1. Ch. 4 REACTIONS,SOLUTIONS Concentration [ ] dilution, molarity (moles/L) Replacement Rxns activity series, solubility Electrolytes Reduction – Oxidation oxidation numbers RXN: Acid-Base, Neutralization PPT; REDOX

2. TERMS Solute The substance that is dissolved in solution Solvent The substance that does the dissolving Solution A homogeneous mixture of 2 or more pure substances

3. MOLARITY --CONCENTRATION Label M mols liter moles of solute liter of solution Symbol Concentration of ACID Concentration of BASE Brackets indicate CONCENTRATION [H + ] [OH - ]

4. 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

5. 290.0 mL of a 0.560 M Fe(OH) 3 solution is required to make a 0.250 M solution by dilution. What is the final volume. Step #1: Identify the parts for equation M 1 = 0.560 M V 1 = 290.0 M 2 = 0.250 M V 2 = X ml Step #2: Set up and solve equation (0.560 M) (290.0 mL) = ( X ml) (0.250 M) 0.250 M Notice: Dilution problems are not volume specific 650 ml

6. 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) = ( X ) (2.0 L) 2.0 L 0.28 M OR

7. Given mass & volume, find molarity Determine the M of 752.0 g Barium Chloride in 575 ml of solution. Step 1: Need to change 575 ml to 0.575 L Step 2: Find formula wt. of CMPD.BaCl 2 = 208.3 g/mol Step 3 : Find moles of CMPD. 752.0 g * 1 mol. 1 208.3 g 3.6 moles Step 4 : Find molarity of solution 6.3 M M = 3.6 moles 0.575 L

8. Given molarity & volume, find moles & mass How many moles of NaCl are in 36.7 ml of a 0.256 M solution? Step 1: Need to change 36.7 ml to 0.0367 L Step 2: Convert M label to mols/L 0.256 mols 1 L Step 3 : Find moles of NaCl Using given concentration & volume, find moles 0.0094 or 9.4 * 10 -3 moles NaCl

9. Continuing, next we need to find the MASS of NaCl in0.0094 moles Step 1: find formula wt. of NaCl 58.5 g /mol Step 2 : Find mass 0.55 g NaCl MOLES MASS Multiply by Divide by

10. PRACTICE PROBLEMS M OLARITY 1. What is the molarity of a solution containing 2.50 moles of KNO 3 dissolved in 5.00 L? 2. How many moles of KCl are present in 100.0 mL of 0.125 M solution? D ILUTION 1. What is the molarity of 50.0 mL of a 0.50 M NaOH solution after it has been diluted to 300.0 mL? 2. If 300.0 mL of water is added to 400.0 mL of a 0.400 M Na 2 CrO 4 solution, what is the molarity of the resulting solution?

11. SINGLE REPLACEMENT FORM: A + B - + E 0 -----  E + B - + A 0 2 reactants: 1 cmpd. & 1 element forms 2 products: 1 new cmpd. & 1 new element DOUBLE REPLACEMENT FORM: A + B - + X + Y - -----  A + Y - + X + B - 2 reactants: 2 compounds form 2 products: 2 new compds.

12. SINGLE REPLACEMENT Reference used for rxn. occurring ACTIVITIES SERIES of METALS Electrochemical Series DOUBLE REPLACEMENT Reference used of rxn. occurring RULES of SOLUBILITY

13. ACID + BASE -----  SALT + H 2 O Nitric Acid + Potassium Hydroxide ---  ?????? HNO 3 (aq) + KOH (aq) ---  KNO 3 (aq) + H 2 O (l) DR Sulfuric Acid + Barium Hydroxide -----  H 2 SO 4 (aq) + Ba(OH) 2 (aq) ---  BaSO 4 (s) + H 2 O (l) 2

14. Zinc + Copper II Sulfate -----  1 st is a reaction going to occur or not? ? ? ? ? What type of RXN? ? ? Which players will trade places? ? ? ? We have an element plus a compound. Single Replacement Check the METAL REACTIVITY list Zn (s) + CuSO 4 (aq) ----  Look at relation between Zn & Cu Zn is more active then Cu, therefore, rxn. occurs ZnSO 4 (aq) + Cu (s) BALANCED???

15. SOLUTIONS AgNO 3 (aq) + NaCl (aq) ----  NaNO 3 (aq) + AgCl (s) Ag + (aq) + NO 3 - (aq) + Na + (aq) + Cl - (aq) -----  Na + (aq) + NO 3 - (aq) + AgCl (s) Ag + NO 3 - Na + Cl - Now, combine both solutions together. What is the expected effect??????

16. Silver + Sulfuric Acid --  ????? Ag (s) + H 2 SO 4 (aq) --  CHECK, “Ag” active metal to replace “H” in an Acid ??? “Ag” is not an active enough metal to replace “H” in an Acid NR Therefore, this is a “No RXN”

17. REDOX – Reduction/Oxidation L ose E lectron O xidize G ain E lectron R educe O xidize I s L ose R educe I s G ain

18. OXIDATION - REDUCTION “REDOX” Reduction: gain electron Oxidation: lose electron charge becomes more neg charge becomes more “+” H 2 (g) + O 2 (g) --------> H 2 O (g) Both H 2 & O 2 are diatomics: charge on each 0 In the cmpd. H is +1 & O is -2 H: from 0 to +1 charge, loses e -, oxidized (reducing agent) O: from 0 to -2 charge, gain e -, reduced (oxidizing agent)

19. Ionic Reaction -- Ionic Equations 3 diff types of equations for same reaction molecuar; total ionic; net ionic Spector ions - not involved in reaction - no  in charge & state STATES STATES : (aq): aqueous; soln (s): solid; precipitate (g): gas (l): liquid: H 2 O Potassium carbonate reacts w/ strontium nitrate to yield ?????????? Type of rxn; molecular, total, net eqns; oxidized/reduced & agents; spectators

20. Type: double replacement; precipitation Spectator: K +1 & NO 3 -1 Reduce, oxidize: none, not a redox rxn

21. Type: DR; acid-base; neutralization Spectator: none, since HCN weak acid Reduce, oxidize: none, not a redox rxn Barium hydroxide and hydrocyanic acid produces ???????

22. Type: DR; acid-base; neutralization Reduce, oxidize: none, not a redox rxn Same reaction but with a strong acid instead Barium hydroxide and hydrochloric acid produces ??????? Spector: Ba +2 & Cl -1, since HCl strong acid

23. Type: Single Replacement; Oxidation Reduction Spectator: SO 4 -2 Oxidize Reduce Red. Agent Ox. Agent Al (0 ---> +3) Al Mn (+2 ---> 0) Mn Aluminum metal & manganese II sulfate produce aluminum sulfate & manganese metal

24. OXIDATION NUMBERS IA:+1 IIA: +2 IIIA: +3 PO 4 -3 : sum “P” charge + “O” charge = -3 P + 4(-2) = -3 P = +5 OH -1 : O = H = -2 +1 -2 + 1 = -1 Carbon monoxide reacts with diiodine pentaoxide yields iodine & carbon dioxide 5 CO (g) + I 2 O 5 (s) -------> I 2 (s) + 5 CO 2 (g) Oxidized: Reduced: Oxidizing Agent: Reducing Agent: +2+5 -2 0 +4-2 C +2 ---> +4; loss 2e- I +5 ---> 0; gain 5 e- ICIC

25. C (s) + CO 2 (g) -------> 2 CO (g) 0 -2 +4 +2 loss 2 e- gain 2 e- Oxidized Reduced Red. Ag Ox. Ag C (s) C in CO 2 C (s) CO 2 0 ----> +2 +4 ----> +2

26. Sn (s) + HNO 3 (aq) -------> Sn(NO 3 ) 2 (aq) + NO 2 (g) + H 2 O (l) Balance REDOX 0 +1 -2 +2 +4 +5 loss 2 e- gain 1 e- Oxidized Reduced Red. Ag Ox. Ag Sn (s) N in HNO 3 Sn (s) N 0 ----> +2 +5 ----> +4

27. PRACTICE PROBLEMS 1. Calcium Acetate (aq) + Aluminum Sulfate (aq) ------  write and balance the molecular, total ionic, & net ionic equations reduced? oxidized????? 3 Ca(C 2 H 3 O 2 ) 2 (aq ) + Al 2 (SO 4 ) 3 (aq) ----> 3 CaSO 4 (s) + 2 Al(C 2 H 3 O 2 ) 3 (aq) 3 Ca +2 + 6 C 2 H 3 O 2 -1 + 2 Al +3 + 3 SO 4 -2 ----> 3 CaSO 4 (s) + 2 Al +3 + 6 C 2 H 3 O 2 -1 Ca +2 (aq ) + SO 4 -2 (aq) ----> CaSO 4 (s) NOT a redox rxn

28. 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: Completely separates into ions ionic cmpds 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)

29. 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 usually dissolves as whole molecule unit C 2 H 8 (aq) ---------> C 2 H 8 (aq) STRONG WEAK NON All ionic cmpds No ionic cmpds Covalent Molecules Strong Acids Weak Acids -non-acids HCl HBr HI Weak Bases -not contain NH 3 HNO 3 H 2 SO 4 NH 3 HClO 3 HClO 4 Strong Bases -1A metal hydroxides -2A heavy metal hydroxides: Ca, Ba, Sr

30. COMPOSITION REACTIONS FORM: A 0 + B 0 -----  A + B - 2 reactants form 1 product DECOMPOSITION REACTIONS FORM: A + B - -----  A 0 + B 0 1 reactants breaks apart into 2 or more product SINGLE REPLACEMENT FORM: A + B - + E 0 -----  E + B - + A 0 2 reactants: 1 cmpd. & 1 element forms 2 products: 1 new cmpd. & 1 new element DOUBLE REPLACEMENT FORM: A + B - + X + Y - -----  A + Y - + X + B - 2 reactants: 2 compounds form 2 products: 2 new compds.