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ACIDS & BASES Acids: - acids are sour tasting - Arrhenius acid: Any substance that when

ACIDS & BASES Acids: - acids are sour tasting - Arrhenius acid: Any substance that when dissolved in water, increases the concentration of hydronium ion (H 3 O + ) - Bronsted-Lowry acid: A proton donor - Lewis Acid: An Electron acceptor Bases:

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ACIDS & BASES Acids: - acids are sour tasting - Arrhenius acid: Any substance that when

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  1. ACIDS & BASES Acids: - acids are sour tasting - Arrhenius acid: Any substance that when dissolved in water, increases the concentration of hydronium ion (H3O+) - Bronsted-Lowry acid: A proton donor - Lewis Acid: An Electron acceptor Bases: - bases are bitter tasting and slippery - Arrhenius base: Any substance that when dissolved in water, increases the concentration of hydroxide ion (OH-) - Bronsted-Lowery base: A proton acceptor - Lewis base: An electron donor

  2. HBr + KOH  H3O+ + NH3  HBr + NH3  NH3 + H2O  

  3. STRONG VS WEAK - completely ionized - partially ionized - strong electrolyte - weak electrolyte - ionic bonds - some covalent bonds STRONG ACIDS:STRONG BASES: HClO4LiOH H2SO4NaOH Hl KOH HBr Ca(OH)2 HCl Sr(OH)2 HNO3Ba(OH)2

  4. table 18.4 THE CONJUGATE PAIRS IN SOME ACID-BASE REACTIONS Conjugate Pair Acid + Base  Base + Acid Conjugate Pair Reaction 1 HF + H2O  F- +H3O+ Reaction 2 HCOOH + CN-  HCOO- + HCN Reaction 3 NH4+ + CO32-  NH3 + HCO3- Reaction 4 H2PO4- + OH-  HPO42- + H2O Reaction 5 H2SO4 + N2H5+  HSO4- + N2H62+ Reaction 6 HPO42- + SO32-  PO43- + NSO3-

  5. Fig. 16.7 CONJUGATE ACID-BASE PAIRS ACIDBASE HCl Cl- H2SO4 HSO4- HNO3 NO3- H+(aq) H2O HSO4- SO42- H3PO4 H2PO4 HF F- HC2H3O2 C2H3O2 H2CO3 HCO3- H2S HS- H2PO4- HPO42- NH4+ NH3 HCO3- CO32- HPO42- PO43- H2O OH- HS- S2- OH- O2- H2 H- 100 percent ionized in H2O strong negligible Base strength increases         Acid strength increases weak weak 100 percent protonated in H2O negligible strong

  6. The strength of an acid depends on how easily the proton, H+, is lost or removed from an H - x bond. Greater Acid Strength: - more polar bonds - larger “x” atom - oxo acids: higher electronegativity - oxo acids: more oxygen atoms - oxo acids: more hydrogen atoms List the following in order of increasing strength: l. HI, HF, HCl 2. H2O, CH4, HF 3. HIO3, HClO3, HBrO3 4. HBrO, HBrO3, HBrO2 5. HI, H2SO4, HClO4, HNO3

  7. A. H - O - Cl > H - O - Br > H - O - l + - + - + - O || B. H - O - Cl << H - O - Cl = O +- +- || O

  8. A C I D S T R E N G T H AcidBase HCl Cl- H2SO4 HSO4- HNO3 NO3- H3O+ H2O HSO4- SO42- H2SO3 HSO3- H3PO4 H2PO4- HF F- CH3COOH CH3COO- H2CO3 HCO3- H2S HS- HSO3- SO32- H2PO4- HPO42- NH4+ NH3 HCO3- CO32- HPO42- PO43- H2O OH- HS- S2- OH- O2- B A S E S T R E N G T H STRONG WEAK NEGLIGIBLE NEGLIGIBLE WEAK STRONG

  9. ACID STRENGTH STRONG ACID More polar  weaker bond larger atoms weaker bond (HF weak HI strong) strength increases down a group OXO Acids HIO < HClO stronger (more electrones) (more O) (more H) 1. HF < HCl < HI strong 2. CH4 < H2O < HF 3. HIO3 < HBrO3 < HClO3 4. HBrO < HBr2O < HBrO3 5. HI < HNO3 < H2SO4 < HClO4

  10. AUTO - IONIZATION A reaction in which two like molecules react to give Ions. 2 H2O  H3O+ + OH- K= [H3O+] [OH-] but [H2O] is essentially [H2O]2 constant  K[H2O]2 = [H3O+] [OH-] Kw = [H3O+] [OH-] Kw = Ion-product constant for water. Kw = 1 x 10-14 at 25°C

  11. Fig. 16.2pH of Some Common Solutions pH [H+] [OH-] pOH --14 1 x 10-14 1 x 10-0 0 --13 1 x 10-13 1 x 10-1 1 --12 1 x 10-12 1 x 10-2 2 --11 1 x 10-11 1 x 10-3 3 --10 1 x 10-10 1 x 10-4 4 -- 9 1 x 10-9 1 x 10-5 5 -- 8 1 x 10-8 1 x 10-6 6 -- 7 1 x 10-7 1 x 10-7 7 -- 6 1 x 10-6 1 x 10-8 8 -- 5 1 x 10-5 1 x 10-9 9 -- 4 1 x 10-4 1 x 10-10 10 -- 3 1 x 10-3 1 x 10-11 11 -- 2 1 x 10-2 1 x 10-12 12 -- 1 1 x 10-1 1 x 10-13 13 -- 0 1 x 100 1 x 10-14 14 M O R E B A S I C NaOH, 0.1 M…………….. Household bleach……….. Household ammonia……. Lime Water……………… Milk of Magnesia……….. Borax……………………. Baking Soda……………. Egg White, Sea Water….. Human blood, Tears…….. M O R E A C I D I C Milk………………………. Saliva……………………… Rain……………………….. Black Coffee………………. Banana……………………. Tomatoes…………………. Wine………………………. Cola, Vinegar…………….. Lemon Juice……………… Gastric Juice……………..

  12. pH I. Kw = [H+] [OH-] take log Log Kw = Log [H+] [OH-] = Log [H+] + log [OH-] p Kw = pH + pOH 14 = pH + pOH II. pOH = -Log [OH-] 1. A 0.0015M NaOH solution has what pH? pOH? [OH-] 2. A soda has pH of 2.94, what is the [H+]? 3. A solution has pOH of 12.7, what is the [H+]?

  13. WEAK ACIDS HA(aq)  H+ (aq) + A- (aq) Ka = [H+] [A-]Ka = acid dissociation constant [HA-] The magnitude of Ka refers to the strength of the acid. Small Ka value = weak acid Q. A student prepared a 0.10M solution of formic acid HCHO2 and measured it’s pH, at 25°C, pH = 2.38 a) calculate Ka b) what percent of acid Ionizes?

  14. GENERAL STEPS FOR CALCULATING THE pH (pOH) OF A WEAK ACID (BASE) Step 1: Write a balanced chemical equation describing the “action”. Step 2: Make a list of given and implied information. Step 3: Write the equilibrium constant equation associated with the balanced chemical equation in Step 1. Step 4: An equilibrium table should be set up since we are dealing with a weak acid (partially dissociated species). The table should describe the changes which occurred in order to establish equilibrium. Step 5: Substitute the equilibrium values from Step 4 into the equilibrium constant equation in Step 3. Solve for x. If the expression can not be solved with basic algebra, try either the quadratic equation or the successive-approximation method. Step 6: Calculate the pH (pOH) using the expression: pH = -Log [H+] or pOH = -Log [OH-]

  15. CALCULATING pH FOR A WEAK ACID Step 1: Calculate the pH of a 0.20 M HCN solution. Step 2: Calculate the percent of HF molecules ionized in a 0.10 M HF solution. Step 3: Compare the above value to the percent obtained for a 0.010 M HF solution.

  16. POLYPROTIC ACIDS H2SO3 H+ + HSO3- Ka1 = 1.7 x 10-2 HSO3-  H+ + SO32- Ka2 = 6.4 x 10-8 Ka1 > Ka2 The solubility of CO2 in pure H2O at 25ºC and 0.1 atm is 0.0037 M. A) What is the pH of a 0.0037 M solution of H2CO3? B) What is the [CO32-] produced?

  17. WEAK BASES B- + H2O  HB+ + OH- K = [HB+] [OH-] [B-] [H2O] Kb = K[H2O] = [HB+] [OH-] [B-] base dissociation constant Q. Calculate [OH-] and pH of a 0.15M NH3 solution.

  18. 1. Calculate the pH of a 0.0850 M HC2H3O2 solution. 2. What is the molarity of an aqueous HCN solution if the pH is 5.7? 3. Calculate the pOH of a 0.351 M aqueous solution of NH3. 4. Calculate the pH of a 0.025M solution of citric acid. Ka (acetic acid) = 1.8 x 10-5 Kb (ammonia) = 1.8 x 10-5 Ka (hydrocyanic) = 4.9 x 10-10 Ka2 (citric acid) = 1.7 x 10-5 Ka1 (citric acid) = 7.4 x 10-4 Ka3 (citric acid) = 4.0 x 10-7

  19. RELATIONSHIP BETWEEN Ka AND Kb A. NH4+ NH3 + H+ B. NH3 + H2O  NH4+ + OH- Ka = [NH3] [H+] [NH4+] Kb = [NH4+] [OH-] [NH3] Add equation A to equation B to get the net reaction: H2O  H+ + OH- C. Equation A + Equation B = Equation C K1 x K2 = K3 KaKb = [NH3] [H+][NH4+] [OH-] = [OH-] [H+] = Kw [NH4+] [NH3]

  20. CALCULATE Kb for F- if Ka = 6.8 x 10-4 Kb = Kw = 1 x 10-14 Ka 6.8 x 10-4 = 1.5 x 10-11

  21. ACID/BASE PROPERTIES OF SALT SOLUTIONS HYDROLYSIS Ions react with water to generate either H+ or OH- A- + H2O  HA + OH- Q. Predict whether Na2 HPO4 will form an acidic or basic solution. Q. Predict whether K2HC7H5O7 will form an acidic or basic solution.

  22. 1. SA/SB neither cation nor anion hydrolyzes 2. SB/WA anion = strong CB  hydrolyzes to produce OH- Basic 3. WB/SA cation = strong CA  hydrolyzes to produce H+ Acidic 4. NH4CN CN- = Kb = 2 x 10-5 NH4+ = Ka = 5.6 x 10-10 CN- hydrolyzes more than NH4+ pH > 7 basic more basic than acidic FeCO3 Fe3+ = 6 x 10-3 Ka CO32- = Ka1 = 4.5 x 10-7 Ka2 = 4.7 x 10-11 Kb1 = 2.2 x 10-8 Kb2 = 2.13 x 10-4 table 18.7 pg. 786 metal ions

  23. SALT SOLUTIONS 1. Salts derived from strong bases and strong acids have pH = 7 NaCl Ca(NO2)2 2. Salts derived from strong bases and weak acids have pH > 7 NaClO Ba(C2H3O2)2 3. Salts derived from weak bases and strong acids have pH < 7 NH4Cl Al(NO3)3 4. Salts derived from weak base and weak acids, pH is dependent on extent NH4CN Fe3(CO3)2 NH4C2H3O2

  24. Table 18.8 THE BEHAVIOR OF SALTS IN WATER Salt Solution pHNature of IonsIon that reacts (Examples)with water Neutral 7.0 Cation of strong base None [NaCl, KBr, Anion of strong acid Ba(NO3)2] Acidic <7.0 Cation of weak base Cation [NH4Cl, NH4NO3, Anion of strong acid CH3NH3Br] Acidic <7.0 Small, highly charged Cation [Al(NO3)3, cation CrCl3, FeBr3] Anion of strong acid Basic >7.0 Cation of strong base Anion [CH3COONa, Anion of weak acid KF, Na2CO3]

  25. CALCULATING Ph OF SALT SOLUTIONS Q. Household bleach is 5% solution of sodium hypochlorite NaClO. Calculate the [OH-] and pH of a 0.70 M NaClO solution. Kb = 2.86 x 10-7

  26. COMMON ION EFFECT A shift of an equilibrium induced by an Ion common to the equilibrium. HC7H5O2 + H2O  C7H5O2- + H3O+ Benzoic Acid Calculate the degree of ionization of benzoic acid in a 0.15 M solution where sufficient HCl is added to make 0.010 M HCl in solution. Compare the degree of ionization to that of a 0.15 M benzoic Acid solution Ka = 6.3 x 10-5

  27. COMMON ION EFFECT HC2H3O2 H+ + C2H3O2- NaC2H3O2strong electrolyte HC2H3O2weak electrolyte Addition of NaC2H3O2 causes equilibrium to shift to the left , decreasing [H+] eq Dissociation of weak acid decreases by adding strong electrolyte w/common Ion. “Predicted from the Le Chatelier’s Principle.” Q. What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution?

  28. 1. Identify the major species in the solution and consider their acidity or basicity HC2H3O2 Na+ C2H3O2- H2O WA neut CB amphoteric spectator 2. Identify important equilibrium Rx H2O <<< HC2H3O2 acidity  pH controlled by HC2H3O2  HC2H3O2 + H2O  H3O+ + C2H3O2- or HC2H3O2  H+ + C2H3O2- but remember [H+]  [C2H3O2-] coz additional NaCl2

  29. Calculate [F-] and pH of a solution containing 0.10 mol of HCl and 0.20 mol of HF in a 1.0 L solution. 1. HCl = SA HF = WA H+ + Cl- HF H2O & eq H+ F- 2. HF  H+ + F-

  30. BUFFERS A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acids or bases are added to it. Buffers contain both an acidic species to neutralize OH- and a basic species to neutralize H3O+. An important characteristic of a buffer is it’s capacity to resist change in pH. This is a special case of the common Ion effect.

  31. Buffer after addition Buffer with equal Buffer after of H3O+ concentrations of addition of OH- conjugate acid & base CH3COOH CH3COO- CH3COO- CH3COOH CH3COO- CH3COOH H3O+   OH-  H2O + CH3COOH  H3O+ + CH3COO- CH3COOH + OH-  CH3COO- + H2O

  32. BUFFER 1. What is the pH of a buffer that is 0.12 M in lactic acid (HC3H5O3) and 0.10 M sodium lactate? Lactic acid Ka = 1.4 x 10-4 2. How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00?

  33. Buffer after Buffer with equal Buffer after addition of concentrations of addition of OH- weak acid and its H+ conjugate base X- HX HX X- HX X- H+ OH- OH- + HX  H2O + X- H+ + X- HX

  34. PROCEDURE FOR CALCULATION OF pH (buffer) Neutralization Add strong acid X- + H3O  HX + H2O Use Ka, [HX] and [X-] to calculate [H+] Buffer containing HA and X- Recalculate [HX] and[X-] pH Neutralization HX + OH-  X- + H2O Add strong base Stoichiometric calculation Equilibrium calculation

  35. ADDITION OF A STRONG ACID OR STRONG BASE TO A BUFFER A buffer is made by adding 0.3 mol of acetic acid and 0.3 mol of sodium acetate to 1.0 L of solution. If the pH of the buffer is 4.74 A. Calculate the pH of a solution after 0.02 mol of NaOH is added B. after 0.02 mol HCl is added.

  36. Henderson - Hasselbach Equation - log [H+] = - log Ka [HX] [X-] = - log Ka - log [HX] [X-] Since - log [H+] = pH and - log Ka = pKa pH = pKa - log [HX] [X-] = pKa + log [X-] [XH] pH = pKa + log [base] [acid] when [base] = [acid] pH = pKa

  37. PROCEDURE FOR CALCULATION OF pH (TITRATION) Neutralization Solution containing weak acid and strong base Calculate [HX] and [X-] after reaction HX + OH- X- + H2O Use Ka, [HX], and [X-] to calculate [H+] pH Stoichiometric calculation Equilibrium calculation

  38. ACID - BASE TITRATION For a strong acid reacting with a strong base, the point of neutralization is when a salt and water is formed  pH = ?. This is also called the equivalence point. Three types of titration curves - SA + SB - WA + SB - SA + WB 1. Calculate the pH if the following quantities of 0.100 M NaOH is added to 50.0 mL of 0.10 M HCl. A. 49.0 mL B. 50.0 mL C. 51.0 mL

  39. STRONG BASE WITH WEAK ACID WA + OH- A- + H2O for each mole of OH- consumer 1 mol WA to produce 1 mol of A- when WA is in excess, need to consider proton transfer between WA and H2O to create A- and H3O+ WA + H2O  A- + H3O+ 1. Stoichiometric calculation allow SB to react with WA, solution product = WA & CB 2. Equilibrium calculation use Ka and equil. to calculate [WA] and CB and H+

  40. Question: If 30.0 ml of 0.200 M acetic acid, HC2H3O2, is titrated with 15.0 ml of 0.100 M sodium hydroxide, NaOH, what is the pH of the resulting solution? Ka for acetic acid is 1.8 x 10-5. Step 1: Write a balanced chemical equation describing the action: HC2H3O2 + OH-  C2H3O2- + H2O Step 2: List all important information under the chemical equation: HC2H3O2 + OH-  C2H3O2- + H20 0.200 M 0.100M 30.0 ml 15.0 ml Step 3: How many moles are initially present? What are we starting with before the titration? n(HC2H3O2)i = MV = (0.030 L)(0.200 M) = 0.006 moles n(OH-)i = MV = (0.015 L)(0.100 M) = 0.0015 moles

  41. Step 4: Since we are dealing with a weak acid, ie., partially dissociated, an equilibrium can be established. So we need to set up a table describing the changes which exist during equilibrium. HC2H3O2 + OH- C2H3O2- + H2O i 0.006 0.0015 0 ---  -0.0015 -0.0015 0.0015 --- eq 0.0045 0 0.0015 --- [HC2H3O2] = n/V = 0.0045/0.045 L = 0.100 M [C2H3O2-] = n/V = 0.0015/0.045 L = 0.033 M Step 5: To calculate the pH, we must first calculate the [H+] HC2H3O2 C2H3O2- + H+ Ka = [C2H3O2-] [H+]/[HC2H3O2] = 1.8 x 10-5 solve for [H+] = Ka[HC2H3O2]/[C2H3O2-] = (1.8 x 10-5)(0.100)/0.033 = 5.45 x 10-5 M Step 6: Calculate the pH from pH = -Log [H+] pH = 4.26

  42. Question: If 30.0 ml of 0.200 M acetic acid, HC2H3O2, is titrated with 0.100 M sodium hydroxide, NaOh, what is the pH at the equivalence point? Ka for acetic acid is 1.8 x 10-5. Step 1: Calculate the number of moles of base used to reach the equivalence point. n(HC2H3O2)i = (0.030 L)(0.200 M) = 0.006 moles there is 1:1 mole ratio between the acid and the base therefore 0.006 moles of base are needed to reach the equivalence point. This corresponds to 60 ml of 0.10M NaOH. The molarity of the base solution titrated is equal to the moles of C2H3O2- produced/total volume of solution: 0.006 moles/0.090 L = 0.067 M

  43. Step 2: At the equivalence point, the solution contains NaC2H3O2, so we may treat this problem similar to the calculation for the pH of a salt solution. NaC2H3O2 + H2O  HC2H3O2 + OH- i 0.067 --- 0 0  -x --- x x eq 0.067-x --- x x Kb = [HOAc] [OH-] / [OAc-] = 5.556 x 10-10 = x* x/0.067 therefore by solving for x = [OH-] = 6.1 x 10-6 M pOH = -Log [OH-] = 5.21 pKw - pOH = pH 14 - 5.21 = 8.79

  44. If 30.0 ml of 0.200 M acetic acid, HC2H3O2, is titrated with 15.0 ml of 0.100 M sodium hydroxide, NaOH, A) What is the pH of the resulting solution? Ka for acetic acid is 1.8 x 10-5. B) What is the pH at the equivalence point?

  45. Calculate pH is titration of HOAc by NaOH after 30 ml of 0.10 M NaOH has been added to 50 ml of 0.100 M HOAc. Step 1: Stachiometry MAVA = 0.10 mol HOAc (.05 L) = nA. = 5 x 10-3 HOAc mol 1 L MBVB = 0.10 mol (.030 L) = nB. = 3 x 10-3 mol NaOH 1 L OH- + HOAc  OAC- + H2O i 3 x 10-3mol 5 x 10-3 0 ---  -3 x 10-3mol -3 x 10-3 3 x 10-3 ---- eq 0 2 x 10-3 3 x 10-3 ---- total volume = 80 mL [HOAc]eq = 2 x 10-3/.08 L = 0.0250 M [OAc-]eq = 3 x 10-3/.08 L = 0.0375 M

  46. Step 2: equilibrium Ka = [H+] [OAc-] = 1.8 x 10-5 [HOAc] [H+] = Ka [HOAc] = 1.8 x 10-5.025 = 1.2 x 10-5 [OAc] .0375 if 25 mL is usedVT = 75 mL NA = 1.25 x 10-3 mol HBz NB = (0.05 mol/L)(.025L) = 1.25 x 10-3 mol OH- [Bz-] = 1.25 x 10-3/.075 = 0.01667 Bz + H2O  HBz + OH- i .01667 --- 0 0  -x --- +x +x eq .01667-x x x Kb = [HBz][OH-] = x2 = 1.54 x 10-10 [Bz-] .01667-x [OH-] = 1.60 x 10-6pOH = 5.80pH = 8.20

  47. We’ve seen what happens when a strong acid is titrated with a strong base but what happens when a weak acid is titrated? What is the fundamental difference between a strong acid and a weak acid? To compare with what we learned about the titration of a strong acid with a strong base, let’s calculate two points along the titration curve of a weak acid, HOAc, with a strong base, NaOH. Q: If 30.0 mL of 0.200 M acetic acid, HC2H3O2, is titrated with 15.0 ml of 0.100 M sodium hydroxide, NaOH, what is the pH of the resulting solution? Ka for acetic acid is 1.8 x 10-5. Step 1: Write a balanced chemical equation describing the action: HC2H3O2 + OH-  C2H3O2 + H2O why did I exclude Na+? Step 2: List all important information under the chemical equation: HC2H3O2 + OH-  C2H3O2 + H2O 0.20 M 0.10M 30mL 15mL

  48. Step 3: How many moles are initially present? What are we starting with before the titration? n(HOAc)i = (0.03 L)(0.200M) = 0.006 moles n(OH-)i = (0.015L)(0.100M) = 0.0015 moles Q: What does this calculation represent? A: During titration OH- reacts with HOAc to form 0.0015 moles of Oac- leaving 0.0045 moles of HOAc left in solution. Step 4: Since we are dealing with a weak acid, ie., partially dissociated, an equilibrium can be established. So we need to set up a table describing the changes which exist during equilibrium. HC2H3O2 + OH-  C2H3O2- + H2O i 0.006 0.0015 0 ---  -.0015 -.0015 0.0015 eq 0.0045 0 0.0015 [HOAc] = NN = 0.0045/0.045 L = 0.100 M [Oac-] = NN = 0.0015/0.045 L = 0.033 M

  49. Step 5: To calculate the pH, we must first calculate the [H+] Q: What is the relationship between [H+] and pH? A: acid-dissociation expression, products over reactants. Q: Which reaction are we establishing an equilibrium acid-dissociation expression for? HC2H3O2 C2H3O2- + H+ Ka = [Oac-] [H+]/[HOAc] = 1.8 x 10-5 solve for [H+] = Ka[HOAc]/[OAc-] = (1.8 x 10-5)(0.100)/0.033 = 5.45 x 10-5 M Step 6: Calculate the pH from pH = -Log [H+] pH = 4.26

  50. So at this point, we have a pH of 4.26, Is this the equivalence point? Is the equivalence point at pH = 7 as with a strong acid titration? Q: By definition, how is the equivalence point calculated? A: moles of base = moles of acid Let’s calculate the pH at the equivalence point. Step 1: Calculate the number of moles of base used to reach the equivalence point. n(HOAc)i = (0.03 L)(0.200 M) = 0.006 moles there is a 1:1 mole ration between the acid and the base therefore 0.006 moles of base are needed. This corresponds to 60 ml of 0.10 M NaOH. The molarity of the base solution titrated is moles of OAc- produced/total volume: 0.006 moles/0.090 L = 0.067 M

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