1. UNIT 8: SOLUTIONS [If you’re not part of the solution…] […you’re part of the precipitate…]

4. INTRODUCTORY INFORMATION Solution : a homogeneous mixture of 2+ substances – Components retain their identities and properties; may be present in any ratio A solution can be any combination of solids, liquids, and/or gases – alloy: mixture of 2 solids – atmosphere: mixture of gases If the solution consists of material(s) dissolved in a liquid… – solvent : the liquid component – solute : dissolved material

5. [CONT.] When H 2 O is the solvent, it is an aqueous solution. Substances that can conduct electricity when dissolved in water are called electrolytes. {More on these later}

6. CONCENTRATION OF SOLUTIONS concentration : a measure of the amount of solute dissolved in a solution

7. CONCENTRATION OF SOLUTIONS Examples of concentration units: A. mass % - you don’t need to worry about this B. ppm or ppb - you don’t need to worry about this D. Molality - (mol/kg) you don’t need to worry about this C. Molarity - (mol/liter)

8. CONCENTRATION OF SOLUTIONS A solution in which no more solvent can be dissolved is called saturated. This is the HIGHEST concentration a solution can have. The only way to manipulate this max. concentration is by changing the temperature of the solution.

9. molarity (M): the number of moles of solute per 1 L of solution This is the unit we will use most often!

10. A “1 molar” (1 M ) solution would contain 1 mole of solute for every liter of solution.

12. Q : What mass of copper(II) sulfate is in 100.0 mL of a 0.25 M solution? First convert 100.0 ml to liters= 0.1 liters M = n  V.25 M = n .1 L.25 x.1 = n.025 moles of CuSO 4.025 x 159.62 = 3.99g CuSO 4

13. PREPARING SOLUTIONS A.)Starting with a solid reagent: – How would you make 500.0 mL of 1.75 M Fe(NO 3 ) 3 ? M = n  V 1.75 M = n .5 liters 1.75 x.5 =.875 moles of Fe(NO 3 ) 3 need to be added to make the solution.

14. PREPARING SOLUTIONS (CONT.) B.)Starting with a more concentrated solution : Q : Mr. H. needs 2.0L of 1.5M HCl. He has a bunch of leftover 6.0M HCl. How should he go about making what he needs? First we need a better tool!

16. Q: Mr. H. needs 2.0L of 1.5M HCl. He has a bunch of leftover 6.0M HCl. How should he go about making what he needs? M 1 V 1 = M 2 V 2 6 x V 1 = 1.5 x 2 V 1 = (1.5 x 2)  6 V 1 =.50 L A: Add 0.50L of the 6.0M HCl with enough water to make 2.0L of solution.

17. Q: Mr. H. needs to make a solution for lab. Each group of two students needs to perform 3 trials using 50.0mL of 3.00 M H 2 SO 4 for each trial. If there are 24 students in the class, how should he prepare enough acid for everyone? (Fully concentrated sulfuric acid is 18.0 M ) 12 groups x 150ml = 1800ml Convert to Liters: 1800ml = 1.8L M 1 V 1 = M 2 V 2 18 x V 1 = 3 x 1.8 V 1 = (3 x 1.8)  18 V 1 =.3 liters or 300ml of concentrated sulfuric acid A: Combine 300mL of concentrated acid to enough distilled water to make 1800mL (1.80L) of solution.