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

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, C 2 H 5 OH, 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 H 2 O Sugar: at room temperature, 200 g can be dissolved in 100 ml of H 2 O

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. saturated unsaturated supersaturated temperature Solute (g) per 100 g H 2 O 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)

27. 92 g of NaNO 3 are added to 100ml of water at 25°C and mixed. What type of solution is it? 80 g of NaNO 3 are added to 100ml of water at 25°C and mixed. What type of solution is it?

28. What is the solubility of NaNO 3 in 100g of H 2 O at 20°C? What is the solubility of NaNO 3 in 200g of H 2 O 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, C 2 H 5 OH, 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 M 1 V 1 = M 2 V 2 M 1 = molarity of stock solution (initial) V 1 = volume of stock solution (initial) M 2 = molarity of dilute solution V 2 = volume of dilute solution

47. M 1 V 1 = M 2 V 2 How many milliliters of aqueous 2.00M MgSO 4 solution must be diluted with water to prepare 100.00 mL of aqueous 0.400M MgSO 4 ? M 1 = 2.00M MgSO 4 M 2 = 0.400M MgSO 4 V 2 = 100.00 mL MgSO 4 V 1 = ?

48. M 1 V 1 = M 2 V 2 Solve for V 1 V 1 = M 2 x V 2 M 1 0.400M x 100.00 mL 2.00M = 20.0 mL

49. M 1 V 1 = M 2 V 2 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

52. If 0.250 L of a 5.00 M HBr solution is diluted to 2.00 L with water what will the resulting concentration be? M1= 5.00 M V1= 0.250 L M2 = ? V2= 2.00 L M2 = M1 x V1 / V2 M2 = (5.00M)(0.250L) / (2.00 L) M2 = 0.625M