1. Chapter 18 Solutions

2. Liquids Miscible means that two liquids can dissolve in each other –water and antifreeze, water and ethanol Partially miscible- slightly –water and ether Immiscible means they can’t –oil and vinegar

3. In general the principle concerning whether substances will mix is "Like dissolves Like"; that is, polar (and ionic) substances will dissolve in polar solvents and non-polar substances will dissolve in non-polar solvents, while non-polar and polar substances will not mix.

4. Solubility in Water : The simple rule to remember is ‘like dissolves like ++ 2-2- ++ Water is a polar solvent and will dissolve polar molecules and substances that contain charged particles. Substances that dissolve in H 2 O are said to be soluble E.g. Sugar, ethanol which are polar most ionic compounds sucrose Site of polarity can form H-bonding

5. Water is polar, the different ends are attracted to the charged ions. 2-2-  + Cl - Na +

6. Substances that don’t dissolve are called insoluble E.g. Petroleum (crude oil), which are non-polar So if you want to dissolve grease which is non-polar, you need to use a non-polar solvent. Petroleum in a non-polar organic molecule

7. Solution formation Nature of the solute and the solvent –Whether a substance will dissolve –How much will dissolve Factors determining rate of solution... –stirred or shaken (agitation) –particles are made smaller –temperature is increased

8. Temperature and Solutions Higher temperature makes the molecules of the solvent move around faster and contact the solute harder and more often. –Speeds up dissolving. Usually increases the amount that will dissolve (exception is gases)

9. How Much? Solubility- The maximum amount of substance that will dissolve at a specific temperature (g solute/100 g solvent) Saturated solution- Contains the maximum amount of solute dissolved Unsaturated solution- Can still dissolve more solute Supersaturated- solution that is holding more than it theoretically can; seed crystal will make it come out;

10. In a saturated solution an equilibrium is reached between dissolving and recrystallization.

12. Concentration is... a measure of the amount of solute dissolved in a given quantity of solvent A concentrated solution has a large amount of solute A dilute solution has a small amount of solute –thus, only qualitative descriptions But, there are ways to express solution concentration quantitatively

13. Concentration = amount of solute amount of solvent (or solution)

14. Molarity (M) = amount of solute in moles amount of solution in liters

15. What is the molarity of a solution with 2.0 moles of NaCl in 250 mL of solution? Molarity= moles/liter = 2.0 mol/.250 L = 8.0 M

16. Calculating molarity Calculate the molarity of Na 2 SO 4 (s) if you have 4.2 moles in 500. L of solution. Molarity = moles/liter = 4.2 mol/ 500. L = 8.4 x 10 -3 M Could also have mol/l or molar as units

17. How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl solution? Moles = (0.75 mol/l)(6.0 l) = 4.5 mol

18. Calculate the mass of Na 2 SO 4 (s) in 1.6 L of a 2.5 M solution. Molarity = moles/volume Moles = (molarity)(volume) = ( 2.5 mol/l) ( 1.6 l) = 4.0 mol moles Then Find mass Mole x molar mass M x V mass

19. Need molar mass: 142.02 g/mol Mass = 568.08g = 570 g

20. Making solutions 10.3 g of NaCl are dissolved in a small amount of water, then diluted to 250 mL. What is the concentration? How many grams of sugar are needed to make 125 mL of a 0.50 M C 6 H 12 O 6 solution?

21. Alternate Measures of Concentration Molar concentration is the accepted method of determining the amount of substance dissolved in a solvent. parts per million (ppm), parts per billion (ppb) parts per trillion (ppt).

22. Part per million ppm = mass of solute x 10 6 mass of solution

23. Part per billion ppb = mass of solute x 10 9 mass of solution

24. Part per trillion ppt = mass of solute x 10 12 mass of solution

25. Molality (independent of temperature) 1 kg of water = 1 liter of water Molarity = moles of solute kg of solvent

26. Molality - What is the molality when 0.750 mol is dissolved in 2.50 L of solvent? –Molality = 0.750 mol 2.50 kg = 0.300m

27. Molality Ex- Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 kg of pure water (the solvent). What would be the molality of the solution?

28. Step One: convert grams to moles. Step Two: divide moles by kg of solvent to get molality. 58.44 grams/mol is the molecular weight of NaCl 58.44g x 1 mol = 1.00 mol. 58.44 g Molality = 1.00 mol 2.0 kg =0.500 mol/kg (or 0.500 m). or 0.500-molal.

29. 1) Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water. 2) 80.0 grams of glucose (C 6 H 12 O 6, mol. wt = 180. g/mol) is dissolved in1.00 kg of water. Calculate the molality. 3) Calcuate the molality when 75.0 grams of MgCl 2 is dissolved in 500.0 g of solvent.

30. 1. We can describe the concentration of a solution by using the mass percent. Mass percent is the mass of solute present in a given mass of solution. Mass Percent = mass of solute  100% Mass of solution = grams of solute  100% (grams of solute) + (grams of solvent)

31. Example 1: We make a solution by dissolving 1.0g of sodium chloride in 48g of water. What is the mass percent of the solute? Mass Percent = 1.0 g solute x100% 49 g solution = 2.0% NaCl

32. Example 2: A solution is prepared by dissolving 1.00g of ethanol, C 2 H 5 OH, with 100.0g of water. What is the mass percent of ethanol in this solution? Mass Percent = 1.00 g ethanol  100% 101 g solution = 0.990% C 2 H 5 OH

33. Example 3: Cow’s milk typically contains 4.5% by mass of the sugar lactose, C 12 H 22 O 11. Calculate the mass of lactose present in 175 g of milk. Mass of solution (milk) = 175g Mass percent of solute (lactose) = 4.5% Mass Percent = grams of solute  100% grams of solution 4.5% = grams of solute  100% 175 g Grams of solute = (4.5%)  (175 g) = 0.045  175g 100% Grams of solute = 7.9g lactose

34. Dilution Adding water to a solution M 1 V 1 = M 2 V 2

35. Dilution The number of moles of solute doesn’t change if you add more solvent! The # moles before = the # moles after M 1 x V 1 = M 2 x V 2 M 1 and V 1 are the starting concentration and volume. M 2 and V 2 are the final concentration and volume. Stock solutions are pre-made to known Molarity

36. Ex:1 - 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M solution. How many mL of 0.800 M can you make? M 1 V 1 = M 2 V 2 (1.50 mol/L)(53.4 mL) = (0.800 mol/L) (x) (0.800 mol/L) (0.800 mol/L) 100. mL = x

37. Ex:2 - 100.0 mL of 2.500 M KBr solution is on hand. You need 0.5500 M. What is the final volume of solution which results? M 1 V 1 = M 2 V 2 (2.500 mol/L)(100.0 mL)=(0.5500mol/L)(x) 454.55 mL = x 454.6 mL = x

38. Ex: 3 - 53.4 mL of a 1.50 M (mol/l) solution of NaCl is on hand, but you need some 0.500 M solution. How many ml of 0.500 M can you make? M 1 V 2 = M 2 V 2 (1.50 mol/L) (53.4 ml) = (0.500 mol/L) (x) 160. ml = x

39. C b V b = C a V b This is the same as the dilution equation, concentration before and volume before is equal to the concentration after and the volume after.

40. Practice a. 2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity? b. You have 150 mL of 6.0 M HCl. What volume of 1.3 M HCl can you make? c. Need 450 mL of 0.15 M NaOH. All you have available is a 2.0 M stock solution of NaOH. How do you make the required solution?