 12.4gAg2CrO4 =

  37. How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate? 2 AgNO3(aq) + K2CrO4(aq) Ag2CrO4(s) + 2 KNO3(aq) 13.3gAg2CrO4 100. mL K2CrO4 =

  38. How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate? 2 AgNO3(aq) + K2CrO4(aq) Ag2CrO4(s) + 2 KNO3(aq) VS

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