1. Chapter 15: Acids & Bases Ridgewood High School scasatelli@ridgewood.k12.nj.us

2. Which are Acids? Bases?  List 4 substances you think are acids What do they have in common?  List 4 substances you think are bases What do they have in common?

3. Properties of Acids  Taste sour  pH values below 7  Turn litmus paper red  React with metals to produce H 2 gas  Change color of indicators  Neutralize bases  Solutions of acids are electrolytes

4. Strong Acids vs. Weak Acids  Strong acids completely dissociate in water HCl  H + + Cl -  Weak acids partially dissociate in water HC 2 H 3 O 2 H + + C 2 H 3 O 2 -

5. Properties of Bases  Feel slippery & taste bitter  pH values above 7  Turn litmus paper blue  Change color of indicators  Neutralize acids  Good electrolytes in solution

6. Strong Bases vs. Weak Bases  Strong bases completely dissociate in water, weak bases do not  Ex. NaOH (strong base) NaOH  Na + + OH -

7. Arrhenius Acids & Bases  Arrhenius acids produce H + ions in water HNO 3  H + + NO 3 - or HNO 3 + H 2 O  H 3 O + + NO 3 -  Note: H + is also written as H 3 O + H 3 O + is called the hydronium ion

8. Arrhenius Acids & Bases  Arrhenius bases produce OH - ions in water Ca(OH) 2  Ca 2+ + 2OH -  Note: OH - ions are called hydroxide ions

9. Arrhenius Acids & Bases  Identify each substance as an acid or a base. Write the dissociation equation. 1. KOH  2. H 2 S  3. H 3 PO 4  4. Mg(OH) 2  5. HF  6. HClO 4 

10. Arrhenius Acids & Bases 1. KOH  K + + OH - (Base) Potassium Hydroxide 2. H 2 S  2H + + S -2 (Acid) Hydrosulfuric Acid 3. H 3 PO 4  3H + + PO 4 -3 (Acid) Phosphoric Acid 4. Mg(OH) 2  Mg +2 + 2OH - (Base) Magnesium Hydroxide 5. HF  H + + F - (Acid) Hydrofluric Acid 6. HClO 4  H + ClO 4 - (Acid) Hypochloric Acid

11. Types of Acids 1. Monoprotic acids: have only 1 H that dissociates when dissolved in water Ex. HCl, HI, HC 2 H 3 O 2 HI  H + + I - 2. Polyprotic acids: have more than 1 H that dissociates when dissolved in water - Ex. H 2 C 2 O 4 (diprotic), H 3 PO 4 (triprotic) H 2 C 2 O 4  2H + + C 2 O 4 2-

12. Polyprotic  Diprotic and Triprotic  Diprotic: compounds that have 2 H + ions EX: H 2 SO 4 : Sulfuric Acid H 2 CO 4 : Carbonic Acid  Triprotic: compounds that have 3 H + ions EX: H 3 PO 4 : Phosphoric Acid H 3 N: Hydronitric Acid

13. Brønsted-Lowry Acids & Bases  Brønsted-Lowry acids: donate H + ions  Brønsted-Lowry bases: accept H + ions Different then Arrhenius Acids and Bases definition because it takes into consideration bases that do not have OH - ions.

14. Brønsted-Lowry Acids & Bases HNO 3 + H 2 O  H 3 O + + NO 3 - acidbase NH 3 + H 2 O  NH 4 + + OH - baseacid Water can act as an acid or base. It’s amphoteric.

15. Conjugate Acid-Base Pairs  Conjugate pairs differ by one H + (or proton) Each pair has 1 acid & 1 base  Acid has extra H +  When one is strong, the other is weak.

16. Conjugate Acid-Base Pairs NH 3 + H 2 O  NH 4 + + OH - baseacid conjugate pair acid base conjugate pair

17. pH: Potential Hydrogen  pH values tells us the concentration of H + ions in solution  Range from 0 to 14 0 7 14 acidicneutral basic below 7 above 7

18. pH  As [H + ] increases, pH decreases (more acidic)  As [H + ] decreases, pH increases (more basic) pH = -log[H + ] or [H + ] = 10 -pH

19. A Little Less Confusing  In other words… The more H + ions you have the more acidic the solution…because you have less OH - ions. The less H + ions you have the more basic the solution is…because you have more H + ions.  pH + pOH = 14

20. pH problems  If [H + ] = 1 x 10 -4 M, what’s the pH? pH = -log(1 x 10 -4 ) pH = 4  If the pH is 11.15, what’s [H + ]? [H + ] = 10 -11.15 [H + ] = 7.08 x 10 -12 M Same mathematics can be applied to OH - ions.

21. Dissociation Constant of Water (K w ) K w = [H + ][OH - ] = 1 x 10 -14 pH + pOH = 14  All aqueous solutions contain H + & OH - ions  Using K w, we can find out [H + ], [OH - ], & pH of a solution  K w = the water constant… K w = 1 x 10 -14

22. K w Problems  If [H + ] in a solution of HCl is 0.003 M, what is [OH - ]? (0.003)[OH - ] = 1 x 10 -14 [OH - ] = 3.3 x 10 -12 M  What is the pH of the solution above? pH=-log[.003] pH= 2.52 (acidic)

23. K w Problems  If [OH - ] in a solution of KOH is 1.8 x 10 -4 M, what is [H + ]? What is the pH? What is the pOH?  [H + ](1.8 x 10 -4 ) = 1 x 10 -14 [H + ] = 5.6 x 10 -11 M pH = -log(5.6 x 10 -11 ) pH = 10.25 (basic) pOH = -log(1.8 x 10 -4 ) pOH = 3.74

24. Neutralization Reactions HCl + KOH  H 2 O + KCl acid base water salt  It is a double replacement reaction that ALWAYS produces water & a salt  H from acid combines with OH from base to make water  Cation & anion left over form a salt A Salt is an ionic compound

25. Neutralization Reactions HCl + Mg(OH) 2  ? acid base HCl + Mg(OH) 2  H 2 O + MgCl 2 acid base water salt Balance equation: 2HCl + Mg(OH) 2  2H 2 O + MgCl 2

26. Neutralization Reaction Problems  Complete the Reactions and identify the acid, base, and salt.  H 2 SO 4 + NH 4 OH   HCl + NaOH 

27. Neutralization Reactions Problems  H 2 SO 4 + 2NH 4 OH  (NH 4 ) 2 SO 4 + 2H 2 O Acid Base Salt  HCl + NaOH  NaCl + H 2 O AcidBase Salt

28. Naming Acids, Bases, & Salts  See Tables 13-1 (acids) & 13-2 (bases) on p.489 of textbook Acids: sometimes begin with hydro; all end with acid  Ex. HCl is hydrochloric acid Bases: many start with element name & end with hydroxide  Ex. KOH is potassium hydroxide

29. Naming Acids, Bases, & Salts  Name of salt is composed of 2 parts 1 st part is like beginning of base’s name 2 nd part comes from the acid used  KCl is potassium chloride (from KOH & HCl)  MgCl 2 is magnesium chloride (from Mg(OH) 2 & HCl)

30. Titrations  It is a technique used to determine the concentration of a solution by reacting it with a solution of known concentration  Acid-base titrations involve neutralization reactions

31. Titrations  The equivalence point (end point) is when the neutralization is complete  An indicator is used to signal this

32. Titration Problems M a V a = M b V b M = molarity V = volume in liters (or mL) a = acid b = base

33. Titration Problems 1. Write balanced neutralization reaction 2. Determine mole ratio between acid & base 3. Use M a V a = M b V b to solve problem

34. Titration Problems 1. If 35 mL of NaOH were added to a 50 mL solution of 0.40 M HCl to reach the equivalence point, what is the molarity of the base? NaOH + HCl  H 2 O + NaCl 1:1 ratio between HCl & NaOH (0.40)(50) = x (35) x =0.57 M NaOH

35. Titration Problems 2. How many milliliters of 0.25 M Ba(OH) 2 must be added to titrate 46 mL of 0.35 M HClO 3 ? Ba(OH) 2 + 2HClO 3  2H 2 O +Ba(ClO 3 ) 2 2:1 ratio of HClO 3 to Ba(OH) 2 (0.35)(46)/2 = (0.25)x x =32.2 mL

36. Dilutions  A dilution is when you make a solution less concentrated by adding solvent (usually water)  Use the same formula as in titration problems M 1 V 1 = M 2 V 2 M is molarity V is volume in liters (or mL)

37. Dilution Problems 1. If 200 mL of a 1.23 M solution is diluted to 500 mL, what is the molarity of the solution? 2. A 0.65 M solution was prepared by diluting 325 mL of a 2.35 M stock solution. What is the volume of the new solution?