1. Acid and Base Equilibrium

2. Some Properties of Acids Produce H 3 O + ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) Taste sour Corrode metals Electrolytes React with bases to form a salt and water pH is less than 7 Turns blue litmus paper to red

3. Some Properties of Bases  Produce OH - ions in water  Taste bitter, chalky  Are electrolytes  Feel soapy, slippery  React with acids to form salts and water  pH greater than 7  Turns red litmus paper to blue “Basic Blue”

4. Arrhenius acid is a substance that produces (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water Acid/Base definitions Definition 1: Arrhenius

5. Acid/Base Definitions  Definition #2: Brønsted – Lowry Acids – proton donor Bases – proton acceptor A “proton” is really just a hydrogen atom that has lost it’s electron!

6. A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor acid conjugate base base conjugate acid

7. ACID-BASE THEORIES The Brønsted definition means NH 3 is a BASE in water — and water is itself an ACID

8. Conjugate Pairs

9. Learning Check! Label the acid, base, conjugate acid, and conjugate base in each reaction: HCl + OH -  Cl - + H 2 O H 2 O + H 2 SO 4  HSO 4 - + H 3 O + AcidAcid AcidAcid BaseBase BaseBase Conj. Base Conj. Acid

10. Acids & Base Definitions Lewis acid - a substance that accepts an electron pair Lewis base - a substance that donates an electron pair Definition #3 – Lewis

11. Lewis Acids & Bases Formation of hydronium ion is also an excellent example. Electron pair of the new O-H bond originates on the Lewis base.Electron pair of the new O-H bond originates on the Lewis base.

12. Lewis Acid/Base Reaction

13. More About Water K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C In a neutral solution [H 3 O + ] = [OH - ] and so [H 3 O + ] = [OH - ] = 1.00 x 10 -7 M Autoionization

14. More About Water H 2 O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION Equilibrium constant for water = K w K w = 1.00 x 10 -14 = [H 3 O + ] [OH - ] at 25 o C Take -logs of both sides 14 = pH + pOH

15. The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H 3 O + (or OH - ) ion. Under 7 = acid 7 = neutral Over 7 = base

16. Calculating the pH pH = - log [H 3 O+] ((Remember that the [ ] mean Molarity or concentration) Example: If [H 3 O+] = 1 X 10 -10 pH = - log 1 X 10 -10 pH = - (- 10) pH = 10 Example: If [H 3 O+] = 1.8 X 10 -5, what is the pH? pH = - log 1.8 X 10 -5 pH = - (- 4.74) pH = 4.74

17. pH calculations – Solving for H 3 O + If the pH of Coke is 3.12, what is the [] ? If the pH of Coke is 3.12, what is the [ H 3 O + ] ? Because pH = - log [] then Because pH = - log [ H 3 O + ] then - pH = log [] - pH = log [ H 3 O + ] Take antilog (10 x ) of both sides and get 10 - pH = [] 10 - pH = [ H 3 O + ] [] = 10 -3.12 = 7.58 x 10 -4 M [ H 3 O + ] = 10 -3.12 = 7.58 x 10 -4 M *** to find antilog on your calculator, look for “Shift” or “2 nd function” and then the log button *** to find antilog on your calculator, look for “Shift” or “2 nd function” and then the log button

18. Strong and Weak Acids/Bases Generally acids and bases are divided into STRONG or WEAK ones.Generally acids and bases are divided into STRONG or WEAK ones. HNO 3, HCl, HBr, HI, HClO 4, H 2 SO 4 ) all others are weak Strong acids – 6 (HNO 3, HCl, HBr, HI, HClO 4, H 2 SO 4 ) all others are weak Strong Bases – group 1 and 2 hydroxides (all others are weak) except Be(OH) 2

19. Strong acids and bases Strong acids and bases Find the pH of these: 1)A 0.15 M solution of Hydrochloric acid 2) A 3.00 X 10 -7 M solution of Nitric acid 3) What is the pH of 0.0034M H 2 SO 4 ? pH = - log [ H 3 O +] pH = - log 0.15 pH = - (- 0.82) pH = 0.82 pH = - log 3 X 10-7 pH = - (- 6.52) pH = 6.52 Strong acids and bases - are 100 % ionized. Strong acids and bases - are 100 % ionized. No equilibrium is set up No equilibrium is set up [Acid] = [H30+] ( 1:1 ratio) [Acid] = [H30+] ( 1:1 ratio) ACID  H A + H 2 O  ACID  H A + H 2 O  H 3 O + + A-  H 2 O  BH + + OH - BASE  B + H 2 O  BH + + OH - Finding pH

20. pH of strong bases What is the pH of the 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10 -3 M) pOH = - log 0.0010 pOH = - log 0.0010 pOH = 3 pOH = 3 pH = 14 – 3 = 11 OR K w = [H 3 O + ] [OH - ] [HO + ] = 1.0 x 10 -11 M [H 3 O + ] = 1.0 x 10 -11 M pH = - log (1.0 x 10 -11 ) = 11.00

21. What is the pH of a 2 x 10 -3 M HNO 3 solution? HNO 3 is a strong acid – 100% dissociation. HNO 3 (aq) + H 2 O (l) H 3 O + (aq) + NO 3 - (aq) pH = -log [H + ] = -log [H 3 O + ] = -log(0.002) = 2.7 Start End 0.002 M 0.0 M What is the pH of a 1.8 x 10 -2 M Ba(OH) 2 solution? Ba(OH) 2 is a strong base – 100% dissociation. Ba(OH) 2 (s) Ba 2+ (aq) + 2OH - (aq) Start End 0.018 M 0.036 M0.0 M pH = 14.00 – pOH = 14.00 - (-log(0.036) )= 12.56 15.4

22.  Weak acids are much less than 100% ionized in water.  Equilibrium is set up. Common weak acids are acetic acid and weak base is ammonia Weak Acids/Bases

23. Equilibria Involving Weak Acids and Bases Consider acetic acid, HC 2 H 3 O 2 (HOAc) HC 2 H 3 O 2(aq) + H 2 O (l)  H 3 O + (aq) + C 2 H 3 O 2 – (aq) Acid Conj. base Acid Conj. base (K is designated K a for ACID) K gives the ratio of ions (split up) to molecules (don’t split up)

24. Ionization Constants for Acids/Bases Acids ConjugateBases Increase strength

25. Equilibrium Constants for Weak Acids Weak acid has K a < 1 Leads to small [H 3 O + ] and a pH of 2 - 7

26. Equilibrium Constants for Weak Bases Kb – base dissociation constant and is temp dependent Weak base has K b < 1 Leads to small [OH - ] and a pH of 12 - 7

27. Relation of K a, K b, [H 3 O + ] and pH

28. Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc][H 3 O + ][OAc - ] [HOAc][H 3 O + ][OAc - ]initialchangeequilib 1.0000 -x+x+x 1.00-xxx

29. Equilibria Involving A Weak Acid Step 2. Write K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. This is a quadratic. Solve using quadratic formula. or you can make an approximation if x is very small! (Rule of thumb: 10 -5 or smaller is ok)

30. Equilibria Involving A Weak Acid Step 3. Solve K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. First assume x is very small because K a is so small. Now we can more easily solve this approximate expression.

31. Equilibria Involving A Weak Acid Step 3. Solve K a approximate expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. x = [ H 3 O + ] = [ OAc - ] = 4.2 x 10 -3 M pH = - log [ H 3 O + ] = -log (4.2 x 10 -3 ) = 2.37

32. Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO 2 H. HCO 2 H + H 2 O  HCO 2 - + H 3 O + HCO 2 H + H 2 O  HCO 2 - + H 3 O + K a = 1.8 x 10 -4 Approximate solution [H 3 O + ] = 4.2 x 10 -4 M, pH = 3.37 [H 3 O + ] = 4.2 x 10 -4 M, pH = 3.37 Exact Solution [H 3 O + ] = [HCO 2 - ] = 3.4 x 10 -4 M [H 3 O + ] = [HCO 2 - ] = 3.4 x 10 -4 M [HCO 2 H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M [HCO 2 H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M pH = 3.47 pH = 3.47

33. Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib 0.01000 -x+x+x 0.010 - xx x

34. Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 2. Solve the equilibrium expression Assume x is small, so x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M and [NH 3 ] = 0.010 - 4.2 x 10 -4 ≈ 0.010 M The approximation is valid!

35. Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - K b = 1.8 x 10 -5 Step 3. Calculate pH [OH - ] = 4.2 x 10 -4 M so pOH = - log [OH - ] = 3.37 Because pH + pOH = 14, pH = 10.63

36. Percent Ionization for Weak Acids Most weak acids ionize < 50% Percent ionization (p) General Weak Acid: HA(aq) + H2O(l)  H3O + (aq) + A - (aq) p varies  depending on concentration: increase [HA ] decreases p This is caused by Le Chatelier’s Principle Remember, for strong acids we assume complete ionization (100%)

37. Examples The pH of a 0.10mol/L methanoic acid (HCOOH) solution is 2.38. Calculate the percent ionization of methanoic acid Ans: 4.2% Calculate the acid ionization constant (Ka )of acetic acid if a 0.1000mol/L solution at equilibrium at SATP has a percent ionization of 1.3% (Hint: ICE table Ans: 1.7x10 -5

38. Relationship Between Ka and Kb for Conjugate Base Pairs Recall: Conjugate Pairs – an acid and base that differ by one hydrogen Lets consider the hypothetical weak acid, HA, and its conjugate base, A - HA (aq) + H 2 O (l)  H 3 O + (aq) + A - (aq)

39. Now consider the hypothetical weak base, A - in water A - (aq) + H 2 O (l)  HA (aq) + OH - (aq) Now let’s put that together HA (aq) + H 2 O (l)  H 3 O + (aq) + A - (aq)  Ka A - (aq) + H 2 O (l)  HA (aq) + OH - (aq)  Kb Add both equations 2H 2 O (l)  H 3 O + (aq) + OH - (aq)  Kw Relationship Between Ka and Kb for Conjugate Base Pairs

40. Recall: Autoionization of water H2O(l) + H2O(l)  H3O + (aq) + OH - (aq) Kw=1.00x10 -14 **must remember this value**

41. Relationship Between Ionization Constants for Conjugate Base Pairs For acids and bases whose chemical formulas differ by only one hydrogen (conjugate pairs) the following apply: Kw = Ka x Kb Kb =Kw/Ka Ka = Kw/Kb Therefore if only the Ka value is available in the table, we can determine the conjugate pairs Kb by using the above equations Note: these equations show the larger the Ka the smaller the Kb Stronger acid  weaker conjugate base Weaker acid  stronger conjugate base

42. Learning check! 1. What is the value of the base ionization constant (Kb) for the acetate ion, C 2 H 3 O 2 - (aq) Ans: 5.6x10 -10 2. Calculate the percent ionization of propanoic acid, HC 3 H 5 O 2(aq), if a 0.050 mol/L solution has a pH of 2.78 Ans. 3.3%

43. Rapid changes in pH can kill fish and other organisms in lakes and streams. Soil pH is affected and can kill plants and create sinkholes