1. 1. Autoionization 1. Autoionization reaction of liquid water 2. pH, pOHpK w 2. pH, pOH, and pK w 3. conjugate acid-base pairs 4. K a, K b, pK a, pK b 4. acid or base strength and the magnitude of K a, K b, pK a, and pK b 5. leveling effect 6. 6. To be able to predict whether reactants or products are favored in an acid-base equilibrium 7. 7. use molecular structure and acid and base strengths 8. 8. use K a and K b values to calculate the percent ionization and pH of a solution of an acid or a base 9. titration 9. calculate the pH at any point in an acid-base titration 10. common ion effects 10. common ion effects and the position of an acid-base equilibrium 11. bufferHenderson- Hasselbalch 11. how a buffer works and how to use the Henderson- Hasselbalch equation to calculate the pH of a buffer Acids and bases

2. MonovalentDivalentTrivalent Hydronium (aqueous)H3O+H3O+ MagnesiumMg 2+ AluminiumAl 3+ Hydrogen (proton)H+H+ CalciumCa 2+ Antimony IIISb 3+ LithiumLi + StrontiumSr 2+ Bismuth IIIBi 3+ SodiumNa + BerylliumBe 2+ PotassiumK+K+ Manganese IIMn 2+ RubidiumRb + BariumBa 2+ CesiumCs + ZincZn 2+ FranciumFr + CadmiumCd 2+ SilverAg + Nickel IINi 2+ AmmoniumNH 4 + Palladium IIPd 2+ ThaliumTl + Platinum IIPt 2+ Copper ICu + Copper IICu 2+ Mercury IIHg 2+ Mercury IHg 2 2+ Iron IIFe 2+ Iron IIIFe 3+ Cobalt IICo 2+ Cobalt IIICo 3+ Chromium IICr 2+ Chromium IIICr 3+ Lead IIPb 2+ Tin IISn 2+ Table of Common Ions Common Positive Ions (Cations)

3. MonovalentDivalentTrivalent HydrideH-H- OxideO 2- NitrideN 3- FluorideFl - PeroxideO 2 2- ChlorideCl - SulfideS 2- BromideBr - SelenideSe 2- IodideI-I- OxalateC 2 O 4 2- HydroxideOH - ChromateCrO 4 2- PermanganteMnO 4 - DichromateCr 2 O 7 2- CyanideCN - TungstateWO 4 2- ThiocynateSCN - MolybdateMoO 4 2- AcetateCH 3 COO - TetrathionateS 4 O 6 2- NitrateNO 3 - ThiosulfateS 2 O 3 2- BisulfiteHSO 3 - SulfiteSO 3 2- BisulfateHSO 4 - SulfateSO 4 2- BicarbonateHCO 3 - CarbonateCO 3 2- Dihydrogen phosphateH 2 PO 4 - Hydrogen phosphateHPO 4 2- PhosphatePO 4 3- NitriteNO 2 - AmideNH 2 - HypochloriteClO - ChloriteClO 2 - ChlorateClO 3 - PerchlorateClO 4 - Table of Common Ions Common Negative Ions (Anions)

4. There are three classes of strong electrolytes. 1Strong Water Soluble Acids Remember the list of strong acids from Chapter 4. 2Strong Water Soluble Bases The entire list of these bases was also introduced in Chapter 4. 3Most Water Soluble Salts The solubility guidelines from Chapter 4 will help you remember these salts. Acids and bases Weak acids and bases ionize or dissociate partially, much less than 100%, and is often less than 10%.

5. Most salts of strong or weak electrolytes can dissolve in water to produce a neutral, basic, or acidic solution, depending on whether it contains the conjugate base of a weak acid as the anion (A – ) or the conjugate acid of a weak base as the cation (BH + ), or possibly both. Salts that contain small, highly charged metal ions produce acidic solutions in H 2 O. The most important parameter for predicting the effect of a metal ion on the acidity of coordinated water molecules is the charge-to-radius ratio of the metal ion. hydrolysis reaction The reaction of a salt with water to produce an acidic or basic solution is called a hydrolysis reaction, which is just an acid-base reaction in which the acid is a cation or the base is an anion. Acids and bases

6. Acid (HCl)Base (NaOH) Arrhenius Brönsted-Lowery Lewis conjugate acid-base pair. Two species that differ by only a proton constitute a conjugate acid-base pair. 1. Conjugate base has one less proton than its acid; A – is the conjugate base of HA 2. Conjugate acid has one more proton than its base; BH + is the conjugate acid of B 3. Conjugates are weaker than strong parents and stronger than weak parents. 4. All acid-base reactions involve two conjugate acid-base pairs. HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl – (aq) parent acid parent base conjugate acid conjugate base amphiprotic hydronium amphoteric Water is amphiprotic: it can act as an acid by donating a proton to a base to form the hydroxide ion, or as a base by accepting a proton from an acid to form the hydronium ion, H 3 O +. Substances that can behave as both an acid and a base are said to be amphoteric. Acids and bases

7. Acids and bases can be defined in different ways: Arrhenius definition: 1. Arrhenius definition: An acid is a substance that dissociates in water to produce H + ions (protons), and a base is a substance that dissociates in water to produce OH – ions (hydroxide); an acid-base reaction involves the reaction of a proton with the hydroxide ion to form water. – Three limitations 1. Definition applied only to substances in aqueous solutions. 2. Definition restricted to substances that produce H + and OH – ions 3. Definition does not explain why some compounds containing hydrogen such as CH 4 dissolve in water and do not give acidic solutions Brønsted–Lowry definition 2. Brønsted–Lowry definition: An acid is any substance that can donate a proton, and a base is any substance that can accept a proton; acid-base reactions involve two conjugate acid-base pairs and the transfer of a proton from one substance (the acid) to another (the base). Not restricted to aqueous solutions, expanding to include other solvent systems and acid-base reactions for gases and solids. Not restricted to bases that only produce OH – ions. Acids still restricted to substances that produce H + ions. Limitation 3 not dealt with. Lewis definition 3. Lewis definition: A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor.

8. Titration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. 4. Net ionic equation H + + OH -  H 2 O H + + OH -  H 2 O 5. At equivalence point moles H + = moles OH - moles H + = moles OH - Titrant Equivalence Point Primary Standard End Point Secondary Standard Titration

9. pH, a Concentration Scale pH: a way to express acidity -- the concentration of H + in solution. Low pH: high [H + ] High pH: low [H + ] Acidic solutionpH < 7 Neutral pH = 7 Neutral pH = 7 Basic solution pH > 7 Basic solution pH > 7 Acidic solutionpH < 7 Neutral pH = 7 Neutral pH = 7 Basic solution pH > 7 Basic solution pH > 7 pH = log (1/ [H + ]) = - log [H + ] AcidFormulapH at half equivalence point AceticCH 3 COOH4.7 NitrousHNO 2 3.3 HydrofluoricHF3.1 HypochlorousHClO7.4 HydrocyanicHCN9

10. You should know the strong acids & bases Everything else is weak.

11. http://www.chem1.com/acad/webtext/abcon/abcon-2.html leveling effect No acid stronger than H 3 O + and no base stronger than OH – can exist in aqueous solution, leading to the phenomenon known as the leveling effect. Any species that is a stronger acid than the conjugate acid of water (H 3 O + ) is leveled to the strength of H 3 O + in aqueous solution because H 3 O + is the strongest acid that can exist in equilibrium with water. In aqueous solution, any base stronger than OH – is leveled to the strength of OH – because OH – is the strongest base that can exist in equilibrium with water It’s all because of Gibbs Free Energy

12. CO 2 +3H 2 O H 2 CO 3 +2H 2 O HCO 3 - +H 3 O + +H 2 O CO 3 2- +2H 3 O + KHKH K1K1 K2K2 CO 2 + CaCO 3 + H 2 O 2HCO 3 - + Ca 2- Atmosphere Biological Calcification (not a reversible reaction) Baking Soda NaHCO 3 Soda Pop CO 2 CO 2 + H 2 O H 2 CO 3 H 2 CO 3 + H 2 O HCO 3 - + H + HCO 3 - + H 2 O CO 3 2- + H + Acids and bases Critter's shells 2.5×10 −4 5.61×10 −11.8317

13. http://www.chem1.com/acad/webtext/abcon/abcon-2.html end point use a pH meter to detect the use an indicator to detect the Acid/Base Equilibrium

14. Autoionization of Water Because water is amphiprotic, one water molecule can react with another to form an OH – ion and an H 3 O + ion in an autoionization process: 2H 2 O (l) ⇋ H 3 O + (aq) + OH – (aq) Equilibrium constant K for this reaction can be written as [H 3 O + ] [OH – ] [H 2 O] 2 1 L of water contains 55.5 moles of water. In dilute aqueous solutions: The water concentration is many orders of magnitude greater than the ion concentrations. The concentration is essentially that of pure water. Recall that the activity of pure water is 1. When pure liquid water is in equilibrium with hydronium and hydroxide ions at 25ºC, the concentrations of hydronium ion and hydroxide ion are equal: [H 3 O + ]=[OH – ] = 1.0 x 10 –7 M [H 3 O + ][OH – ] = 1.0 x 10 –14 M = K w [H 3 O + ]=[OH – ] = 1.0 x 10 –7 M [H 3 O + ][OH – ] = 1.0 x 10 –14 M = K w pH = pOH = 7 pH + pOH = pK w = 14 pH = pOH = 7 pH + pOH = pK w = 14 K c = K c [H 2 O] 2 = K w = [H 3 O + ][OH – ] = 1.0 x 10 –14

15.  A 1 M solution is prepared by dissolving 1 mol of acid or base in water and adding enough water to give a final volume of exactly 1 L.  If the actual concentrations of all species present in the solution were listed, it would be determined that none of the values is exactly 1 M because a weak acid or a weak base always reacts with water to some extent.  Only the total concentration of both the ionized and unionized species is equal to 1 M.  The analytical concentration (C) is defined as the total concentration of all forms of an acid or base that are present in solution, regardless of their state of protonation.  Thus; a 1 M solution has an analytical concentration of 1 M, which is the sum of the actual concentrations of unionized acid or base and the ionized form. Ionization Constants for Weak Acids and Bases When is a 1M solution not a 1 M solution?

16. The equation for the ionization of acetic acid is: The equilibrium constant for this ionization is expressed as: Ionization Constants for Weak Monoprotic Acids and Bases We can define a new equilibrium constant for weak acid equilibrium, K a, the acid ionization constant, using this definition. –The symbol for the ionization constant is K a. –The larger the K a value the stronger the acid and the higher the equilibrium [H + ] –The larger the K b value the stronger the base and the higher the equilibrium [OH – ]

17. Acid and Base strengths can be compared using K a and K b values. The larger the K a or K b value the more product favored the dissociation. An acid-base equilibrium always favors the side with the weaker acid and base. In an acid-base reaction the proton always reacts with the strongest base until totally consumed before reacting with any weaker bases. Any substance whose anion is the conjugate base of a weak acid weaker than OH - reacts quantitatively with water to form more hydroxide ions. Step 1. NaCH 3 COO → Na + + CH 3 COO - Acetate ion is the conjugate base of acetic acid, a weak acid. Step 2. CH 3 COO - + H 2 O CH 3 COOH + OH - Hydrolysis: Hydrolysis: Aqueous solutions of salts that dissociate into both: 1.A strong conjugate acid and a strong conjugate base are neutral (KNO 3 ). 2.A strong conjugate acid and a weak conjugate base are acidic (HCl). 3.A strong conjugate base and a weak conjugate acid are basic (NaOH). 4.A weak conjugate base and a weak conjugate acid can be neutral, basic or acidic: The comparison of the values of K a and K b determine the pH of these solutions. a.K base = K acid make neutral solutions (NH 4 CH 3 OO) b. K base > K acid make basic solutions (NH 4 ClO) c. K base < K acid make acidic solutions (CH 3 ) 3 NHF stronger acid + stronger base weaker acid + weaker base H2OH2O Acids, Bases, and ionization constants

18. The ionization constant values for several acids are given below. –Which acid is the strongest? –Are all of these acids weak acids? –What is the relationship between K a and strength? –What is the relationship between pK a and strength? –What is the relationship between pH and strength? AcidFormulaK a valuepK a value -log K a pH of 1M analytical [HA] AceticCH 3 COOH1.8 x 10 -5 4.72.4 NitrousHNO 2 4.5 x 10 -4 3.31.7 HydrofluoricHF7.2 x 10 -4 3.11.6 HypochlorousHClO3.5 x 10 -8 7.53.7 HydrocyanicHCN4.0 x 10 -10 9.44.7 Ionization Constants for Weak Monoprotic Acids and Bases

19. Ionization Constants for Weak Monoprotic Acids and Bases the MATH – To determine the ionization constant you need the analytical concentration of the acid or base, one must be able to measure the concentration of a least one of the species in the equilibrium constant expression in order to determine the value of K a or K b. – Two common ways to obtain the concentrations 1.By measuring the electrical conductivity of the solution, which is related to the total concentration of ions present 2.By measuring the pH of the solution, which gives [H + ] or [OH – ]

20. Use the concentrations that were just determined in the ionization constant expression to get the value of K a. R HY H 3 O + + Y - I C E In a 0.12 M solution of a weak monoprotic acid, HY, the acid is 5.0% ionized. Calculate the ionization constant for the weak acid. Ionization Constants for Weak Monoprotic Acids the MATH

21. The pH of a 0.10 M solution of a weak monoprotic acid, HA, is found to be 2.97. What is the value for its ionization constant? Use the [H 3 O + ] and the stoichiometry of the ionization reaction to determine concentrations of all species. Calculate the ionization constant from this information. R HA H 3 O + + A - I C E Simplifying Assumption: Is the change significant? Later we will find that in general, if the K a /[] is < 1x10 -3 you can apply the simplifying assumption. Ionization Constants for Weak Monoprotic Acids the MATH

22. Calculate the concentrations of the various species in 0.15 M acetic acid, CH 3 COOH, solution. 1.It is always a good idea to write down the ionization reaction and the ionization constant expression. 2.Next we combine the basic chemical concepts with some algebra to solve the problem RCH 3 COOH H 3 O + + CH 3 COO - I C E Ionization Constants for Weak Monoprotic Acids the MATH

23. Substitute these algebraic quantities into the ionization expression. Solve the algebraic equation, using a simplifying assumption that is appropriate for all weak acid and base ionizations. Ionization Constants for Weak Monoprotic Acids the MATH Complete the algebra and solve for the concentrations of the species. Note that the properly applied simplifying assumption gives the same result as solving the quadratic equation does.

24. Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution. K a = 4.0 x 10 -10 for HCN R HCN H 3 O + + CN - I C E Substitute these algebraic quantities into the ionization expression. Solve the algebraic equation, using the simplifying assumption that is appropriate for all weak acid and base ionizations. Ionization Constants for Weak Monoprotic Acids the MATH

25. Ionization Constants for Weak Monoprotic Acids Let’s look at the percent ionization of two previous weak acids as a function of their ionization constants. Note that the [H + ] in 0.15 M acetic acid is more than 200 times greater than for 0.15 M HCN. SolutionKaKa [H + ]pH% ionization 0.15 M acetic acid1.8 x 10 -5 1.6 x 10 -3 2.801.1 0.15 M HCN4.0 x 10 -10 7.7 x 10 -6 5.110.0051 % ionization = x 100% [unionized HY] [ionized HY]

26. All of the calculations and understanding we have at present can be applied to weak acids and weak bases. Calculate the concentrations of the various species in 0.15 M aqueous ammonia. K b = 1.8 E -5 R NH 3 NH 4 + + OH - I C E Ionization Constants for Weak Monoprotic Bases the MATH

27. The pH of an aqueous ammonia solution is 11.37. Calculate the molarity (original concentration) of the aqueous ammonia solution R NH 3 NH 4 + + OH - I C E Examination of the last equation suggests that our simplifying assumption can be applied. In other words (x-2.3x10 -3 )  x. –Making this assumption simplifies the calculation. Ionization Constants for Weak Monoprotic Bases the MATH

28. Polyprotic acids contain more than one ionizable proton, and the protons are lost in a stepwise manner. The fully protonated species is always the strongest acid because it is easier to remove a proton from a neutral molecule than from a negatively charged ion; the fully deprotonated species is the strongest base. Acid strength decreases with the loss of subsequent protons, and the pK a increases. The strengths of the conjugate acids and bases are related by pK a + pK b = pK w, and equilibrium favors formation of the weaker acid-base pair. Polyprotic Acids

29. Many weak acids contain two or more acidic hydrogens. –Examples include H 3 PO 4 and H 3 AsO 4. The calculation of equilibria for polyprotic acids is done in a stepwise fashion. –There is an ionization constant for each step. Consider arsenic acid, H 3 AsO 4, which has three ionization constants. 1K a1 = 2.5 x 10 -4 2K a2 = 5.6 x 10 -8 3K a3 = 3.0 x 10 -13 Notice that the ionization constants vary in the following fashion: This is a general relationship. –For weak polyprotic acids the K a1 is always > K a2, etc.

30. Polyprotic Acids The first ionization step for arsenic acid is: The second ionization step for arsenic acid is: The third ionization step for arsenic acid is:

31. Polyprotic Acids The MATH Calculate the concentration of all species in 0.100 M arsenic acid, H 3 AsO 4, solution. 1Write the first ionization step and represent the concentrations. Approach this problem exactly as previously done. R H 3 AsO 4 H 3 O + + H 2 AsO 4 - I C E The simplifying assumption cannot be used. Using the quadratic equation x =

32. 2.Next, write the equation for the second step ionization and represent the concentrations and work as before. R H 2 AsO 4 - H 3 O + + HAsO 4 -2 I C E The simplifying assumption can be used. Polyprotic Acids The MATH

33. 3.Finally, repeat the entire procedure for the third ionization step. R HAsO 4 -2 H 3 O + + AsO 4 -3 I C E The simplifying assumption can be used. 3.0x10 -13 /5.6x10 -8 = 5.4x10 -6 <1.0x10 -3 Polyprotic Acids The MATH

34. Polyprotic Acids A comparison of the various species in 0.100 M H 3 AsO 4 solution follows. SpeciesConcentration H 3 AsO 4 0.095 M H+H+ 4.9 x 10 -3 M H 2 AsO 4 - 4.9 x 10 -3 M HAsO 4 2- 5.6 x 10 -8 M AsO 4 3- 3.4 x 10 -18 M OH - 2.0 x 10 -12 M When a strong base is added to a solution of a polyprotic acid, the neutralization reaction occurs in stages. 1. The most acidic group is titrated first, followed by the next most acidic, and so forth 2. If the pK a values are separated by at least three pK a units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton

35. Polyprotic Acids

36. The ionization equilibrium of a weak acid (HA) is affected by the addition of either the conjugate base of the acid (A – ) or a strong acid (a source of H + ); LeChâtelier’s principle is used to predict the effect on the equilibrium position of the solution Common-ion effect Common-ion effect—the shift in the position of an equilibrium on addition of a substance that provides an ion in common with one of the ions already involved in the equilibrium; equilibrium is shifted in the direction that reduces the concentration of the common ion The Common Ion Effect The Common Ion Effect is important to 1.Buffers 2.Solubility Equilibrium

37. The Common Ion Effect and Buffer Solutions The Derivation of a Powerful Shortcut The general expression for the ionization of a weak monoprotic acid is: The generalized ionization constant expression for a weak acid is: If we solve the expression for [H + ], this relationship results: By making the assumption that the concentrations of the weak acid and the salt are reasonable, taking the logarithm of both sides, multiplying both sides by –1, and replacing the negative logarithms the expression reduces to:

38. The Common Ion Effect and Buffer Solutions Simple rearrangement of this equation and application of algebra yields the Henderson-Hasselbach equation

39. Buffers are characterized by the following: 1.the pH range over which they can maintain a constant pH—depends strongly on the chemical properties of the weak acid or base used to prepare the buffer (on K) 2.buffer capacity 2.buffer capacity, is the number of moles of strong acid or strong base needed to change the pH of 1 Liter of buffer solution by 1 pH unit. a.depends solely on the concentration of the species in the buffered solution (the more concentrated the buffer solution, the greater its buffer capacity) b.A general estimate of the buffer capacity is 40% of the sum of the molarities of the conjugate acid and conjugate base 3.observed change in the pH of the buffer is inversely proportional to the concentration of the buffer

40. The Henderson-Hasselbach equation is one method to calculate the pH of a buffer given the concentrations of the salt and acid. A special case exists for the Henderson-Hasselbalch equation when [base]/[acid] = some power of 10, regardless of the actual concentrations of the acid and base, where the Henderson- Hasselbalch equation can be interpreted without the need for calculations : [base]/[acid] = 10 x log 10 x = x in general pH = pK a + x Examples: 1. when [base] = [acid], [base]/[acid] = 1 or 10 0, log 1 = 0, pH = pK a, (corresponds to the midpoint in the titration of a weak acid or base) 2. when [base]/[acid] = 10 or 10 1, log 10 = 1 then pH = pK a + 1 3. when [base]/[acid] =.001 or 10 -2, log 10 = -2 then pH = pK a -2 Henderson-Hasselbalch equation is valid for solutions whose concentrations are at least 100 times greater than the value of their K a ’s Henderson-Hasselbalch - Caveats and Advantages

41. Buffer Solutions There are two common kinds of buffer solutions: I.Commonly, solutions made from a weak acid plus a soluble ionic salt of the conjugate base of the weak acid. II.Less common, solutions made from a weak base plus a soluble ionic salt of the conjugate acid of the weak base. Both of the above may also be prepared by starting with a weak acid (or weak base) and add half as many moles of strong base (acid)

42. This is an equilibrium problem with a starting concentration for both the cation and anion. After calculating the concentration of H + and the pH of a solution that is 0.15 M in both acetic acid sodium acetate yields: RCH 3 COOH H 3 O + + CH 3 COO - I C E One example of the type I of buffer system is: The weak acid - acetic acid CH 3 COOH The soluble ionic salt - sodium acetate NaCH 3 COO CH 3 COOH H 3 O + + CH 3 COO - NaCH 3 COO →Na + + CH 3 COO - ~100% Buffer Solutions Weak Acids Plus Salts of Their Conjugate Bases

43. Solution[H + ]pH 0.15 M CH 3 COOH1.6 x 10 -3 2.80 0.15 M CH 3 COOH & 0.15 M NaCH 3 COO buffer 1.8 x 10 -5 4.74 [H + ] is ~90 times greater in pure acetic acid than in buffer solution. Note that the pH of the buffer equals the pK a of the buffering acid. Alternatively you might have noticed: [base]/[acid] = 1 = 10 0 log 10 0 = 0 pH = pK a + 0 = pK a = 4.74 [H + ] = K a = 1.8E-5 Buffer Solutions Weak Acids Plus Salts of Their Conjugate Bases

44. Weak Bases plus Salts of Their Conjugate Acids We can derive a general relationship for buffer solutions that contain a weak base soluble and an ionic salt of the conjugate acid of the weak base similar to the acid buffer relationship. –The general ionization equation for weak bases is: Henderson-Hasselbach equation

45. This is an equilibrium problem with a starting concentration for both the cation and anion. After calculating the concentration of OH- and the pOH of the solution that is 0.15 M in aqueous ammonia, NH 3, and 0.30 M in ammonium nitrate, NH 4 NO 3 yeilds: R NH 3 NH 4 + + OH - I C E NH 4 NO 3 → NH 4 + + NO 3 - ~100% NH 3 NH 4 + + OH - Buffer Solutions: Weak Bases Plus Salts of Their Conjugate Acids One example of the type II of buffer system is: The weak base – ammonia NH 3 The soluble ionic salt – ammonium nitrate NH 4 NO 3

46. A comparison of the aqueous ammonia concentration to that of the buffer described above shows the buffering effect. Solution[OH - ]pH 0.15 M NH 3 1.6 x 10 -3 M11.20 0.15 M NH 3 & 0.15 M NH 4 NO 3 buffer 9.0 x 10 -6 M8.95 The [OH-] in aqueous ammonia is 180 times greater than in the buffer. The pK a of the base is 9.26 How effective is this buffer system? Buffer Solutions Weak Bases Plus Salts of Their Conjugate Acids

47. Buffering Action If 0.020 mole of gaseous HCl is added to 1.00 liter of a buffer solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the HCl. 1Calculate the pH of the original buffer solution. Substitute the quantities determined in the previous relationship into the ionization expression for ammonia. R NH 3 NH 4 + + OH - I C E NH 4 NO 3 → NH 4 + + NO 3 - ~100% NH 3 NH 4 + + OH -

48. NH 3 + H + NH 4 + Buffering Action 2Next, calculate the concentration of all species after the addition of the gaseous HCl. –The HCl will react with some of the ammonia and change the concentrations of the species. –This is another limiting reactant problem. R NH 3 NH 4 + + OH - I C E HCl → H + + Cl - ~100%

49. Buffering Action 3Using the concentrations of the salt and base and the Henderson- Hassselbach equation, the pH can be calculated. 4Finally, calculate the change in pH.

50. Buffering Action 1.If 0.020 mole of NaOH is added to 1.00 liter of solution that is 0.100 M in aqueous ammonia and 0.200 M in ammonium chloride, how much does the pH change? Assume no volume change due to addition of the solid NaOH. NH 4 + + OH - NH 3 R NH 3 NH 4 + + OH - I C E NaOH → Na + + OH - ~100%

51. Buffering Action 3.Finally, calculate the change in pH. 2.Using the concentrations of the salt and base and the Henderson- Hassselbach equation, the pH can be calculated.

52. Buffering Action Notice that the pH changes only slightly in each case. Original Solution Original pH Acid or base added New pH  pH 1.00 L of solution containing 0.100 M NH 3 and 0.200 M NH 4 Cl 8.95 0.020 mol NaOH9.08+0.13 0.020 mol HCl8.81-0.14

53. Preparation of Buffer Solutions Calculate the concentration of H + and the pH of the solution prepared by mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions. Determine the amounts of acetic acid and sodium hydroxide prior to the acid-base reaction. NaOH and CH 3 COOH react in a 1:1 mole ratio. After the two solutions are mixed, Calculate total volume. The concentrations of the acid and base are: Substitution of these values into the ionization constant expression (or the Henderson-Hasselbach equation) permits calculation of the pH.

54. Preparation of Buffer Solutions For biochemical situations, it is sometimes important to prepare a buffer solution of a given pH. Starting with a solution that is 0.100M in aqueous ammonia prepare 1.00L of a buffer solution that has a pH of 9.15 using ammonium chloride as the source of the soluble ionic salt of the conjugate weak acid. The Henderson-Hasselbalch equation is used to determine the ratio of the conjugate acid base pair pOH can be determined from the pH: pKb can be looked up in a table: [base] concentration is provided: Solve for [acid]: Does this result make sense? NH 4 Cl NH 4 + + Cl - ~100% NH 3 NH 4 + + OH - H2OH2O

55. Titration Curves Strong Acid/Strong Base Titration Curves These graphs are a plot of pH vs. volume of acid or base added in a titration. As an example, consider the titration of 100.0 mL of 0.100 M perchloric acid with 0.100 M potassium hydroxide. –In this case, we plot pH of the mixture vs. mL of KOH added. –Note that the reaction is a 1:1 mole ratio.

56. Strong Acid/Strong Base Titration Curves (An M 1 V 1 M 2 V 2 problem) Before any KOH is added the pH of the HClO 4 solution is 1.00. Remember perchloric acid is a strong acid that ionizes essentially 100%. mL KOHmMol KOH † mMol HClO 4 mL soln. Ұ pH 0010 * 1001.00 2028‡8‡ 1201.18 5055‡5‡ 1501.48 9091‡1‡ 1902.28 100100‡0‡ 2007.00 ‡ mMol HClO 4 = ml HClO 4 - mMol KOH *mMol HClO 4 = ml HClO 4 ∙ HClO 4 M pH = -log[H + ] mMol H + = mMol HClO 4 [H + ] = mMol HClO 4 / mL soln. Ұ mL soln = mL HClO 4 + mL KOH = 100 mL + total base added † mMol KOH = ml KOH∙NaOH M 1:1 mole ratio

57. Strong Acid/Strong Base Titration Curves We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve.

58. The pH changes much more gradually around the equivalence point in the titration of a weak acid or a weak base. [H + ] of a solution of a weak acid (HA) is not equal to the concentration of the acid (HA) [H + ] depends on both its K a and the analytical concentration of the acid (HA). Only a fraction of a weak acid dissociates, so [H + ] is less than [HA]; therefore, the pH of a solution of a weak acid is higher than the pH of a solution of a strong acid of the same concentration. Weak Acid/Strong Base Titration Curves

59. As an example, consider the titration of 100.0 mL of 0.100 M acetic acid, CH 3 COOH, (a weak acid) with 0.100 M KOH (a strong base). –The acid and base react in a 1:1 mole ratio. Before the equivalence point is reached, both CH 3 COOH and KCH 3 COO are present in solution forming a buffer. –The KOH reacts with CH 3 COOH to form KCH 3 COO. A weak acid plus the salt of a weak acid’s conjugate base form a buffer. Hypothesize how the buffer production will effect the titration curve.

60. Weak Acid/Strong Base Titration Curves (a RICE problem) 1.Determine the pH of the acetic acid solution before the titration is begun. 2.Solve the algebraic equation for each addition of strong base, using a simplifying assumption that is appropriate for all weak acid and base ionizations. At the equivalence point, the solution is 0.500 M in KCH 3 COO, the salt of a strong base and a weak acid which hydrolyzes to give a basic solution. 3.The solution cannot have a pH=7.00 at equivalence point. –Both processes make the solution basic. Concentrations must now be calculated using the equation for K b. Remember that K w = K a K b and that pH + pOH = 14

61. Weak Acid/Strong Base Titration Curves mL KOHmMol OH † [CH 3 COOH][CH 3 COO - ][H 3 O + ]mL soln. Ұ pH 000.1001.34E-02 1001.87 202.00.08001.67E-028.64E-051204.06 505.00.05003.33E-022.70E-051504.57 909.00.01004.74E-023.80E-061905.42 100105.30E-065.00E-021.89E-092008.72 110100.0005.00E-022.10E-1321012.68 After the equivalence point is reached, the pH is determined by the excess KOH just as in the strong acid/strong base example.

62. Weak Acid/Strong Base Titration Curves We have calculated only a few points on the titration curve. Similar calculations for remainder of titration show clearly the shape of the titration curve. Identity of the weak acid or base being titrated strongly affects the shape of the titration curve. The shape of titration curves as a function of the pK a or pK b shows that as the acid or base being titrated becomes weaker (its pK a or pK b becomes larger), the pH change around the equivalence point decreases significantly.

63. Weak Acid/Weak Base Titration Curves Weak Acid/Weak Base Titration curves have very short vertical sections. Visual indicators cannot be used. The solution is buffered both before and after the equivalence point. Comparison of the respective K a and K b values can be used to determine the pH of the equivalence points of these titrations. a.K base = K acid neutral solutions b. K base > K acid basic solutions c. K base < K acid acidic solutions

64. Acid-Base Indicators The point in a titration at which chemically equivalent amounts of acid and base have reacted is called the equivalence point. The point in a titration at which a chemical indicator changes color is called the end point. A symbolic representation of the indicator’s color change at the end point is: The equilibrium constant expression for an indicator would be expressed as:

65. Acid-Base Indicators If the preceding expression is rearranged the range over which the indicator changes color can be discerned.