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Solutions

Solutions. Definitions. Solution - homogeneous mixture. Solute - substance that dissolves in a solvent and is said to be soluble. Solvent - present in greater amount and dissolves solute.

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Solutions

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  1. Solutions

  2. Definitions • Solution - homogeneous mixture Solute - substance that dissolves in a solvent and is said to be soluble Solvent - present in greater amount and dissolves solute

  3. Universal Solvent is Water b/c almost everything dissolves in waterMost common solvent among liquids is water Solute - KMnO4 Solvent - H2O

  4. Solvation- process of dissolving solute particles are surrounded by solvent particles First... solute particles are separated and pulled into solution (microscopic level) Then...

  5. Solvation- process of dissolving If you shake or stir the solution it increases the rate of solvation by breaking up solute and surrounding it with solvent What else can speed up the rate of solvation?

  6. “Like Dissolves Like” Oil on your clothes?

  7. NONPOLAR NONPOLAR POLAR POLAR “Like Dissolves Like” Non polar solutes are more soluble in nonpolar solvents

  8. NONPOLAR NONPOLAR POLAR POLAR “Like Dissolves Like” Polar dissolves Polar Ex. water dissolves salts (ionic & polar molecules)

  9. Solvation • Soap/Detergent • polar “head” with long nonpolar “tail” • dissolves nonpolar grease in polar water

  10. - + - - + + acetic acid salt sugar Electrolyte- solution that conducts electricity Non- Electrolyte Weak Electrolyte Strong Electrolyte solute exists as ions and molecules solute exists as ions only solute exists as molecules only DISSOCIATION IONIZATION View animation online.

  11. UNSATURATED SOLUTION more solute dissolves SATURATED SOLUTION no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form Solubility- the amount of solute that dissolves in given volume of solvent concentration

  12. UNSATURATED (dilute) Small amt. of solute dissolved SATURATED SOLUTION (no more solute dissolves) SUPERSATURATED Increase temp. to dissolve more solute Solubility- the amount of solute that dissolves in given volume of solvent concentration

  13. Solubility • Solubility • maximum grams of solute that will dissolve in 100 g of solvent at a given temperature • varies with temp • based on a saturated solution

  14. Definitions • Soluble- any substance that dissolves in something else • Insoluble- Any substance that does not dissolve

  15. Definitions • Miscible – one liquid dissolves in another liquid (ex. 2 polar liquids will be miscible) • Immiscible – one liquid does not dissolve in another liquid (ex. oil and water) don’t mix b/c oil nonpolar and water polar

  16. Definitions • Alloy one metal dissolved in another

  17. Solubility Curve • Solubility Curve • shows the dependence of solubility on temperature

  18. Solubility • Solids are more soluble at... • high temperatures. • Gases are more soluble at... • low temperatures & • high pressures (Henry’s Law). • EX: soda can

  19. Henry’s Law & Soft Drinks • Soft drinks contain “carbonated water” carbon dioxide gas dissolved in water • Gas in liquid is more soluble if pressure above liquid is high (under pressure) • When bottle opened pressure of CO2 decreases and solubility of CO2 also decreases, according to Henry’s Law. • Result, bubbles of CO2 escape from solution.

  20. Solution and Concentration Concentration- a measure of how much solute is dissolved in the solution

  21. Solution and Concentration Concentration- a measure of how much solute is dissolved in the solution Range from dilute saturated  supersaturated

  22. ways of expressing concentration • Molarity(M):moles solute / Liter solution • Mass percent(mass solute / mass of solution) x 100 • Molality (m)- moles solute / Kg solvent

  23. Molarity(M) =moles solute Liter solution

  24. Calculate the molarity of the following solution 125 ml which contains 2.50 g NaCl

  25. Calculate the molarity of the following solution 125 ml which contains 2.50 g NaCl Molarity = moles of solute Liter of solution

  26. Calculate the molarity of the following solution 125 ml which contains 2.50 g NaCl Molarity = moles of solute Liter of solution First change grams to moles 2.50 g NaCl x 1 mole NaCl = 0.0428 moles NaCl 58.44 g NaCl

  27. Calculate the molarity of the following solution 125 ml which contains 2.50 g NaCl Molarity = moles of solute Liter of solution First change grams to moles 2.50 g NaCl x 1 mole NaCl = 0.0428 moles NaCl 58.44 g NaCl M= 0.0428 moles NaCl = 0.342 M NaCl 0.125 L

  28. Calculate the mass of the solute in the solution 750 cm cubed of CaCl2 solution which is 0.500 M Molarity = moles of solute Liter of solution First change Molarity to moles

  29. Calculate the mass of the solute in the solution 750 cm cubed of CaCl2 solution which is 0.500 M (0.500 moles/1L) Molarity = moles of solute Liter of solution First change Molarity to moles 0.500 moles = moles CaCl2 1 Liter .750 L

  30. Calculate the mass of the solute in the solution 750 cm cubed of CaCl2 solution which is 0.500 M(0.500moles/1L) Molarity = moles of solute Liter of solution First change Molarity to moles 0.500 moles = moles CaCl2 1 Liter .750 L 0.500 moles x .750 L = moles CaCl2 1 Liter

  31. Calculate the mass of the solute in the solution 750 cm cubed of CaCl2 solution which is 0.500 M Molarity = moles of solute Liter of solution First change Molarity to moles 0.500 moles = moles CaCl2 1 Liter .750 L 0.500 moles x .750 L = moles CaCl2 1 Liter 0.500 moles x .750 L = 0.375 moles CaCl2 1 Liter

  32. Calculate the mass of the solute in the solution 750 cm cubed of CaCl2 solution which is 0.500 M Molarity = moles of solute Liter of solution First change Molarity to moles 0.500 moles = moles CaCl2 1 Liter .750 L 0.500 moles x .750 L = moles CaCl2 1 Liter 0.500 moles x .750 L = 0.375 moles CaCl2 1 Liter Next convert moles to grams 0.375 moles CaCl2 x 110.98 g CaCl2 = 41.6 g CaCl2

  33. Calculate how many liters of solution can be made from A 2.00 M solution using 60.0 g NaOH Molarity = moles of solute Liter of solution

  34. Calculate how many liters of solution can be made from A 2.00 M solution using 60.0 g NaOH Molarity = moles of solute Liter of solution First change grams of NaOH to moles 60.00 g NaOH x 1 mole NaOH = 1.50 moles NaOH 40.00 g NaOH

  35. Calculate how many liters of solution can be made from A 2.00 M solution using 60.0 g NaOH Molarity = moles of solute Liter of solution First change grams of NaOH to moles 60.00 g NaOH x 1 mole NaOH = 1.50 moles NaOH 40.00 g NaOH Next change Molaririty to moles/liter 2.0 M = 2moles/1L 2.00 moles = 1.5 moles NaOH 1 Liter ? L

  36. Calculate how many liters of solution can be made from A 2.00 M solution using 60.0 g NaOH Molarity = moles of solute Liter of solution First change grams of NaOH to moles 60.00 g NaOH x 1 mole NaOH = 1.50 moles NaOH 40.00 g NaOH Next change Molaririty to moles/liter 2.00 moles = 1.5 moles NaOH 1 Liter ? L ? L = 1.5 moles NaOH x 1 Liter 2.0 moles

  37. Calculate how many liters of solution can be made from A 2.00 M solution using 60.0 g NaOH Molarity = moles of solute Liter of solution First change grams of NaOH to moles 60.00 g NaOH x 1 mole NaOH = 1.50 moles NaOH 40.00 g NaOH Next change Molaririty to moles/liter 2.00 moles = 1.5 moles NaOH 1 Liter ? L ? L = 1.5 moles NaOH x 1 Liter 2.0 moles L= 0.750 liter

  38. Concentration: Molarity Example Molarity(M) =moles solute Liter solution • How would you prepare the following solution? 1.00 L of a 1.80 M aqueous solution of MgCl2

  39. Concentration: Molarity Example • How would you prepare the following solution? 1.00 L of a 1.80 M aqueous solution of MgCl2 1.80 M = 1.8 mol MgCl2 1.00 L H20

  40. Concentration: Molarity Example • How would you prepare the following solution? 1.00 L of a 1.80 M aqueous solution of MgCl2 1.80 M = 1.8 mol MgCl2 1.00 L H20 1.8 mol MgCl2 x 95.21 g MgCl2 = 171.38 g MgCl2 1 mol MgCl2 Take 171.38 g MgCl2 add water until you reach 1 liter

  41. Concentration: Molarity Example • How would you prepare the following solution? 1 L of a 2.05 M aqueous solution of MgCl2 2.05 M = 2.05 mol MgCl2 1.0 L H20 2.05 mol MgCl2 x 95.21 g MgCl2 = 195.18 g MgCl2 1 mol MgCl2 Take 195 g MgCl2 add water until you reach 1 liter

  42. Concentration: Molarity Example 0.435 g of KMnO4 in 250. mL of solution, what is the molarity of KMnO4? Molarity(M) =moles solute Liter solution

  43. Concentration: Molarity Example 0.435 g of KMnO4 in 250. mL of solution, what is the molarity of KMnO4? first step convert the mass of material to moles. 0.435 g KMnO4 • 1 mol KMnO4 = 0.00275 mol KMnO4 158.0 g KMnO4

  44. Concentration: Molarity Example 0.435 g of KMnO4 in 250. mL of solution, what is the molarity of KMnO4? 0.435 g KMnO4 • 1 mol KMnO4 = 0.00275 mol KMnO4 158.0 g KMnO4 Now convert 250 ml to L 250 mL = 0.250 L

  45. Concentration: Molarity Example 0.435 g of KMnO4 in 250. mL of solution, what is the molarity of KMnO4? 0.435 g KMnO4 • 1 mol KMnO4 = 0.00275 mol KMnO4 158.0 g KMnO4 Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 M 0.250 L solution

  46. Concentration: Molarity Example 0.435 g of KMnO4 in 250. mL of solution, what is the molarity of KMnO4? As is almost always the case, the first step is to convert the mass of material to moles. 0.435 g KMnO4 • 1 mol KMnO4 = 0.00275 mol KMnO4 158.0 g KMnO4 Now that the number of moles of substance is known, this can be combined with the volume of solution — which must be in liters — to give the molarity. Because 250. mL is equivalent to 0.250 L . Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 M 0.250 L solution

  47. Suppose you have 0.500 M sucrose stock solution. How do you prepare 250 mL of 0.348 M sucrose solution ? Concentration 0.500 M Sucrose 250 mL of 0.348 M sucrose Dilution Lowering the concentration by adding solvent (water) The amount of solute remains constant before and after the dilution but there is more solvent: moles BEFORE = moles AFTER C1V1 = C2V2 A bottle of 0.500 M standard sucrose stock solution is in the lab. Give precise instructions to your assistant on how to use the stock solution to prepare 250.0 mL of a 0.348 M sucrose solution.

  48. Dilution Formula and Calculations Mi x Vi = Mf x Vf How to make dilutions from concentrated stock Solve for VOLUME What volume of 6.00M HCl must be diluted to obtain 5.00L of 1.50M HCl?Initial concentration of HCl Mi = 6.00MFinal concentration of HCl Mf = 1.50MFinal volume of solution Vf = 5.00L6.00 M x Vi = 1.50 M x 5.00 L Vi = 1.25 LTo make the desired quantity of diluted HCl, the chemist should begin with 1.25L of the concentrated solution and add enough water to bring the volume up to 5.00 L.

  49. Dilution Formula and Calculations Solve for MOLARITY If you add 25 ml of water to 100 ml of KCl with a .30 M KCl what will the molarity of the diluted solution be?Initial concentration of KCl Mi = 0.30 M Initial volume of KCl = 100 ml Final concentration of KCl Mf = ? MFinal volume of solution Vf = 100 + 25 = 125 ml 0.30 M x 100 ml = ?M x 125 ml 0.30 M x 100 ml= ?M 125 mlMf = 0.24 M Mi x Vi = Mf x Vf

  50. Factors Affecting Solubility 1. Nature of Solute / Solvent. 2. Temperature - 3. Pressure Factor (gas) 4. Size of Particle 5. Stirring mixture

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