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Autoionization reaction of liquid water pH, pOH , and p K w conjugate acid-base pairs

Acids and bases. Autoionization reaction of liquid water pH, pOH , and p K w conjugate acid-base pairs acid or base strength and the magnitude of K a , K b , p K a , and p K b leveling effect To be able to predict whether reactants or products are favored in an acid-base equilibrium

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Autoionization reaction of liquid water pH, pOH , and p K w conjugate acid-base pairs

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  1. Acids and bases Autoionization reaction of liquid water pH, pOH, and pKw conjugate acid-base pairs acid or base strength and the magnitude ofKa, Kb, pKa, and pKb leveling effect To be able to predict whether reactants or products are favored in an acid-base equilibrium usemolecular structure and acid and base strengths use Ka and Kb values to calculate the percent ionization and pH of a solution of an acid or a base calculate the pH at any point in an acid-base titration common ion effects and the position of an acid-base equilibrium how a buffer works and how to use the Henderson-Hasselbalch equation to calculate the pH of a buffer

  2. Table of Common Ions Common Positive Ions (Cations)

  3. Table of Common Ions Common Negative Ions (Anions)

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

  5. Acids and bases • 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 H2O. • 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. •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.

  6. Acids and bases • 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-basepairs. • HCl (aq)+ H2O (l) H3O+(aq)+ Cl–(aq) parent acid parent base conjugate acid conjugate base 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, H3O+. Substances that can behave as both an acid and a base are said to be amphoteric.

  7. Acids and bases can be defined in different ways: • 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 CH4 dissolve in water and do not give acidic solutions • 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. • 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-H2O 5. At equivalence point 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. pH = log (1/ [H+]) = - log [H+] Low pH: high [H+] High pH: low [H+] Acidic solution pH < 7 Neutral pH = 7 Basic solution pH > 7

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

  11. It’s all because of Gibbs Free Energy No acid stronger than H3O+ 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 (H3O+) is leveled to the strength of H3O+ in aqueous solution because H3O+ 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 http://www.chem1.com/acad/webtext/abcon/abcon-2.html

  12. K2 KH K1 CO2+3H2O H2CO3+2H2O HCO3-+H3O++H2O CO32-+2H3O+ Acids and bases CO2 + H2O H2CO3 H2CO3 + H2O HCO3- + H+ HCO3- + H2O CO32- + H+ Baking Soda NaHCO3 Soda Pop CO2 Atmosphere .8317 5.61×10−11 2.5×10−4 CO2 + CaCO3 + H2O 2HCO3- + Ca2- Biological Calcification (not a reversible reaction) Critter's shells

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

  14. Autoionization of Water • Because water is amphiprotic, one water molecule can react with another to form an OH– ion and an H3O+ ion in an autoionization process: 2H2O(l)⇋H3O+ (aq) + OH– (aq) • Equilibrium constant K for this reaction can be written as [H3O+] [OH–] [H2O]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: [H3O+]=[OH–] = 1.0 x 10–7 M [H3O+][OH–] = 1.0 x 10–14 M = Kw pH = pOH = 7 pH + pOH = pKw = 14 Kc = Kc [H2O]2 = Kw = [H3O+][OH–] = 1.0 x 10–14

  15. Ionization Constants for Weak Acids and Bases When is a 1M solution not a 1 M solution? • 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.

  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, Ka, the acid ionization constant, using this definition. • The symbol for the ionization constant is Ka. • The larger the Ka value the stronger the acid and the higher the equilibrium [H+] • The larger the Kb value the stronger the base and the higher the equilibrium [OH–]

  17. Acids, Bases, and ionization constants • • Acid and Base strengths can be compared using Ka and Kb values. The larger the Ka or Kb 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. NaCH3COO → Na+ + CH3COO- • Acetate ion is the conjugate base of acetic acid, a weak acid. • Step 2. CH3COO- + H2O CH3COOH + OH- • Hydrolysis: Aqueous solutions of salts that dissociate into both: • A strong conjugate acid and a strong conjugate base are neutral (KNO3). • A strong conjugate acid and a weak conjugate base are acidic (HCl). • A strong conjugate base and a weak conjugate acid are basic (NaOH). • A weak conjugate base and a weak conjugate acid can be neutral, basic or acidic: • The comparison of the values of Ka and Kb determine the pH of these solutions. • Kbase = Kacidmake neutral solutions (NH4CH3OO) • Kbase > Kacidmake basic solutions (NH4ClO) • Kbase < Kacidmake acidic solutions (CH3)3NHF stronger acid + stronger base weaker acid + weaker base H2O

  18. Ionization Constants for Weak Monoprotic Acids and Bases • 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 Ka and strength? • What is the relationship between pKa and strength? • What is the relationship between pH and strength?

  19. Ionization Constants for Weak Monoprotic Acids and Basesthe 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 Ka or Kb. • – Two common ways to obtain the concentrations • By measuring the electrical conductivity of the solution, which is related to the total concentration of ions present • By measuring the pH of the solution, which gives [H+] or [OH–]

  20. Ionization Constants for Weak Monoprotic Acids the MATH 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. • Use the concentrations that were just determined in the ionization constant expression to get the value of Ka.

  21. Ionization Constants for Weak Monoprotic Acids the MATH • 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 [H3O+] and the stoichiometry of the ionization reaction to determine concentrations of all species. Simplifying Assumption: Is the change significant? Later we will find that in general, if the Ka/[] is < 1x10-3 you can apply the simplifying assumption. • Calculate the ionization constant from this information.

  22. Ionization Constants for Weak Monoprotic Acidsthe MATH Calculate the concentrations of the various species in 0.15 M acetic acid, CH3COOH, solution. • It is always a good idea to write down the ionization reaction and the ionization constant expression. Next we combine the basic chemical concepts with some algebra to solve the problem

  23. Ionization Constants for Weak Monoprotic Acidsthe MATH • 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. • 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. Ionization Constants for Weak Monoprotic Acidsthe MATH • Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution. Ka= 4.0 x 10-10 for HCN • 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.

  25. 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. Ionization Constants for Weak Monoprotic Acids [ionized HY] % ionization = x 100% [unionized HY]

  26. Ionization Constants for Weak Monoprotic Basesthe MATH • 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. Kb = 1.8E-5

  27. Ionization Constants for Weak Monoprotic Basesthe MATH • The pH of an aqueous ammonia solution is 11.37. Calculate the molarity (original concentration) of the aqueous ammonia solution • 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.

  28. Polyprotic Acids • 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 pKa increases. • The strengths of the conjugate acids and bases are related by pKa + pKb= pKw, and equilibrium favors formation of the weaker acid-base pair.

  29. Polyprotic Acids • Many weak acids contain two or more acidic hydrogens. • Examples include H3PO4 and H3AsO4. • The calculation of equilibria for polyprotic acids is done in a stepwise fashion. • There is an ionization constant for each step. • Consider arsenic acid, H3AsO4, which has three ionization constants. • Ka1 = 2.5 x 10-4 • Ka2 = 5.6 x 10-8 • Ka3 = 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 Ka1 is always > Ka2, 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, H3AsO4, solution. • Write the first ionization step and represent the concentrations. Approach this problem exactly as previously done. The simplifying assumption cannot be used. Using the quadratic equation x=

  32. Polyprotic Acids The MATH • Next, write the equation for the second step ionization and represent the concentrations and work as before. The simplifying assumption can be used.

  33. Polyprotic Acids The MATH • Finally, repeat the entire procedure for the third ionization step. The simplifying assumption can be used. 3.0x10-13/5.6x10-8 = 5.4x10-6<1.0x10-3

  34. Polyprotic Acids • A comparison of the various species in 0.100 M H3AsO4 solution follows. • 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 pKa values are separated by at least three pKa units, then the overall titration curve shows well-resolved “steps” corresponding to the titration of each proton

  35. Polyprotic Acids

  36. The Common Ion Effect • 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—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 is important to Buffers Solubility Equilibrium

  37. The general expression for the ionization of a weak monoprotic acid is: The generalized ionization constant expression for a weak acid is: The Common Ion Effect and Buffer SolutionsThe Derivation of a Powerful Shortcut • 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: • 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) • 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. • depends solely on the concentration of the species in the buffered solution (the more concentrated the buffer solution, the greater its buffer capacity) • A general estimate of the buffer capacity is 40% of the sum of the molarities of the conjugate acid and conjugate base • observed change in the pH of the buffer is inversely proportional to the concentration of the buffer

  40. Henderson-Hasselbalch - Caveats and Advantages • Henderson-Hasselbalch equation is valid for solutions whose concentrations are at least 100 times greater than the value of their Ka’s • 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] = 10x • log 10x = x • in general pH = pKa + x • Examples: • when [base] = [acid], [base]/[acid] = 1 or 100, log 1 = 0, pH = pKa, • (corresponds to the midpoint in the titration of a weak acid or base) • 2. when [base]/[acid] = 10 or 101, log 10 = 1 then pH = pKa + 1 • 3. when [base]/[acid] = .001 or 10-2, log 10 = -2 then pH = pKa -2

  41. Buffer Solutions There are two common kinds of buffer solutions: • Commonly, solutions made from a weak acid plus a soluble ionic salt of the conjugate base of the weak acid. • 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. Buffer Solutions Weak Acids Plus Salts of Their Conjugate Bases • One example of the type I of buffer system is: • The weak acid - acetic acid CH3COOH • The soluble ionic salt - sodium acetate NaCH3COO CH3COOH H3O+ + CH3COO- ~100% NaCH3COO →Na+ + CH3COO- • 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:

  43. Buffer Solutions Weak Acids Plus Salts of Their Conjugate Bases • Alternatively you might have noticed: • [base]/[acid] = 1 = 100 • log 100 = 0 • pH = pKa + 0 = pKa= 4.74 • [H+] = Ka = 1.8E-5 • [H+] is ~90 times greater in pure acetic acid than in buffer solution. • Note that the pH of the buffer equals the pKa of the buffering acid.

  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. Buffer Solutions: Weak Bases Plus Salts of Their Conjugate Acids • One example of the type II of buffer system is: • The weak base – ammonia NH3 • The soluble ionic salt – ammonium nitrate NH4NO3 • 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, NH3, and 0.30 M in ammonium nitrate, NH4NO3 yeilds: NH3 NH4 ++OH- NH4NO3→NH4 ++NO3- ~100%

  46. Buffer Solutions Weak Bases Plus Salts of Their Conjugate Acids • A comparison of the aqueous ammonia concentration to that of the buffer described above shows the buffering effect. • The [OH-] in aqueous ammonia is 180times greater than in the buffer. • The pKa of the base is 9.26 How effective is this buffer system?

  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. • Calculate the pH of the original buffer solution. NH4NO3→NH4 ++NO3- ~100% NH3 NH4 ++OH- • Substitute the quantities determined in the previous relationship into the ionization expression for ammonia.

  48. Buffering Action • Next, 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. HCl→H++Cl- ~100% NH3 + H+ NH4 +

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

  50. Buffering Action • 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. NaOH →Na++OH- ~100% NH4 + + OH- NH3

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