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Unit 17 Acids and Bases Chapter 14. Arrhenius Model Acids produce H + in aqueous solution while bases produce OH - Bronsted-Lowry Model Acid is a proton.

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Presentation on theme: "Unit 17 Acids and Bases Chapter 14. Arrhenius Model Acids produce H + in aqueous solution while bases produce OH - Bronsted-Lowry Model Acid is a proton."— Presentation transcript:

1 Unit 17 Acids and Bases Chapter 14

2 Arrhenius Model Acids produce H + in aqueous solution while bases produce OH - Bronsted-Lowry Model Acid is a proton (H+) donor. Base is proton acceptor Show how water can act as an acid or a base Hydronium ion- H 3 O + Section 14.1: The Nature of Acids and Bases

3 Conjugate acid/base pair- two substances related to each other by the donating and accepting of a single proton Think think of the reaction between an acid and a base as competition for the H + between the two bases (the original base and the CB)

4 Acid Dissociation in Water: HA + H 2 O ↔ H 3 O + + A - or HA + H 2 O ↔ H + + A - Acid dissociation constant (K a ): K a = [H 3 O + ][A - ]/[HA] or K a = [H+][A-]/[HA] *Remember that pure solids or pure liquids are not included in equilibrium expressions- this is the reason that water is not present in the equation *Although water is not included in the equilibrium expression, it does play a role in ionization of the acid Example: pg. 673 (#28 both CB/CA and Ka)

5 Strong acid- almost all the original HA is dissociated at equilibrium (→) strong acid yields a weak CB The CB is much weaker than water for the affinity for H + HA + H 2 O→H 3 O + + A - The equilibrium lies far to the right for a strong acid K a is large because the equilibrium lies to the right (product favored) All [HA] o = [H + ] eq and [HA] eq =0 because we assume 100% dissociation Section 14.2: Acid Strength

6 Weak acid- acid for which very little dissociates (↔ or ←) weak acid yields a strong CB The CB is much stronger than water for the affinity for H+ HA + H 2 O←H 3 O + + A - The equilibrium lies far to the left for a weak acid Ka is small because the equilibrium lies to the left (reactant favored) All [HA] eq > [H + ] eq because we assume little dissociation

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8 Diprotic acid- acid that has 2 acidic protons H 2 SO 4 + H 2 O → HSO 4 -1 + H 3 O + (strong for 1 st H + ) HSO 4 -1 + H 2 O ↔ SO 4 2- + H 3 O + (weak acid-HSO 4 + ) Monoprotic acids- acids with one proton Oxyacids- acidic proton is attached to an oxygen atom- most acids (HNO 3, H 3 PO 4, etc) Organic Acids- acids with a carbon atom backbone Contain carboxyl group (COOH)- this is the acidic proton usually these acids are weak- acetic acid Amphoteric- substances that behave either as acids or as bases (H 2 O and NH 3 are the most common)

9 Autoionization- transfer of a proton from one water molecule to another to produce a hydroxide ion and a hydronium ion H 2 O + H 2 O  H 3 O + + OH - NH 3 + NH 3 ↔ NH 4 + + NH 2 - Ion-product constant (dissociation constant or equilibrium constant for water) K w = [H + ][OH - ] = 1.0 x 10 -14 @ 25°C (WEAK acid b/c small K a )

10 In ANY aqueous solution at 25 degrees Celsius, no matter what it contains, the value of K w is always 1.0 x 10 -14 There are three possibilities: Neutral solution where [H + ] = [OH - ] Acidic Solution where [H + ] > [OH - ] Basic Solution where [H + ] < [OH - ] 1.0 x 10 -14 = [H + ] [OH - ] Reminder: You can compare the strengths of acids based on their K a values!! Examples: pgs. 673-674 (#32, 36, 38) HW: 14.1 (pg. 673 # 27, 29) and 14.2 (pg. 674 #33-37 odd)

11 Sig Figs for logarithms- the number of decimal places in the log is equal to the number of significant figures in the original number Ex. [H + ]= 1.0 x 10 -9 (2 sig figs) then the log[H + ]=2.00 (2 decimals) Equations: pH = -log[H + ] pOH = -log[OH-] the pH changes by 1 for every power of 10 change Also, the pH decreases as [H + ] increases pK = -log K K w = [H + ] [OH - ] [H + ] = 10 -pH [OH - ] = 10 -pOH pH + pOH = 14 (at 25°C) Example pg. 674 (#40) HW pg. 674 (#39-45 odd) Section 14.3: The pH Scale

12 2 Things to Focus on for Acid-Base Calculations -solution components and their chemistry. -which components are significant and which can be ignored Major species- solution components present in relatively large amounts -For strong acids, it is IONS and WATER (because the acid has dissociated) If the [Strong Acid] > 1.0 x 10 -7, then [H + ] from autoionization of water can be ignored because the [H + ] produced is negligible. - so 0.10 M HCl = [H + ] = 0.10 M, therefore the pH = -log (0.10) = 1.00 If the [Strong Acid] < 1.0 x 10 -7, then [H + ] from the strong acid can be ignored because the amount dissociated is negligible. -so [H + ] = 1.0 x 10 -7 (from water) and pH = 7.00 Example pg. 674 (#48, 50, 52) HW pg. 674-675 (#47-51 odd) Section 14.4: Calculating the pH of Strong Acid Solutions

13 What are the steps for calculating the pH of a weak acid solution? Find the major species in the solution. Determine which of the major species supplies the hydrogen ions. IF K a > K w, the H + comes from the weak acid List the initial concentrations. Initials for the IONS will usually be 0 Let x be the change in the concentration of the substance required to reach equilibrium. “-x” for WEAK ACID and “+x” for the IONS Assume “-x” is negligible for the weak acid Solve “x” will usually be the amount of H + dissociated Use the 5% rule to verify whether approximations are valid x/[HA] o * 100% < 5% (if not, you can’t assume “-x” is negligible for the weak acid Section 14.5: Calculating the pH of Weak Acid Solutions

14 Percent dissociation- also percent ionization amount dissociated initial concentration x 100% What happens to percent dissociation when a weak acid is diluted? the percent dissociation increases When water is added, Q is less than K a, so the system must adjust to the right to reach the new equilibrium position, so percent dissociation increases when the acid is diluted. Example: What is the pH of 1.00 M HF? What is the percent dissociation? HW: pg. 675 (#53, 55, 59-69 odd)

15 Strong bases- complete dissociation of OH compounds (100% soluble) - Group 1, calcium, strontium, and barium -KOH → K + + OH - -Ca(OH) 2 → Ca 2+ + OH - Many bases do not contain the hydroxide ion. How do these substances act as bases? they increase the concentration of hydroxide ion because of their reaction with water. NH 3 +H 2 O ↔ NH 4 + + OH - What do many of these bases without hydroxide ions have in common? A lone pair of electrons is located on a nitrogen atom in many cases Section 14.6: Bases

16 Equations: K b = [BH + ][OH - ] / [B] K b x K a = K w HW: pg 675-676 (#71-83 odd, 87-91 odd) K b – refers to the reaction of a base with water to form the conjugate acid and the hydroxide ion Weak bases- bases with small K b values- where the base competes with the OH - for the H + ion - B + HOH ↔ BH + + OH - -Calculations are similar to those for acids

17 Polyprotic acids- acids that furnish more than one proton- dissociates in a stepwise manner- one proton at a time Triprotic acids- acids with 3 protons- like phosphoric acid. Describe the relative sizes of dissociation constants for polyprotic acids. (K a 1 > K a 2> K a 3) (see pg. 650-651) This means that the first proton is easiest to remove and that it is more difficult to remove successive protons- this happens because the negative charge on the acid increases, making it more difficult to remove a positively charged proton only the first dissociation step makes an important contribution to [H+] this simplifies pH calculations Section 14.7 Polyprotic Acids

18 Why is sulfuric acid unique among strong acids? It is the only diprotic strong acid - its first dissociation step and a weak acid in its second step. General Rules: For concentrated solutions, 1M or higher, only use K a1 for calculations. For dilute concentrations- the second step does make a significant contribution and the quadratic equation needs to be used -Use all K values for each successive step HW: pg. 676 (# 93-97)

19 Salt- an ionic compound What type of salts produce neutral solutions when dissolved in water? -salts that consist of the cations of strong bases and the anions of strong acids -KCl, NaNO 3, NaCl, KNO 3 Section 14.8: Acid-Base Properties of Salts

20 A salt that consist of a cation of a strong base and the anion of a weak acid A solution of sodium acetate contains sodium and acetate and water. NaC 2 H 3 O 2 (aq) + H 2 O (l) → NaOH (aq) + HC 2 H 3 O 2 (aq) Weak Acid Na + + C 2 H 3 O 2 - + H 2 O (l) → Na + + OH - + HC 2 H 3 O 2 (aq) C 2 H 3 O 2 - + H 2 O (l) → OH - + HC 2 H 3 O 2 (aq) BASIC What kind of salt yields a basic solution?

21 A salt that consist of a cation of a weak base and the anion of a strong acid NH 4 Cl (aq) + H 2 O (l) → NH 4 OH (aq) + HCl (aq) NH 4 + + Cl - + H 2 O (l) → NH 3 (g) + HOH (l) + H + + Cl - NH 4 + → NH 3 (g) + H + Acidic Salts with highly charged metal ions also produce acidic solutions. (Al +3, W +6, etc) The high charge on the metal ion polarizes the O-H bonds in the attached water molecules, making the hydrogens in these water molecules more acidic than those in free water molecules. The higher the charge on the metal ion, the stronger the acidity of the hydrated ion. What kind of salt yields a acidic solution?

22 How do you determine whether a solution will be acidic, basic, or neutral when there is a salt made up of 2 ions that can affect the pH of aqueous solutions? Compare the K a value for the acidic ion with the K b value for the basic ion. If the K a > K b, the solution will be acidic. If the K b > K a value, the solution will be basic. Equal K a and K b values mean that the solution is neutral Equations: K a x K b =K w Example: pg. 676 (# 105) HW: pg. 676 (#99, 101, 109, 111 (ab, de))

23 What are the two main factors that determine whether a molecule containing a bond between an atom and hydrogen will behave as a Bronsted-Lowry acid? 1. strength of the bond (weak bonds, more acidic) 2. polarity of the bond (more polar, more acidic) C-H bonds are strong and nonpolar, so there is no tendency to donate protons. What is the trend in strength of oxyacids? -The strength of the acid grows as the # of oxygen atoms increases. -This occurs because very electronegative oxygen atoms are used to draw electrons away, which polarizes and weakens the O-H bond -A proton is easiest to remove from a molecule with the largest number of attached oxygen atoms. -This is the same reason that hydrated metal ions become acidic. -The greater the electronegativity of the attached ion, the stronger the acid because the EN strong atom pulls electrons away from others and weakens bonds. Section 14.9: The effect of Structure on Acid-Base Properties

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25 Substances that contain H-O-X bonds can behave as acids or bases. What determines which behavior will occur? -If X has a high electronegativity, the O-X bond will be covalent and strong, meaning that a proton will be released. -If X has a low electronegativity, the O-X bond will be ionic and can be broken in polar water and OH - is released Acidic oxides: when a covalent oxide dissolves in water, an acidic solution forms SO 3 + H 2 O → H 2 SO 4 Basic oxides: when an ionic oxide (metal oxide) dissolves in water and a basic solution results CaO + H 2 O → Ca(OH) 2 HW: pg.677 (113-118) Section 14.10: Acid-Base Properties of Oxides

26 Table 14.9 Comparison of Electronegativity of X and Ka Value for a Series of Oxyacids

27 Lewis acid- electron-pair acceptor OR it has an empty atomic orbital that it can use to accept or share an electron pair from a molecule that has a lone pair of electrons (usually BORON) Lewis base- electron-pair donor OR molecule that has a lone pair of electrons (usually oxygen, nitrogen, or fluorine) Lewis model for acids and bases is beneficial because it covers many reactions that don’t involve Bronsted-Lowry aids HW: pg. 677 (#119, 123) Section 14.11: The Lewis Acid-Base Model

28 Acid-Base Equilibria Sections 15.1-15.5 (Unit 18A)

29 15.1 Common Ion Effect Suppose we have a solution containing the weak acid hydrofluoric acid (K a =7.2x10 -4 ) and its salt NaF. Remember that when a strong electrolyte dissolves, it completely dissociates. –What is the common ion? –How does this ion effect pH?

30 HF (aq)↔ H + (aq) +F - (aq) Added F - ions from NaF Equilibrium shift. Form more HF. Fewer H + ions present. Less acidic.

31 NH 3 (aq) + H 2 O↔ NH 4 + (aq) +OH - (aq) Added NH 4 + ions from NH 4 Cl Equilibrium shift. Form more NH 3 and H 2 O. Fewer OH - ions present. Less basic.

32 Equilibrium Calculations Use the same procedures for finding pH of a solution with a weak acid or base as last unit (use ICE). The only difference is the initial concentration of the common ion will NOT be zero. Example 15.23 on pg. 740

33 15.2 Buffered Solutions Buffered solutions resist a change in pH regardless of whether OH - or H + is added. –Most common example is our blood, which maintains a constant pH so that cells can survive. –Contain a weak acid and its common ion salt, such as HF and NaF. –OR –Contain a weak base and its common ion salt, such as NH 3 and NH 4 Cl.

34 How does buffering work? Buffer solution of a weak acid (HA) and its conjugate base (A - ).

35 K a = [H + ][A - ] or [H + ]= K a [HA] [HA] [A - ] –If you add protons, H + + A - → HA –A - converted to HA ; [HA]/[A - ] increases; [H + ] increases; pH increases. –****If [A - ] and [HA] are large compared to [H + ] added, then there will be little change in the pH.

36 Addition of OH - to Buffered Solutions –If amounts of HA and A - originally present are large compared to OH -, change in [HA]/[A - ] is small. Therefore, [H + ] and pH remain constant.

37 K a = [H + ][A - ] or [H + ]= K a [HA] [HA] [A - ] –If you add hydroxide ions, OH - + HA → A - + H 2 O –HA is converted to A - ; [HA]/[A - ] decreases; [H + ] decreases; pH decreases (but only slightly if [HA] and [A - ] are large).

38 1.Buffered solutions are simply weak acids or bases with a common ion. 2.pH calculations for buffered solutions are exactly the same procedures as last unit. 3.When a strong acid or base is added to a buffered solution, deal with stoichiometry first, then equilibrium calculations.

39 Buffer Calculations Calculating [H + ]: [H + ]= K a [HA] [A - ] Henderson-Hasselbalch Equation -log[H + ]=-log(K a ) –log ([HA]/[A - ]) pH=pK a -log([HA]/[A - ]) pH=pK a +log([A - ]/[HA]) pH=pK a +log(base/acid)

40 See the Summary of the Most Important Characteristics of Buffered Solutions on page 692. Examples 15.37 HW: 41 and 43 on pages 740-741

41 15.3 Buffering Capacity Buffering capacity is the amount of protons or hydroxide ions the buffer can absorb without a significant change in pH. –pH depends on the [A - ]/[HA] ratio –Buffering capacity depends on the magnitudes of [HA] and [A - ] Larger concentrations allow for more absorbtion of H + and OH - with little change in pH

42 Optimal Buffering –When [HA]=[A - ], so that [A - ]/[HA]=1. –pK a =pH (or as close as possible)

43 Examples 15.46 on pg. 741 HW: 45 and 49* on pg. 741

44 15.4 Titrations and pH Curves Titration: used to determine the amount of acid or base in a solution, involving a solution of known concentration (titrant), indicator, and unknown solution. pH curve or titration curve: a plot of the pH of an acid- base titration as a function of the amount of titrant added.

45 Strong Acid-Strong Base Titrations Net Ionic Reaction: H + (aq) + OH - (aq)→ H 2 O (l) Computing [H + ]: –Determine mmol of H + at that point and divide by total volume of solution –1 mmol= (1 mol/1000) = 10 -3 mol *used because titrations involve small quantities –Molarity = mol solute = (mol solute/1000) = mmol solute L solution (L soltn/1000) mL soltn - Mmol =mL x molarity

46 Case Study 50.0 mL of 0.200 M HNO 3 is titrated with 0.100 M NaOH. We will calculate the pH of the solution at selected points during the course of the titration. –A. No NaOH has been added. [H + ]=0.200 M and pH=0.699

47 –B. 10.0 mL of 0.100 M NaOH has been added. [H + ]=9.00 mmol/(50.0+10.0)= 0.15 M and pH=0.82 –C. 20.0 mL (total) of 0.100 M NaOH has been added. [H + ]=8.00 mmol/(50.0+20.0)= 0.11 M and pH=0.942 H + +OH - → H2OH2O Before50.0 mL x 0.200 M =10.0 mmol 10.0 mL x 0.100 M =1.00 mmol Reaction-1.00 mmol 1.00 mmol After9.00 mmol left0 mmol1.00 mmol H + +OH - → H2OH2O Before50.0 mL x 0.200 M =10.0 mmol 20.0 mL x 0.100 M =2.00 mmol Reaction-2.00 mmol 2.00 mmol After8.00 mmol left0 mmol2.00 mmol

48 –D. 50.0 mL (total) of 0.100 M NaOH has been added. Proceed exactly as for points B and C. pH=1.301

49 –E. 100.0 mL (total) of 0.100 M NaOH has been added. Stoichiometric point or equilvalence point (pH=7.00) –F. 150.0 mL (total) of 0.100 M NaOH has been added. [OH - ]=5.00 mmol/(50.0+150.0)= 0.025 M, pOH=1.60, pH=12.40 H + +OH - → H2OH2O Before50.0 mL x 0.200 M =10.0 mmol 100.0 mL x 0.100 M =10.0 mmol Reaction-10.0 mmol 10.0 mmol After0 mmol 10.0 mmol H + +OH - → H2OH2O Before50.0 mL x 0.200 M =10.0 mmol 150.0 mL x 0.100 M =15.0 mmol Reaction-10.0 mmol 10.0 mmol After0 mmol5.0 mmol left10.0 mmol

50 –G. 200.0 mL (total) of 0.100 M NaOH has been added Proceed the same as point F. pH = 12.60

51 Characteristics of Strong-Strong pH Titration Curves pH changes are gradual until the titration is close to the equivalence point, due to the large amount of original H + or OH - ions depending on the titration. At the equivalence point, pH=7.00. Pg. 742 (#53)

52 Titrations of Weak Acids with Strong Bases Calculating the pH curve: 1.Stoichiometry problem (BRA): The rxn of OH - with the weak acid runs to completion, and [remaining acid] and [formed conjugate base] are determined. 2.Equilibrium problem (ICE): The position of the weak acid equilibrium is determined; pH is calculated.

53 Case Study 50.0 mL of 0.10 M acetic acid (K a =1.8x10 -5 ) is titrated with 0.10 M NaOH. We will calculate pH at various points. A. No NaOH has been added. –Complete this the same as last unit. –pH=2.87 (check it yourself) B. 10.0 mL of 0.10 M NaOH has been added. OH - +CH 3 COOH → CH 3 COO - + H 2 O Before10.0 mL x 0.10 M =1.0 mmol 50.0 mL x 0.10 M =5.0 mmol 0 mmol Reaction-1.0 mmol +1.0 mmol After0 mmol4.0 mmol left1.0 mmol

54 Since only CH 3 COOH and CH 3 COO - are the only species left in solution: –K a =1.8x10 -5 = [H + ][CH 3 COO - ] = (0.017)x [CH 3 COOH] 0.067 –X=7.2 x 10 -5 M = [H + ] –pH=4.14 CH 3 COOH → CH 3 COO - + H + Initial 4.0 mmol (50.0+10.0) mL 1.0 mmol (50.0 +10.0) mL 0 Change-x+x Equilibrium0.067-x ≈ 0.0670.017+x ≈ 0.017x

55 –C. 25.0 mL (total) of 0.10 M NaOH has been added This is the halfway point. [CH 3 COOH]=[CH 3 COO - ] [H + ]=K a pH=pK a –D. 50.0 mL (total) of 0.10 M NaOH has been added. This is the equivalence point because 5.0 mmol of OH - will react with the 5.0 mmol of CH 3 COOH originally present. Species left is CH 3 COO -. –CH 3 COO - + H 2 O (l) ↔ CH 3 COOH (aq) + OH - –Find K b. –Determine pH.

56 –E. 60.0 mL (total) of 0.10 M NaOH has been added At this point, excess OH - has been added. pH is determined by the excess OH - because it is a strong base [OH - ]=1.0 mmol / (50.0+60.0)mL= 9.1x10 -3 M pOH=2.04 pH=11.96 –HW: pg 742 (#55) OH - +CH 3 COOH → CH 3 COO - + H 2 O Before60.0 mL x 0.10 M =6.0 mmol 50.0 mL x 0.10 M =5.0 mmol 0 mmol Reaction-5.0 mmol +5.0 mmol After1.0 mmol in excess4.0 mmol left5.0 mmol

57 Comparisons: Near the beginning, pH increases more rapidly than strong/strong. Weak curve levels off near the halfway point due to buffering. Weak curve, the pH at the equivalence point > 7 due to basicity of the conjugate base. Shapes of both curves are the same after the equivalence point. *Remember that equivalence point is determined by stoichiometry (enough titrant has been added to react exactly with acid or base), not the pH.

58 The amount of acid present determines the equivalence point, not its strength. pH value at the equivalence point is affected by the acid strength. –Weak acid=high pH Notice the difference in the shapes of the curves.

59 Titrations of Weak Bases with Strong Acids 1 st : Think about the major species in solution. 2 nd : Decide whether a reaction occurs that runs to completion and do stoichiometric calculations. 3 rd : Choose the dominant equilibrium and calculate pH. -See Case Study on page 709

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61 Note that the pH at the equivalence point < 7 since the solution contains the weak acid NH 4 +.

62 15.5 Acid-Base Indicators 2 methods for determining equivalence points: –Use a pH meter to plot the titration curve. The center of the vertical region indicates the equivalence point. –Use an acid-base indicator, which marks the end point by changing colors. Although equivalence point ≠ end point, careful selection of the indicator will ensure error is negligible.

63 Common Acid-Base Indicators are Complex Molecules HIn: represents the acid form of a complex molecule that is a particular color or colorless. In - : represents the conjugate base of the complex molecule that shows as a different color.

64 Assume that the ratio of [In - ]/[HIn] must be 1/10 for the human eye to detect a color change. a.) Yellow acid form of bromthymol blue; b.) a greenish tint is seen when the solution contains 1 part blue and 10 parts yellow; c.) blue basic form

65 Henderson-Hasselbalch equation useful to determine pH at the indicator color change. –Acid Solution being titrated. pH=pK a + log ([In - ]/[HIn]) pH=pK a + log (1/10)= pK a -1 FOR ACID –pH=pKa - 1 –Basic solution being titrated (indicator initially in the form of In - ) pH=pK a + log ([In - ]/[HIn]) pH=pK a + log(10/1)=pK a +1 FOR BASE –pH=pKa + 1

66 The useful pH range for an indicator is pK a (indicator)± 1.

67 Choosing an Indicator The indicator end point (when it changes colors) should be as close as possible to the titration equivalence point. –What indicator would you use for the titration of 100.00 mL of 0.100 M HCl with 0.100 M NaOH? Equivalence point occurs at pH of 7.00 (strong/strong) Titrating an acid; pH=pK a -1 –pK a =pH+1=8 –K a =1x10 -8

68 Indicators for strong acid-strong base titrations –Color changes will be sharp, occurring with the addition of a single drop of titrant. –There is a wide choice of suitable indicators.

69 Titration of Weak Acids –Smaller vertical area around the equivalence point. –Less flexibility in choices of indicators. –Choose an indicator whose useful pH range has a midpoint as close as possible to the equivalence point.

70 Examples 15.65, 15.69 on pages 742-743 THE END!!


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