1. Concentration of Solutions

2. Review: Solutions are made up of 1)Solute - substance dissolved or present in lesser proportion 2) Solvent - substance that is the dissolving medium - what the solute is dissolved in - many times this is water

3. Solutions can be classified as 1)Electrolyte solutions - solutions that conduct electricity (solute is either an ionic compound or forms ions when it dissolves) 2)Nonelectrolyte solutions - solutions that do not conduct electricity (solute is a molecular compound that does not form ions as it dissolves)

4. Components that make up mixtures can be 1)miscible - will form solutions in most any proportions (only liquids or gases) 2)immiscible substances - will not form solutions - example: oil and water

5. There is a limit to the amount of solute that can be dissolved in a given amount of solvent - this is the "solubility" of the solute. Solubility of a solute can be expressed in the following terms: 1)soluble – The substance mostly dissolves 2)insoluble – Very little of the substance dissolves 3)slightly soluble – In between: some dissolves but it may not be enough to affect the properties of the solution.

6. Concentration The concentration of a solution refers to the amount of solute dissolved in the solvent Qualitative terms used are dilute (not much solute) and concentrated (alot of solute) - concentrated does not mean pure.

7. Molarity Molarity is one way to measure the concentration of a solution. moles of solute volume of solution in liters Molarity (M) = Molarity is the most used - other units include: molality, normality, formality, mole fraction, % weight, % volume (proof)

8. Making a Solution…

9. Sample Problem #1 3.85 g of NaCl is dissolved in enough water to make 81.0 mL of sol’n. Calculate the molarity of the solution. 3.85 g NaCl 1 mol NaCl ------------------ X ------------------- =.812M NaCl.0810 L sol’n 58.5 g NaCl

10. Sample Problem #2 How many grams of NaCl are required to make 450 mL of.500 M soln?.500 moles NaCl 58.5 g NaCl.45 L soln X ---------------------- X ---------------- = 13 g NaCl 1 L soln 1 mole NaCl

11. Sample Problem #3 How many mL of a.250 M NaCl soln can be prepared using 7.51 g NaCl? 1 mole NaCl 1 L soln 1000 mL 7.51 g NaCl X-------------- X ----------------X -------- = 514mL NaCl soln 58.5 g NaCl.250 mole NaCl 1 L

12. Titration

13. Sample Problem #4 25.0 mL of.325 molar hydrochloric acid (HCl) completely neutralizes 35.0 mL of a calcium hydroxide solution. What is the molarity of the calcium hydroxide solution? 25.0 mL35.0 mL.325 M ? M 2 HCl + Ca(OH) 2  2 H 2 O + CaCl 2.0250 L HCl.325 mole HCl 1 mole Ca(OH) 2 ------------------ X ---------------- X ---------------- =.116 M Ca(OH) 2.0350 L Ca(OH) 2 1 L HCl 2 mole HCl

14. Sample Problem #5 How many mL of.525 M nitric acid (HNO 3 ) solution would completely neutralize 22.5 mL of.275 M Ca(OH) 2 base solution?.525 M.275 M ? mL 22.5 mL 2 HNO 3 + Ca(OH) 2 ----> 2 H 2 O + Ca(NO 3 ) 2.275 mole Ca(OH) 2 2 mole HNO 3 1 L HNO 3.0225 L Ca(OH) 2 X -------------------- X ------------------- X ----------------- 1 L Ca(OH) 2 1 mole Ca(OH) 2.525 moles HNO 3 =.0236 L or 23.6 mL HNO 3

15. Dilutions

16. Sample Problem #6 20.0 mL of.250 M HCl solution is added to 30.0 mL of.150 M HCl solution. What is the concentration of the resulting solution?.0200 L X.250 moles/L =.00500 moles HCl.0300 L X.150 moles/L =.00450 moles HCl ------------ -------------------------.0500 L.00950 moles HCl total total moles total molarity = ------------------ total Liters.00950 moles HCl M = -------------------------- =.190 M HCl.0500 L soln.

17. Sample Problem #7 10.0 mL of.375 M HCl solution is diluted by adding water to a new volume of 50.0 mL of solution. What is the concentration of the resulting solution? total moles total molarity = ------------------- total Liters.0100 L X.375 moles/L =.00375 moles HCl.00375 moles HCl M = ---------------------------- =.0750 M HCl.0500 L soln.

18. Sample Problem #7 - 2 10.0 mL of.375 M HCl solution is diluted by adding water to a new volume of 50.0 mL of solution. What is the concentration of the resulting solution? M 1 V 1 = M 2 V 2 (.375M)(10.0mL) = M(50.0mL) M =.0750 M

19. Sample Problem #8 Indicate the concentration of each ion present in the solution formed by mixing 44.0 mL of 0.100 M Na 2 SO 4 and 25.0 mL of 0.150 M KCl. Na 2 SO 4 + KCl  X (No reaction!) Total Volume = 44.0mL + 25.0mL = 69.0mL=.0690L

21. Sample Problem #9 1.00 g of aluminum reacts with 75.0 mL of 0.300 M ZnI 2 solution. How many grams of zinc are produced? 0.300 M 1.00 g 75.0 mL x g 2 Al (s) + 3 ZnI 2 (aq)  2 AlI 3 (aq) + 3 Zn (s) 0.300 mole ZnI 2 2 mole Al 27.0 g Al.0750 L soln x ------------------ x --------------- x ----------- = 0.405 g Al 1 L soln 3 mole ZnI 2 1 mole Al Al is excess or ZnI 2 is limiting 0.300 mole ZnI 2 3 mole Zn 65.4 g Zn.0750 L soln x ----------------- x ------------- x ------------- = 1.47 g Zn 1 L soln 3 mole ZnI 2 1 mole Zn

22. Sample Problem #10 75.0 mL of 0.300 M ZnI 2 soln is added to 125 mL of 0.450 M AgNO 3 soln. How many grams of the precipitate AgI are produced? 0.300 M 0.450 M 75.0 mL 125 mL x g ZnI 2 (aq) + 2 AgNO 3 (aq)  2 AgI (s) + Zn(NO 3 ) 2 (aq) 0.300 mole ZnI 2 2 mole AgNO 3 1 L soln 0.0750 L soln x --------------- x --------------- x ------------ = 0.100 L AgNO 3 1 L soln 1 mole ZnI 2 0.450 mole AgNO 3 excess AgNO 3 or ZnI 2 is limiting 0.300 mole ZnI 2 2 mole AgI 235 g AgI 0.0750 L soln x ----------------- x --------------- x -------------= 10.6 g AgI 1 L soln 1 mole ZnI 2 1 mole AgI