1. HETEROGENEOUS:Diff. subst; not uniform in appearance; diff. composition & properties of each retained HOMOGENEOUS:Uniform in appearance;1 set of unique properties SOLUTIONS:Aqueous, homogeneous mixture MISCIBLE:Mutually soluble SOLUTE:Subst. being dissolved SOLVENT: Subst. doing the dissolving Like Dissolves LikePolar solvent dissolves Polar solute Nonpolar solvent dissolves Nonpolar solute Similar intermolecular forces form solns w/each other As “C” chain increases, Polarity decreases ( of course, depending upon structure ) HYDRATES: Cmpds. w/ H 2 O molecules attached ANNHYDROUS:Cmpds. w/o H 2 O molecules attached

2. EFFECTS 1. TEMP: Solids-- temp, ~ solubilityGases-- temp, solubility 2. PRESSURE: GASES-- P, solubility UNITS -- CONCENTRATION 2. Weight/vol %, (w/v)% g, mL 3. Vol.vol %, (v/v)% labels any, both must be same 4. ppm, 10 6 ppb, 10 9

3. Molarity, M Quanity: moles Vol: L EX. Calculate the M of a solution when 863.72 g of potassium chromate is dissolved in 750 mL of water. K 2 CrO 4 -------> formula weight = 194.2 g/mol 750 mL =.750 L Required:

4. wt/vol % Examples:

5. Prepare soln, dissolve 24.0 g NaCl in 152.0 ml H 2 O H 2 O: 152.0 ml = 152.0 g mass solute: 24.0 g tot. mass soln: 24.0 + 152.0 = 176.0 g 13.6 % *100

6. Potassium cyanide is a lethal poison for rats. A lethal dosage is 10 mg/Kg of body weight. What is the ppm concentration? ppm = (Mass Solute, g / Mass Soln, g ) * 10 6 ppm = 10 mg/Kg * ( 1 g / 1000 mg) * (1 Kg / 1000 g) ppm=.00001 * 10 6 ppm = 10 ppm

7. How many grams are needed to prepare 50.0 mL of 8.0% (w/v) KCl? Need to find “mass of solute”, rearrange w/v% equation Mass Solute = (Vol. solution, mL) * (w/v%) / 100% Mass Solute = (50.0 mL) * (8.0%) / 100 = 400% / 100 = 4.0 g KCl

8. ELECTROLYTES: Subst. that produces ions when in H 2 O; conducts electrical current 1. All ionic subst. ARE 2. Most covalent ARE NOT ; acids ARE DISSOCIATE: Separate into ions STRONG ELECTROLYTE: Ionize completely in H 2 O NaCl (aq) ---------> Na +1 (aq) + Cl -1 (aq) H 2 SO 4 (aq) ---------> 2 H +1 (aq) + SO 4 -2 (aq)

9. WEAK ELECTROLYTE: Partially ionize in H 2 O HC 2 H 3 O 2 (aq) ---------> HC 2 H 3 O 2 (aq) + H +1 (aq) + C 2 H 3 O 2 -1 (aq) NONELECTROYLTE: Not produce ions in H 2 O C 2 H 6 (aq) ---------> C 2 H 6 (aq) ELECTROYLTES IN THE BODY 1 Eq (Equivalents) 0.001 Eq = 1 mEq 1000 mEq = 1 Eq

10. SOLUTIONS Solution vs. Pure Subst. Solution has higher boiling pt. & lower freezing pt. than Pure Subst. COLLIGATIVEPROPERTIES COLLIGATIVE PROPERTIES Behavior of soln dependent on # of solute particles dissolved in soln Vapor Pressure Boiling Pt. Elevation Freezing Pt. Depression Pure Subst > soln soln > Pure Subst

11. DILUTION A single solution (homogeneous mixture) of a known concentration needs to be diluted to a lower concentration. This is accomplished by adding a known quantity of water (distilled) to the original solution volume. The terms used indicate: M o ; V o Initial – Original -- Beginning concentration & volume M 1 ; V 1 ; M i ; V i concentration & volume M d ; V d Final – Diluted -- Ending concentration & volume M 2 ; V 2 ; M f ; V f concentration & volume Equation Equation : M 1 * V 1 = M 2 * V 2

12. What volume of a 0.560 M Fe(OH) 3 solution is required to make 650 ml of a 0.250 M solution by dilution? Step #1: Identify the parts for equation M 1 = 0.560 M V 1 = X M 2 = 0.250 M V 2 = 650 ml Step #2: Set up and solve equation (0.560 M) ( X ) = (0.25 M) (650 ml) 0.560 M Notice: Dilution problems are not volume specific 290.2 ml

13. 1.25 L of H 2 O is added to 750 ml of 0.75 M HCl solution. What is the resulting molarity? Step #1: Identify parts for equation M 1 = 0.75 M V 1 = 750 ml M 2 = X V 2 = Look at the problem, what does it say!!! 1.25 L is added to …. So, final vol. is amount Start + Added 2.0 L or 2000.0 ml or 0.75 L Step #2: Set up & solve equation (0.75 M) (0.75 L) = 2.0 L X = 0.28 M OR