2. Acid-Base Equilibria

3. Arrhenius Definition Acids are substances that produce hydrogen ions when dissolved in water. Acids are substances that produce hydrogen ions when dissolved in water. HCl → H + + Cl - Bases are substances that produce hydroxide ions when dissolved in water. Bases are substances that produce hydroxide ions when dissolved in water. NaOH → Na + + OH - Problem: NH 3 (ammonia) when dissolved in water forms NH 4 OH, a weak base, but NH 3 could not be an Arrhenius base based on traditional definition because NH 3 does not have a hydroxide to donate. Problem: NH 3 (ammonia) when dissolved in water forms NH 4 OH, a weak base, but NH 3 could not be an Arrhenius base based on traditional definition because NH 3 does not have a hydroxide to donate.

4. A Brønsted–Lowry acid… …must have a removable (acidic) proton. A Brønsted–Lowry base… …must have a pair of nonbonding electrons. Brønsted-Lowry Definitions

5. According to this theory, an acid is a proton (hydrogen ion, H + ) donor and a base is a proton (hydrogen ion, H + ) acceptor. …Consider introducing HCl (g) into water EOS Brønsted-Lowry Definitions

6. What Happens When an Acid Dissolves in Water? Water acts as a Brønsted– Lowry base and extracts a proton (H + ) from the acid, becoming a proton acceptor. As a result, the conjugate base of the acid and a hydronium ion (H 3 O + ) are formed.

7. NH 3 works under this definition as a base. Let’s see how. NH 3 works under this definition as a base. Let’s see how. Notice that water now behaves as an acid

8. If it can be either an acid or base…...it is amphiprotic (amphoteric). HCO 3 − HSO 4 − H2OH2OH2OH2O

9. Conjugate Acids and Bases: From the Latin word conjugare, meaning “to join together.” From the Latin word conjugare, meaning “to join together.” Reactions between acids and bases always yield their conjugate bases and acids. Reactions between acids and bases always yield their conjugate bases and acids. If the reaction proceeds in the forward direction, then HNO 2 acts as an acid by donating a hydrogen ion (proton). However, if the reaction were to then go backwards, then NO 2 -1 would act as a base by accepting a hydrogen ion (proton) Conjugate acid and base pairs are related by a hydrogen ion on either side of the equation

10. The conjugate acid of a base is the base PLUS the attached proton and the conjugate base of an acid is the acid MINUS the proton

11. Identify the acid, base, conjugate acid, conjugate base and pairs Identify the acid, base, conjugate acid, conjugate base and pairs HCN + NH 3  CN - +NH 4 + HCN + NH 3  CN - +NH 4 + Acid – HCN Acid – HCN Base – NH 3 Base – NH 3 Conjugate Base - CN - Conjugate Base - CN - Conjugate Acid - NH 4 + Conjugate Acid - NH 4 +

12. Acid and Base Strength A strong acid/base undergoes complete ionization A strong acid/base undergoes complete ionization Reaction goes to completion Reaction goes to completion Full dissociation. Equilibrium lies completely to the product side Full dissociation. Equilibrium lies completely to the product side LARGE value of Kc LARGE value of Kc Ex: HCl (aq)   H 3 O + (aq) +Cl - (aq) Ex: HCl (aq)   H 3 O + (aq) +Cl - (aq) Ex: NaOH (aq)   Na + (aq) +OH - (aq) Ex: NaOH (aq)   Na + (aq) +OH - (aq) The conjugate base/acid therefore is extremely weak The conjugate base/acid therefore is extremely weak Ex: Cl- has VERY poor ability to attract protons to itself Ex: Cl- has VERY poor ability to attract protons to itself

13. Acid and Base Strength Weak acids/bases have very little ionization. Very little (partial) dissociation in water Equilibria lies mostly to the LEFT. SMALL value of Kc Their conjugate bases/acidss are weak to exceedingly strong As acid/base strength decreases, the conjugate base/acid strength increases

14. Acid and Base Strength Substances with negligible acidity do not dissociate in water. Their conjugate bases are exceedingly strong. CH 4 + H 2 O   CH 3 - + H 3 O + CH 3 - is a VERY strong base due to its extreme attraction for H +

15. Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl (aq) + H 2 O (l)  H 3 O + (aq) + Cl − (aq) H 2 O is a much stronger base than Cl −, so the equilibrium lies so far to the right K is not measured (K>>1). ASSUMPTION: 100% dissociation. [H + ] or [H 3 O + ] = [ACID given] ** important when calculating pH

16. Acid and Base Strength Acetate ion is a stronger base than H 2 O, Predict K>>>1K>1K<1K<<<1 equilibrium favors the left side (K<<<1). HC 2 H 3 O 2 (aq) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 − (aq)

17. Consider the four diagrams where in each case, HA is an acid, H+ is a hydrogen ion, A- is an anion, and water molecules Consider the four diagrams where in each case, HA is an acid, H+ is a hydrogen ion, A- is an anion, and water molecules HAH + A-H2OA-H2OA-H2OA-H2O 1.Which diagram represents a relatively concentrated weak acid? Explain1.Which diagram represents a relatively concentrated weak acid? Explain 2.Which diagram represents a relatively concentrated strong acid? Explain2.Which diagram represents a relatively concentrated strong acid? Explain 3.Assign relative strengths and concentrations to the two remaining diagrams. Explain.

18. Carefully note that the concentration of H+ and OH- are dependent upon TWO, separate factors. Carefully note that the concentration of H+ and OH- are dependent upon TWO, separate factors. Strength of acid or base, i.e, degree of dissociation/ionization Strength of acid or base, i.e, degree of dissociation/ionization Amount of water present, i.e. concentration of sample solution Amount of water present, i.e. concentration of sample solution It is possible to have a dilute, strong acid and a weak, concentrated acid with the SAME hydronium ion concentration It is possible to have a dilute, strong acid and a weak, concentrated acid with the SAME hydronium ion concentration Therefore, the same pH value Therefore, the same pH value

19. Examples of Acids and Bases Weak AcidsOrganic (carboxylic) acids, such as butanoic, propanoic, ethanoic and methanoic Strong AcidsHCl, HBr, HI, HClO 4, HNO 3, H 2 SO 4 Weak BasesAmmonia, and organic bases such as amines and pyridines Strong BasesGroup 1 and Group 2 hydroxides

20. Autoionization of Water As we have seen, water is amphoteric. As we have seen, water is amphoteric. In pure water, a few molecules act as bases and a few act as acids. In pure water, a few molecules act as bases and a few act as acids. This is referred to as autoionization. This is referred to as autoionization. H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH − (aq)

21. pH (power of Hydrogen) Many of the concentration measurements in acid-base problems are given to us in terms of pH (power of H + ) and pOH (power of OH - ). Many of the concentration measurements in acid-base problems are given to us in terms of pH (power of H + ) and pOH (power of OH - ). p (anything) = -log (anything) p (anything) = -log (anything) pH = -log [H + ] pH = -log [H + ] pOH = -log [OH - ] pOH = -log [OH - ] pK a = -log K a pK a = -log K a

22. pH (power of H) pH scale is used to indicate the strength of an acid or base SAMPLE. pH scale is used to indicate the strength of an acid or base SAMPLE. Traditional range 0-14 Traditional range 0-14 Negative pH is possible Negative pH is possible 0- <7 Acidic 0- <7 Acidic 7> - 14 Basic 7> - 14 Basic 7 Neutral 7 Neutral

23. Recall: For a strong acid/base, ionization is considered to be 100%. Therefore Recall: For a strong acid/base, ionization is considered to be 100%. Therefore H + or OH - concentration can be determined directly from the stoichiometric ratio in the balanced equation and the concentration of acid or base. H + or OH - concentration can be determined directly from the stoichiometric ratio in the balanced equation and the concentration of acid or base. Ex: HNO 3 is a strong acid. If 0.10 M HNO3 dissociates, [H + ] is ALSO 0.10 M HNO 3  H + + NO3- 0.10 M0.10 M0.10M Ex: HNO 3 is a strong acid. If 0.10 M HNO3 dissociates, [H + ] is ALSO 0.10 M HNO 3  H + + NO3- 0.10 M0.10 M0.10M Calculate pH of above sample Calculate pH of above sample pH = -log[H + ] pH = -log[H + ] pH = -log [0.10] pH = -log [0.10] pH = 1.00 pH = 1.00

24. pH pH= -log[H + ] pH= -log[H + ] Used because [H + ] is usually very small Used because [H + ] is usually very small As pH decreases, [H + ] increases exponentially As pH decreases, [H + ] increases exponentially

25. Practice Calculate pH of each of the following solutions Calculate pH of each of the following solutions 0.0030 M HCl 0.0030 M HCl 0.030 M HCl 0.030 M HCl 0.30 M HCl 0.30 M HCl 3.0 M HCl 3.0 M HCl Carefully note that since pH is based on a log scale, there is a 10 fold change in concentration associated with a pH change of 1 unit (EXPONENTIAL) Carefully note that since pH is based on a log scale, there is a 10 fold change in concentration associated with a pH change of 1 unit (EXPONENTIAL)

26. pH of 3 = 10 times more concentrated than pH of ???, and ???times more concentrated than pH of 5 pH of 3 = 10 times more concentrated than pH of ???, and ???times more concentrated than pH of 5 pH – cont. pH of 3 = 10 times more concentrated than pH of 4, and 100 times more concentrated than pH of 5

27. Determining [conc] from pH/pOH EOS pH is a measure of the strength of an acid sample; low pH = stronger acid pOH is a measure of the strength of a base sample; low pOH = stronger base (also means HIGH pH pH = –log[H 3 O + ] and [H 3 O + ] = 10 (–pH) pOH = –log[OH – ] and [OH – ] = 10 (–pOH)

28. Practice: Predict the following from highest to lowest [H + ]. Then determine [H + ] if pH is 1.58 9.5 14 -0.87 4.2 Practice: Predict the following from highest to lowest [H + ]. Then determine [H + ] if pH is 1.58 9.5 14 -0.87 4.2

29. Additionally, at 298 K, 14 = pH + pOH Additionally, at 298 K, 14 = pH + pOH Therefore, ALL values (pH, pOH, OH - or H + ) can be determined based on ANY one known value Therefore, ALL values (pH, pOH, OH - or H + ) can be determined based on ANY one known value

30. Practice Calculate the pH of a 0.010 M solution of Lithium hydroxide Calculate the pH of a 0.010 M solution of Lithium hydroxide Calculate the hydrogen ion concentration in a solution with a pH of 4.32 Calculate the hydrogen ion concentration in a solution with a pH of 4.32 Calculate the pH of a solution made by dissolving 2.00g KOH in water to a total volume of 250. mL Calculate the pH of a solution made by dissolving 2.00g KOH in water to a total volume of 250. mL

31. pH of weak acids Weak acids do not completely dissociate in water. Weak acids do not completely dissociate in water. Equilibrate!! Equilibrate!! Problems solved similar to equilibrium problems. Problems solved similar to equilibrium problems. Recall: What is the definition of a weak acid? Incomplete dissociation What does the pH and/or strength of sample depend on? amount of acid dissociated [H + ] AND amount of water present.

32. HA (aq) + H 2 O (l)   H 3 O + (aq) + A - (aq) IX00 C-x+x EX - x0 + x We do NOT know how much acid actually dissociated Therefore, CANNOT use pH = - log [H + ] because CANNOT assume the [H + ] = [acid]. Carefully note that since ionization is minimal, value of x (dissociation) is small. Considered NEGLIGIBLE in terms of (X – x)

33. Additionally Additionally [H + ] = [A - ] [H + ] = [A - ] H 2 O (pure liquid) does not appear in K a (Acid dissociation constant) expression H 2 O (pure liquid) does not appear in K a (Acid dissociation constant) expression Therefore, the following Ka expression can be derived Therefore, the following Ka expression can be derived K a = OR And pK a = -log K a And pK a = -log K a [H + ][A - ] [HA] Mathematically equal. MUST NOT use as the EQUATION on ap frq. SHOW the ACTUAL equation. [H + ] 2 [HA] K a =

34. Practice Problems If ethanoic acid has a pK a 0f 4.74, what is its K a ? If ethanoic acid has a pK a 0f 4.74, what is its K a ? If propanoic acid has a Ka of 1.38 X 10 -5, what is it’s pK a ? If propanoic acid has a Ka of 1.38 X 10 -5, what is it’s pK a ? Which is a stronger acid, propanoic or ethanoic? Which is a stronger acid, propanoic or ethanoic? What is the pH of a 1.00 M solution of ethanoic acid? What is the pH of a 1.00 M solution of ethanoic acid? Calculate the K a value of a 0.100 M solution of a weak acid with a [H+] of 1.75 X 10 -3 M Calculate the K a value of a 0.100 M solution of a weak acid with a [H+] of 1.75 X 10 -3 M Calculate the K a of a solution of 0.250 M of a weak acid with a pH of 5.11 Calculate the K a of a solution of 0.250 M of a weak acid with a pH of 5.11

35. ORGANIC LESSON Carboxylic acid (organic acid) Carboxylic acid (organic acid) General makeup General makeup R-COOH R-COOH Example: ethanoic Example: ethanoic Ethane (C 2 H 6 ): remove methyl (CH 3 ) and add –COOH Ethane (C 2 H 6 ): remove methyl (CH 3 ) and add –COOH Formula: CH 3 COOH Formula: CH 3 COOH Example: propanoic Example: propanoic propane: (C 3 H 8 ): remove methyl (CH 3 ) and add –COOH propane: (C 3 H 8 ): remove methyl (CH 3 ) and add –COOH Formula: C 2 H 5 COOH Formula: C 2 H 5 COOH Determine formula for methanoic and butanoic acids. Draw Structure. Determine formula for methanoic and butanoic acids. Draw Structure.

36. Strength of carboxylic acid depends upon the stability of the anion (called: ___oate) formed when labile (mobile) proton is lost. Strength of carboxylic acid depends upon the stability of the anion (called: ___oate) formed when labile (mobile) proton is lost. All carboxylic acids that dissociates, give the following equilibrium, where R (hydrocarbon chain) can vary All carboxylic acids that dissociates, give the following equilibrium, where R (hydrocarbon chain) can vary RCOOH  RCOO - +H + RCOOH  RCOO - +H + Anions may form, releasing H + to form acidic solution Anions may form, releasing H + to form acidic solution Stability of anion depends on size of R group. Stability of anion depends on size of R group. Smaller R group = FEWER electrons “pumped” to COO- anion = HIGHER the stability of anion = STRONGER acid = LARGER Ka = SMALLER pKa Smaller R group = FEWER electrons “pumped” to COO- anion = HIGHER the stability of anion = STRONGER acid = LARGER Ka = SMALLER pKa Explain in terms of polarity: Explain in terms of polarity: Higher polarity of a smaller particle (methanoate) will cause greater attraction to water, therefore higher chance of H + ionizing. Lower polarity of a larger non-polar R group (butanoate) will have far less attraction to water, therefore smaller chance of H + ionizing. Higher polarity of a smaller particle (methanoate) will cause greater attraction to water, therefore higher chance of H + ionizing. Lower polarity of a larger non-polar R group (butanoate) will have far less attraction to water, therefore smaller chance of H + ionizing. RECALL: carboxylic acids up to 4 carbons are soluble/acidic RECALL: carboxylic acids up to 4 carbons are soluble/acidic

37. Carefully note: as carboxylic acid R group increases in size, acid strength decreases. Carefully note: as carboxylic acid R group increases in size, acid strength decreases. Ka decreases, pKa increases. Ka decreases, pKa increases.

38. Other related molecules Other related molecules Similar effects are observed when comparing relative strenghts of halogen substituted carboxylic acids. Similar effects are observed when comparing relative strenghts of halogen substituted carboxylic acids. As the number of chlorine atoms present increases, so does the acid strength. As the number of chlorine atoms present increases, so does the acid strength. The high electronegativity of Cl atoms attract electron density away from COO - The high electronegativity of Cl atoms attract electron density away from COO -

39. High Halogen Electronegativity = high Ka = Stronger Acid = Low pKa High Halogen Electronegativity = high Ka = Stronger Acid = Low pKa Why? Why? Greater attraction for electrons, so flourine pulls electrons away from the COO -, allowing easier removal of H + Greater attraction for electrons, so flourine pulls electrons away from the COO -, allowing easier removal of H +

40. K b : Equilibrium Constant for Weak Bases Similar to weak acid equilibria, weak base ionization is incomplete Similar to weak acid equilibria, weak base ionization is incomplete Cannot directly determine amount of base dissociated based on Molarity of solution Cannot directly determine amount of base dissociated based on Molarity of solution Cannot use pOH = -log [Base] Cannot use pOH = -log [Base] Must determine and use ACTUAL [OH - ] Must determine and use ACTUAL [OH - ] Same assumptions as weak acid equilibria calculations Same assumptions as weak acid equilibria calculations Assume x (dissociated ion) is negligible relative to X (base molarity) Assume x (dissociated ion) is negligible relative to X (base molarity) Assume [cation] = [OH - ] Assume [cation] = [OH - ]

41. NH 3 + H 2 O  NH 4 + + OH - NH 3 + H 2 O  NH 4 + + OH - Therefore: Kb = Kb = Often written as Kb = Kb =And pK b = -log K b pK b = -log K b [OH - ][NH 4 + ] [NH 3 ] [OH - ] 2 [NH 3 ] DO NOT USE ON AP FRQ. Use ACTUAL Kb expression. CLEARLY SPECIFY that you are ASSUMING [OH-] = [Cation] CLEARLY SPECIFY that you are ASSUMING [OH-] to be negligible, therefore using [Base] as given

42. Practice

43. Autoionization of Water Although pure water is essentially covalent, a small amount of self-ionization occurs Although pure water is essentially covalent, a small amount of self-ionization occurs Water autoionizes, by donating a proton to another water molecule: Water autoionizes, by donating a proton to another water molecule: H 2 O+ H 2 O  H 3 O + + OH - or H 2 O  H + + OH - No individual ion remains ionized for long No individual ion remains ionized for long At 298 K, only about 2 out of every 10 9 molecules are ionized at any given moment At 298 K, only about 2 out of every 10 9 molecules are ionized at any given moment

44. Autoionization of Water Equilibrium constant, Kw, written as Equilibrium constant, Kw, written as K w = [H + ][OH - ] At 298 K, K w = 1.0 X 10 -14 At 298 K, K w = 1.0 X 10 -14 Since [H + ] = [OH - ], therefore 1.0 X 10 -14 = [H + ][OH - ] Since [H + ] = [OH - ], therefore 1.0 X 10 -14 = [H + ][OH - ] and [H + ] = [OH - ] = = 1.0 X 10 -7 M and [H + ] = [OH - ] = = 1.0 X 10 -7 M Therefore, pH = -log [1.0 X 10 -7 M] = 7 Therefore, pH = -log [1.0 X 10 -7 M] = 7 √1.0 X 10 -14

45. Determine pKw Determine pKw pK w = -log K w = -log [1.0 X 10 -14 ] = 14 pK w = -log K w = -log [1.0 X 10 -14 ] = 14 Since we know that K w = [H + ][OH - ]therefore K w = K a · K b therefore pK w = pK a + pK b therefore 14 = pH + pOH *** KNOW THIS Since we know that K w = [H + ][OH - ]therefore K w = K a · K b therefore pK w = pK a + pK b therefore 14 = pH + pOH *** KNOW THIS

46. Temperature dependent Temperature dependent Carefully note that at temperatures other than 298 K, Kw value may be different, HOWEVER, [H + ] = [OH - ] Carefully note that at temperatures other than 298 K, Kw value may be different, HOWEVER, [H + ] = [OH - ] pH may be different from 7 pH may be different from 7 HOWEVER, still considered NEUTRAL due to [H + ] = [OH - ] (no excess H + or OH - ) HOWEVER, still considered NEUTRAL due to [H + ] = [OH - ] (no excess H + or OH - )

47. Autoionization of Water H 2 O  H + + OH - Since this is specifically for the autoionization of water we use K w as the equilibrium constant. Since this is specifically for the autoionization of water we use K w as the equilibrium constant. At 25 o C: At 25 o C: K w = [H + ][OH - ] = 1.0 x 10 -14 [H + ] > [OH - ]  acid [H + ] > [OH - ]  acid [H + ] < [OH - ]  base [H + ] < [OH - ]  base [H + ] = [OH - ]  neutral [H + ] = [OH - ]  neutral

48. Autoionization of Water H 2 O  H + + OH - K w = [H + ][OH - ] = 1.0 x 10 -14 At this point it is very useful to remember that pH = - log [H + ] At this point it is very useful to remember that pH = - log [H + ] Since K w is a constant, if we are given [OH - ] it is a very simple matter to calculate pH Since K w is a constant, if we are given [OH - ] it is a very simple matter to calculate pH Since pH, pOH, [H + ], [OH - ], pKa, pKb, pKw and K w are ALL related, knowing ANY of these values is sufficient to calculate ALL others. Since pH, pOH, [H + ], [OH - ], pKa, pKb, pKw and K w are ALL related, knowing ANY of these values is sufficient to calculate ALL others.

49. Titrations Titration: experimental technique to perform a stoichiometrically balanced neutralization reaction. Titration: experimental technique to perform a stoichiometrically balanced neutralization reaction. Accurately graduated glassware (volumetric flasks, graduated glass pipets and burets) is used in quantitative manner Accurately graduated glassware (volumetric flasks, graduated glass pipets and burets) is used in quantitative manner Helps determine unknown concentration of an acid or a base Helps determine unknown concentration of an acid or a base

50. Titrations Terms to Know: Terms to Know: Titrant:solution added FROM the buret Titrant:solution added FROM the buret Titrate: solution that titrant is added to. Titrate: solution that titrant is added to. Equivalence point: 100% neutralization. [H + ] = [OH - ] Equivalence point: 100% neutralization. [H + ] = [OH - ] End point: Change in color of indicator End point: Change in color of indicator Half Equivalence point: 1/2 total amount of titrant added. Buffer solution pH = pK a Half Equivalence point: 1/2 total amount of titrant added. Buffer solution pH = pK a

51. Titrations As acid or base is added, there is As acid or base is added, there is Very little change in pH Very little change in pH pH change of less than 1.5 units expected upto the point where 90% of acid/base is neutralized pH change of less than 1.5 units expected upto the point where 90% of acid/base is neutralized At equivalence point, when 100% of acid or base is neutralized, RAPID change in pH is observed At equivalence point, when 100% of acid or base is neutralized, RAPID change in pH is observed Summarized using titration curve plots Summarized using titration curve plots

52. Strong acid-Strong base Titration At equivalence point, pH = 7 At equivalence point, pH = 7 Carefully note: IF and ONLY IF concentrations of monoprotic acid and base are equal, then VOLUME of acid and base will also be equal. Carefully note: IF and ONLY IF concentrations of monoprotic acid and base are equal, then VOLUME of acid and base will also be equal. Generally, expect questions where EITHER concentration OR volume of acid OR base is UNKNOWN, and you are asked to solve for it. Generally, expect questions where EITHER concentration OR volume of acid OR base is UNKNOWN, and you are asked to solve for it. DUMB QUESTION: In the above example, which is the titrate and which is the titrant? How would you know if labels were not given?

53. Weak acid-Strong Base Titration PREDICT: How will a weak acid-strong base curve differ from strong acid-strong base curve? PREDICT: How will a weak acid-strong base curve differ from strong acid-strong base curve? Starts at a HIGHER pH Starts at a HIGHER pH Plateaus at relatively HIGH basic pH Plateaus at relatively HIGH basic pH Equivalence point ABOVE 7. Why? Equivalence point ABOVE 7. Why? Weak acids doesn’t ionize fully. Removal of dissociated H + due to neutralization FORCES equilibrium shift towards dissociation of acid. Reaction rate SLOWS down as 100% neutralization approaches. Need EXCESS base to FULLY ionize the weak acid for complete neutralization

54. Strong acid-Weak base Titration PREDICT: How will a strong acid-weak base curve differ from strong acid- strong base curve? PREDICT: How will a strong acid-weak base curve differ from strong acid- strong base curve? Starts at LOW pH Starts at LOW pH Equivalence point BELOW 7 Equivalence point BELOW 7 Plateaus at relatively LOW basic pH Plateaus at relatively LOW basic pH

55. Weak acid-Weak base Titration Starts at relatively HIGH acid pH Starts at relatively HIGH acid pH Plateaus at relatively LOW basic pH Plateaus at relatively LOW basic pH No sharp change in pH at equivalence point No sharp change in pH at equivalence point Equivalence point likely at pH = 7 Equivalence point likely at pH = 7 May be variable based on relative strength of acid and base. May be variable based on relative strength of acid and base.

56. Polyprotic acid neutralization Multiple equivalence points Multiple equivalence points Recall: Acid strength decreases with EACH hydrogen ion removal Recall: Acid strength decreases with EACH hydrogen ion removal H 3 PO 4 > H 2 PO 4 -1 > HPO4 -2 H 3 PO 4 > H 2 PO 4 -1 > HPO4 -2

57. Indicators Most acids and bases form colorless solutions Most acids and bases form colorless solutions Driving force of neutralization reaction (formation of water) does not produce an observable change. Driving force of neutralization reaction (formation of water) does not produce an observable change. Need a method of determining when equivalence point has been reached. Need a method of determining when equivalence point has been reached. Indicator: chemical that changes color at various pH’s Indicator: chemical that changes color at various pH’s Often weak acids, where ionized and unionized form have different colors. Often weak acids, where ionized and unionized form have different colors. Ideal indicators change color over a small, given range of pH Ideal indicators change color over a small, given range of pH

58. Indicators Choose an indicator that changes color at a pH value as close to the equivalence point as possible Choose an indicator that changes color at a pH value as close to the equivalence point as possible End point should correspond closely to equivalence point. End point should correspond closely to equivalence point. Suitable Indicators: Suitable Indicators: Strong acid-Strong basemost Strong acid-Strong basemost Weak Acid-Strong basephenolphthalein Weak Acid-Strong basephenolphthalein Strong acid-Weak basemethyl orange Strong acid-Weak basemethyl orange Weak acid-Weak basenone due to lack of sharp change in pH (use pH meter) Weak acid-Weak basenone due to lack of sharp change in pH (use pH meter)

59. Buffer Solutions A buffer solution is one that resists changes in pH, when either a small amount of acid or base is added to it. A buffer solution is one that resists changes in pH, when either a small amount of acid or base is added to it. General composition General composition Weak acid and one of its salts (i.e. its conjugate base) Weak acid and one of its salts (i.e. its conjugate base) Weak base and one of its salts (i.e its conjugate acid) Weak base and one of its salts (i.e its conjugate acid) Example: ethanoic acid and sodium ethanoate Example: ethanoic acid and sodium ethanoate

60. Ethanoic acid acts as the acid in the buffer solution. Absorbs base CH 3 COOH + OH -  CH 3 COO - + H 2 O Ethanoic acid acts as the acid in the buffer solution. Absorbs base CH 3 COOH + OH -  CH 3 COO - + H 2 O Ethanoate ion acts as the base. Absorbs acid CH 3 COO - + H 3 O +  CH 3 COOH + H 2 O Ethanoate ion acts as the base. Absorbs acid CH 3 COO - + H 3 O +  CH 3 COOH + H 2 O Process of “mopping up” of added acids and bases allows pH to remain relatively unchanged. Process of “mopping up” of added acids and bases allows pH to remain relatively unchanged.

61. Capacity of buffers Capacity defined as the ability to continue reacting as extra acid/base is added. Depends upon Capacity defined as the ability to continue reacting as extra acid/base is added. Depends upon Concentration of components Concentration of components Higher the concentration, the more acid or base it can absorb, the higher its capacity Higher the concentration, the more acid or base it can absorb, the higher its capacity

62. pH of Buffers pH of buffer is determined by the pH of buffer is determined by the pK a or pK b of the weak acid or base, and pK a or pK b of the weak acid or base, and Ratio of concentrations of each component Ratio of concentrations of each component To calculate pH of buffer solution, use Henderson-Hasselbach equation for acidic buffer To calculate pH of buffer solution, use Henderson-Hasselbach equation for acidic buffer for basic buffer for basic buffer [salt] [acid] pH = pK a + log [salt] [base] pOH = pK b + log

63. = BUFFER SOLUTION Carefully note a crucial application of buffer solutions (expect ** AP ** question) Carefully note a crucial application of buffer solutions (expect ** AP ** question) Weak acid-strong base or weak base-strong acid titrations produce a buffer region on the graph Weak acid-strong base or weak base-strong acid titrations produce a buffer region on the graph Gradual addition of Strong base to weak acid results in Gradual addition of Strong base to weak acid results in Some weak acid neutralization Some weak acid neutralization Production of the salt of weak acid Production of the salt of weak acid Production of water Production of water

65. When 1/2 the acid is neutralized, what is the ratio of acid to salt? When 1/2 the acid is neutralized, what is the ratio of acid to salt? 1:1 1:1 Plugging into Henderson-Hasselbach, we get log (1) = 0 Plugging into Henderson-Hasselbach, we get log (1) = 0 Therefore when acid to salt ratio 1:1, then pH = pKa Therefore when acid to salt ratio 1:1, then pH = pKa True at ½ equivalence point True at ½ equivalence point

66. Practice Ethanoic acid has a pKa of 4.75. Find the pH of the solution that results from the addition of 40.0 mL of 0.040 M NaOH to 50.0 mL of 0.075 M ethanoic acid Ethanoic acid has a pKa of 4.75. Find the pH of the solution that results from the addition of 40.0 mL of 0.040 M NaOH to 50.0 mL of 0.075 M ethanoic acid Methanoic acid has a Ka = 1.60 X 10-4. Calculate the pH of the final solution when 23.90 mL of 0.100 M sodium h Methanoic acid has a Ka = 1.60 X 10-4. Calculate the pH of the final solution when 23.90 mL of 0.100 M sodium h

67. Acid dissociation constant = K a Acid dissociation constant is the equilibrium constant for an acid. Acid dissociation constant is the equilibrium constant for an acid. HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) K a = [H 3 O + ][A - ] [HA] K a = [H 3 O + ][A - ] [HA] H 3 O + is often written H +, ignore the water in equation. H 3 O + is often written H +, ignore the water in equation. R

68. Acid dissociation constant K a HA (aq) H + (aq) + A - (aq) HA (aq) H + (aq) + A - (aq) K a = [H + ][A - ] [HA] K a = [H + ][A - ] [HA] We can write the expression for any acid. We can write the expression for any acid. Strong acids dissociate completely. Strong acids dissociate completely. Equilibrium far to right. Equilibrium far to right. Conjugate base is weak. Conjugate base is weak.

69. Write the acid dissociation/ionization reaction and its K a (omit water) Write the acid dissociation/ionization reaction and its K a (omit water) HC 2 H 3 O 2 for Acetic Acid HC 2 H 3 O 2 for Acetic Acid

70. K b = Base dissociation constant B (aq) + H 2 O (l)  BH + (aq) + OH - (aq) B (aq) + H 2 O (l)  BH + (aq) + OH - (aq) K b =[BH + ][OH - ] K b =[BH + ][OH - ] [B] [B]

71. Strong Acids Equilibrium lies far to the right Equilibrium lies far to the right Almost 100% dissociation Almost 100% dissociation HA  H + + A - 1%99% Gives a weak conjugate base (weaker than water) Gives a weak conjugate base (weaker than water) Large K a Large K a

72. Weak acids Equilibrium lies to the left Equilibrium lies to the left Very little H + Very little H + HA  H + + A - 99%1% Strong conjugate base (stronger than water) Strong conjugate base (stronger than water) Small K a Small K a

73. Back to Pairs Strong acids Strong acids K a is large K a is large [H + ] is >>> to [HA] [H + ] is >>> to [HA] A - is a weaker base than water A - is a weaker base than water Weak acids K a is small [H + ] <<< [HA] A - is a stronger base than water

74. Weak Acids and Bases For weak acids and bases, equations can be written to describe equilibrium conditions EOS

75. For strong acids, water has a leveling effect; that is, when the strong acids are dissolved in water, they all completely ionize to the hydronium ion Acid–Base Strength (cont’d) K a values are used to compare the strengths of weak acids;  K,  strength

76. Strong Bases Large K b Large K b Almost completely dissociated Almost completely dissociated Conjugate acid is weak Conjugate acid is weak Alkali metal hydroxides NaOH, LiOH, KOH Alkali metal hydroxides NaOH, LiOH, KOH Ba(OH) 2 Ba(OH) 2 Ca(OH) 2 or Sr(OH) 2 Ca(OH) 2 or Sr(OH) 2

77. K w, K a, K b, pH, Oh my! How does this all relate and tie together??? The symbols are actually very basic: K w is the equilibrium constant for water K w is the equilibrium constant for water K a is the acid-dissociation constant K a is the acid-dissociation constant K b is the base-dissociation constant K b is the base-dissociation constant pH (power of hydrogen) is the –log of the concentration of hydrogen pH (power of hydrogen) is the –log of the concentration of hydrogen S

78. K w, K a, K b, Oh my! Given an acid reaction: HA (aq) + H 2 O (l)  H + (aq) + A - (aq) K a = [H + ] [A - ] [HA] [HA] Given a base reaction: B (aq) + H 2 O (l)  HB + (aq) + OH - (aq) K b = [HB + ] [OH - ] [B] [B]

79. K w, K a, K b, Oh my! Here we have a conjugate acid-base pair: NH 4 + (aq)  NH 3 (aq) + H + (aq) NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq) K a = [NH 3 ] [H + ] [NH 4 + ] [NH 4 + ] K b = [NH 4 + ] [OH - ] [NH 3 ] [NH 3 ]

80. Mathematical Approach We can add the two equations together: We can add the two equations together: NH 4 + (aq)  NH 3 (aq) + H + (aq) NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq) NH 3(aq) + H 2 O (l) + NH 4 + (aq)  NH 3 (aq) + H + (aq) + NH 4 + (aq) + OH - (aq) Cancel out the duplicate terms on opposite sides of the arrows… Cancel out the duplicate terms on opposite sides of the arrows… NH 3(aq) + H 2 O (l) + NH 4 + (aq)  NH 3 (aq) + H + (aq) + NH 4 + (aq) + OH - (aq) You are now left with: You are now left with: H 2 O  H + + OH -

81. Mathematical Approach If we try to multiply K a and K b. this is what we get: If we try to multiply K a and K b. this is what we get: K a x K b K a x K b [NH 3 ] [H + ] x [NH 4 + ] [OH - ]= [NH 4 + ] [NH 3 ] [NH 4 + ] [NH 3 ]Therefore… K a x K b = [H + ] x [OH - ] K a x K b = 1.0 x 10 -14 = K w

82. Watch This! Because [H 3 O + ] [OH − ] = K w = 1.0  10 −14, we know that −log [H 3 O + ] + −log [OH − ] = −log K w = 14.00 or, in other words, pH + pOH = pK w = 14.00

83. Mathematical Approach K W = [H + ][OH - ] = 1.0 x 10 -14 K a × K b = K w In many books you will find a table for K a they rarely are given for K b since it is such an easy calculation

84. Relationships K w = [H + ][OH - ] K w = [H + ][OH - ] -log K w = -log([H + ][OH - ]) -log K w = -log([H + ][OH - ]) -log K w = -log[H + ]+ -log[OH - ] -log K w = -log[H + ]+ -log[OH - ] pK w = pH + pOH pK w = pH + pOH K w = 1.0 x10 -14 K w = 1.0 x10 -14 14.00 = pH + pOH 14.00 = pH + pOH Given any one of these we can find the other three: Given any one of these we can find the other three: [H + ], [OH - ], pH and pOH

85. Lewis Acids Lewis acids are defined as electron-pair acceptors. Includes cations and incomplete octets. Atoms with an empty valence orbital can be Lewis acids.

86. Lewis Bases Lewis bases are defined as electron-pair donors. Should have a lone pair of electrons. Anything that could be a Brønsted–Lowry base is a Lewis base. Lewis bases can interact with things other than protons, however.

87. Lewis Acids and Bases Boron triflouride wants more electrons, because it has an incomplete octet. Boron triflouride wants more electrons, because it has an incomplete octet. BF F F :N H H H B F F F N H H H

88. pH of strong acids Finding the pH of strong acids is easy since there is almost complete dissociation Finding the pH of strong acids is easy since there is almost complete dissociation Common strong acids include: Common strong acids include: HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 4 ASSUMPTION: [H+] = [acid given]

89. Example ASSUMPTION: [H+] = [acid given] ASSUMPTION: [H+] = [acid given] HCl (aq) + H 2 O (l)  H 3 O + (aq ) + Cl - (aq) HCl (aq) + H 2 O (l)  H 3 O + (aq ) + Cl - (aq) 0.0004% at 99.996% at equilibrium equilibrium

90. Example Determine the pH of a 0.5 M HBr solution Determine the pH of a 0.5 M HBr solution First,write the equilibrium reaction First,write the equilibrium reaction For strong acids, we assume 100% dissociation, therefore For strong acids, we assume 100% dissociation, therefore [H+] = [HBr]=0.5 M Determine pH pH = - log [H+] Determine pH pH = - log [H+] pH = - log [0.5] pH = - log [0.5] pH = 0.3 pH = 0.3