1. Helpful Definitions Solutions: homogeneous mixture of two or more substances physically mixed together in a uniform way. Solute: substance being dissolved. Solvent: part of a solution doing the dissolving. Soluble: when a substance dissolves in another substance. Solubility: the ability of a substance to dissolve in another substance.

2. Concentration of Solute The amount of solute in a solution is given by its concentration. Molarity ( M ) = moles solute liters of solution

3. Units of Molarity 0.0170 M HCl = 0.0170 moles HCl 1 L HCl solution 6.0 M HCl= 6.0 moles HCl 1 L HCl solution

4. Example problem #1 A solution has a volume of 250mL and contains 0.70 mol NaCl. What is its molarity? a).0028 M b)175 M c) 2.8 M mol L M

5. Example problem #1 A solution has a volume of 250mL and contains 0.70 mol NaCl. What is its molarity? given: 250 mL = 0.25L mol= 0.70 mol NaCl M = mol = 0.70 mol NaCl = 2.8M NaCl L 0.25L

6. 40.00 g NaOH 1 mol NaOH How many moles solute are required to make 1.35 L of 2.50 M solution? mol = M L What mass sodium hydroxide is this? 3.38 mol = 2.50 mol (1.35 L) L = 3.38 mol NaOH = 135 g NaOH Example problem #2

7. Example problem #3 A KOH solution with a volume of 400 mL contains 2 mole KOH. What is the molarity of the solution? a) 8 M b) 5 M c) 2 M Drano

8. Example problem #3 A KOH solution with a volume of 400 mL contains 2 moles of KOH. What is the molarity of the solution? M = 2 mole KOH = 5 M 0.4 L Drano

9. Example problem #4 NaOH is used to open stopped sinks, to treat cellulose in the making of nylon, and to remove potato peels commercially. If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution? M = 0.20M NaOH

10. 40.00 g NaOH 1 mol NaOH 4.0 g NaOH = 0.10mol NaOH Example problem #4 M = mol = 0.10 mol NaOH = 0.20 M NaOH L 0.500L

11. Example problem #5 How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? a)12 g b)48 g c) 300 g

12. Molarity Conversion Factors A solution is a 3.0 M NaOH.. Write the molarity in the form of conversion factors. 3.0 moles NaOH and 1 L NaOH soln 1 L NaOH soln 3.0 moles NaOH

13. Example problem #6 Stomach acid is a 0.10 M HCl solution. How many moles of HCl are in 1500 mL of stomach acid solution? a) 15 moles HCl b) 1.5 moles HCl c) 0.15 moles HCl

14. Example problem #6 1500 mL acid soln x 1 L = 1.5 L acid soln 1000 mL 1.5 L acid soln x 0.10 mole HCl = 0.15 mole HCl 1 L acid soln (Molarity factor)

15. 0.342 mol 5.65 L Example problem #7 Find molarity if 58.6 g barium hydroxide are in 5.65 L solution. 171.3 g Ba(OH) 2 1 mol Ba(OH) 2 Step 2). What is the molarity of a 5.65 L solution containing 0.342 mol solute? Step 1). How many moles barium hydroxide is this? = 0.0605 M Ba(OH) 2 X mol Ba(OH) 2 = 58.6 g Ba(OH) 2 = 0.342 mol Ba(OH) 2 M = mol L M =

16. Dilution Dilution is the process of decreasing the concentration of a stock solution by adding more solvent to the solution. A stock solution is a concentrated solution that will be diluted to a lower concentration for actual use. M1V1=M2V2 M 1 = molarity of the stock solution M 2 = molarity of the diluted solution V 1 = volume of stock solution V 2 = volume of diluted solution

17. Dilution *notice that the number of moles are the same *but the volume of solvent has changed M M M M

18. Dilution Preparation of a desired solution by adding water to a concentrate. Moles of solute remain the same.

19. Dilution Example #1 A stock solution of 1.00M of NaCl is available. How many milliliters are needed to make a 100.0 mL of 0.750M? What we know: M 1 = 1.00M V 1 = ? M2= 0.750M V2= 100.0 mL M 1 V 1 =M 2 V 2 75.0 mL

20. Dilution Example #2 Concentrated HCl is 12M. What volume is needed to make 2.0L of a 1.0M solution? What we know: M 1 = 12 M V 1 = ? M 2 = 1.0 M V 2 = 2.0 L Plug in the values you have into the equation to solve for the missing value. M 1 V 1 =M 2 V 2 0.17 L

21. Dilution Example #3 What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

22. 461 g PbI 2 Strategy: 1 Pb(NO 3 ) 2 (aq) + 2 KI (aq)  1 PbI 2 (s) + 2 KNO 3 (aq) 89 g PbI 2 1 mol PbI 2 2 mol KI (1) Find mol KI needed to yield 89 g PbI 2. (2) Based on (1), find volume of 4.0 M KI solution. What volume of 4.0 M KI solution is required to yield 89 g PbI 2 ? 89 g ? L 4.0 M 1 L KI soln 4.0 mol KI = 0.098 L of 4.0 M KI

23. How many mL of a 0.500 M CuSO 4 solution will react with excess Al to produce 11.0 g Cu? __CuSO 4 (aq) + __Al (s)  CuSO 4 (aq) + Al (s)  Cu(s) + Al 2 (SO 4 ) 3 (aq)3231 ? mol11 g __Cu(s) + __Al 2 (SO 4 ) 3 (aq) 11 g Cu 1 mol Cu 63.5 g Cu 3 mol CuSO 4 3 mol Cu 0.500 mol CuSO 4 1000 mL 1 L = 346 mL 1 L CuSO 4 soln