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Understanding Solutions: Homogeneous Mixtures and Solubility

Learn about solutions, homogeneous mixtures of substances in a single physical state, their composition, solubility, and different concentration measurements such as volume percent and mass percent.

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Understanding Solutions: Homogeneous Mixtures and Solubility

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  1. Solutions Chapter 14

  2. solution • Homogeneous mixture of 2 or more substances in a single physical state • particles in a solution are very small • particles in a solution are evenly distributed • particles in a solution will not separate

  3. solute • The substance that is dissolved • examples: sugar, salt

  4. solvent • Substance that does the dissolving • example: water, ethanol • Aqueous solutions-use water as solvent

  5. Like dissolves like • A solute will dissolve best in a solvent with similar intermolecular forces. • If the intermolecular forces are too different the solute will not dissolve in that solvent.

  6. Calculating the strength of a solution Often the strength of a solution can be expressed in terms of percent.

  7. Percent Solutions can be calculated 2 ways. • % by volume • This compares the volume of solute to the total volume of solution. • % by mass • This compares the mass of solute to the total mass of solution.

  8. Volume Percent Volume of solute present in a total volume of solution. Volume Percent (v/v) = volume of solute / volume of solution x 100%

  9. Calculating volume percent A solution is prepared by dissolving 36 ml of ethanol in water to a final volume of 150 ml what is the solution’s volume percent? % (v/v) ethanol = 36 ml ethanol / 150 ml total x 100% Volume Percent ethanol = 24 %

  10. Volume percent If 15.0ml of acetone is diluted to 500ml with water what is the % (v/v) of the prepared solution? % (v/v) = 15.0ml / 500ml x 100% % v/v = 3.0% acetone

  11. Mass percent • Way to describe solutions composition • mass of solute present in given mass of solution mass percent = mass of solute mass of solution grams of solute grams of solute + grams of solvent X 100 X 100

  12. Mass percent • A solution is prepared by dissolving 1.0g of sodium chloride in 48 g of water. The solution has a mass of 49 g, and there is 1.0g of solute (NaCl) present. Find the mass percent of solute.

  13. Mass Percent • A solution is prepared by mixing 1.00g of ethanol, C2H5OH, with 100.0 g of water. Calculate the mass percent of ethanol in this solution.

  14. Solubility • The extent to which a solute will dissolve • expressed in grams of solute per 100g of solvent • ‘likes dissolve likes’

  15. Not every substance dissolves in every other substance • soluble- capable of being dissolved • salt • insoluble- does not dissolve in another • oil does not dissolve in water

  16. Solubility & liquids • Miscible- two liquids that dissolve in each other completely • immiscible- liquids that are insoluble in one another • oil & vinegar

  17. The compositions of the solvent and solute will determine if the substance will dissolve • stirring • temperature • surface area of the dissolving particles

  18. A solution is prepared by mixing 2.8 g of sodium chloride with 100 g of water. What is the mass percent of NaCl? • What is the volume percent alcohol when you add sufficient water to 700mL of isopropyl alcohol to obtain 1000mL of solution?

  19. saturated solution • contains the maximum amount of solute for a given quantity of solvent • no more solute will dissolve

  20. unsaturated solution • contains less solute than a saturated solution • could use more

  21. Supersaturated solution • Solution contains more solute than it can ‘hold’ • too much

  22. Dilute solution- contains a small amount of solute • Concentrated solution- contains large amount of solute

  23. Solubility • Table salt: at room temperature, 37.7 g can be dissolved in 100 ml of H2O • Sugar: at room temperature, 200 g can be dissolved in 100 ml of H2O

  24. Solubility Curve

  25. Solubility curve • Determines solubility of substances at specific temperatures • with raising temperature solids increase in solubility • with increase in temperature gases decrease in solubility • ex: fish die

  26. on the line- saturated (can not hold anymore) above the line- supersaturated (holding more than it can) below the line- unsaturated (can hold more solute) supersaturated Solute (g) per 100 g H2O saturated unsaturated temperature

  27. 92 g of NaNO3 are added to 100ml of water at 25°C and mixed. What type of solution is it? • 80 g of NaNO3 are added to 100ml of water at 25°C and mixed. What type of solution is it?

  28. What is the solubility of NaNO3 in 100g of H2O at 20°C? • What is the solubility of NaNO3 in 200g of H2O at 20°C?

  29. Concentration of solutions • Concentration of a solution is the amount of solute in a given amount of solvent • most common measurements of concentration are: • molarity • (mole fraction)

  30. Concentration of solutions • Concentration of a solution is the amount of solute in a given amount of solvent • most common measurements of concentration are: • molarity • (mole fraction) – not discussed in this class

  31. Molarity • Number of moles of solute per volume of solution in liters moles of solute • molarity (M) = liters of solution mol L

  32. Molarity • Calculate the molarity of a solution prepared by dissolving 11.5g of solid NaOH in enough water to make 1.50 L of solution. • Given: • mass of solute = 11.5 g NaOH • vol of solution = 1.50 L • molarity is moles of solute per liters of solution

  33. Molarity • Convert mass of solute to moles (using molar mass of NaOH). Then we can divide by volume • molar mass of solute = 40.0 g 11.5 g NaOH x 1 mol NaOH 40.0 g NaOH 0.288 mol NaOH 1.50 L solution = 0.288 mol NaOH = 0.192 M NaOH

  34. Molarity • Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl into enough water to make 26.8 mL of solution. • Given: mass of solute (HCl) = 1.56 g volume of solution = 26.8 mL

  35. Molarity • Molarity is moles per liters • we have to change 1.56 g HCl to moles HCl and then change 26.8 mL to liters • molar mass of HCl = 36.5 g • 1.56 g HCl x 1 mol HCl 36.5 g HCl = 0.0427 mol HCl = 4.27 x 10-2 mol HCl

  36. Molarity • Change the volume from mL to liters • 1 L = 1000 mL • 26.8 mL x 1 L 1000 mL = 0.0268 L = 2.68 x 10-2 L

  37. molarity • Finally, divide the moles of solute by the liters of solution • molarity = 4.27 x 10-2 mol HCl 2.68 x 10-2 L = 1.59 M HCl

  38. molarity • Calculate the molarity of a solution prepared by dissolving 1.00 g of ethanol, C2H5OH, in enough water to give a final volume of 101mL. Molarity = moles of solute/ L of solution Moles of ethanol MM ethanol = 46.08 g/mol 1.00 g ethanol / 46.08 g/mol= 0.0217 mol

  39. Solution volume = 101 ml convert to liters 101 ml / 1000ml per liter = 0.101 L Molarity (M) = moles / L M = 0.0217 moles / 0.101 L Molarity = 0.215 M ethanol

  40. molarity • One saline solution contains 0.90 g NaCl in exactly 1.0 L of solution. What is the molarity of the solution? • Calculate the moles of NaCl • MM NaCl = 58.44 g / mol

  41. 0.90 g NaCl / 58.44 g / mol = 0.015 moles • Volume = 1.0 L • Molarity = 0.015 moles / 1.0 L • Molarity = 0.015 M NaCl

  42. molarity • A solution has a volume of 250 mL and contains 7.0 x 10⁻¹ mol NaCl. What is its molarity? • Convert volume to liters • 250 ml / 1000 ml per L = 0.25 L • M = 7.0 x 10⁻¹ / 0.25 L • Molarity of NaCl = 2.8 M

  43. Finding moles to calculate grams How many grams of solute is needed to prepare 300. ml of 3.2 M KCl solution? Use the molarity relationship the find the number of mol. moles = L x M Convert volume to L 300. ml x (1 L/1000ml) = 0.300 L Calculate mol moles = 0.300 L x 3.2M = 0.96 mol Convert mol to grams 0.96 mol KCl x (74.5g/mol) = 72 g

  44. Finding volume • How many liters of 0.442 M MgS can be made with 27.3 g of MgS? MM of MgS = 56 g/mol. • Use the relationship L = mol / M • Convert g to mol 27.3g x (1 mol/56 g) = 0.488 mol Calculate liters L = 0.488 mol/ 0.442 M = 1.10 L

  45. dilution • Diluting a solution: • reduces the number of moles of solute per unit volume • the total number of moles of solute in solution does not change

  46. Diluting solutions M1V1 = M2V2 • M1 = molarity of stock solution (initial) • V1 = volume of stock solution (initial) • M2 = molarity of dilute solution • V2 = volume of dilute solution

  47. M1V1 = M2V2 • How many milliliters of aqueous 2.00M MgSO4 solution must be diluted with water to prepare 100.00 mL of aqueous 0.400M MgSO4? • M1 = 2.00M MgSO4 • M2 = 0.400M MgSO4 • V2 = 100.00 mL MgSO4 • V1 = ?

  48. M1V1 = M2V2 • Solve for V1 • V1 = M2 x V2 M1 0.400M x 100.00 mL 2.00M = 20.0 mL

  49. M1V1 = M2V2 • How many milliliters of a solution of 4.00M KI are needed to prepare 0.250 L of 0.760M KI? • Mı=4.00M Vı=? • M2=0.760M • V2=0.250 L

  50. V1=M2xV2/M1 • V1=(0.760M)(0.250 L)/4.00M • V1=0.0475 L

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