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Acids and Bases

Acids and Bases. Acids and Bases. The concepts acids and bases were loosely defined as substances that change some properties of water. One of the criteria that was often used was taste. Substances were classified salty-tasting sour-tasting sweet-tasting bitter-tasting

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Acids and Bases

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  1. Acids and Bases

  2. Acids and Bases • The concepts acids and bases were loosely defined as substances that change some properties of water. • One of the criteria that was often used was taste. • Substances were classified • salty-tasting • sour-tasting • sweet-tasting • bitter-tasting • Sour-tasting substances would give rise to the word 'acid', which is derived from the Greek word oxein, which mutated into the Latin verb acere, which means 'to make sour' • Vinegar is a solution of acetic acid. Citrus fruits contain citric acid.

  3. Acids • React with certain metals to produce hydrogen gas. • React with carbonates and bicarbonates to produce carbon dioxide gas Bases • Have a bitter taste • Feel slippery. • Many soaps contain bases.

  4. Properties of Acids • Produce H+ (as H3O+) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) • Taste sour • Corrode metals • Good Electrolytes • React with bases to form a salt and water • pH is less than 7 • Turns blue litmus paper to red “Blue to Red A-CID”

  5. Properties of Bases • Generally produce OH- ions in water • Taste bitter, chalky • Are electrolytes • Feel soapy, slippery • React with acids to form salts and water • pH greater than 7 • Turns red litmus paper to blue “BasicBlue”

  6. Arrhenius Definition Arrhenius Acid - Substances in water that increase the concentration of hydrogen ions (H+). Base - Substances in water that increase concentration of hydroxide ions (OH-). Categorical definition – easy to sort substances into acids and bases Problem – many bases do not actually contain hydroxides

  7. Bronsted-Lowry Definition Acid - neutral molecule, anion, or cation that donates a proton. Base - neutral molecule, anion, or cation that accepts a proton.HA + :B HB+ + :A- Ex HCl + H2O  H3O+ + Cl- Acid Base Conj Acid Conj Base • Operational definition - The classification depends on how the substance behaves in a chemical reaction

  8. Conjugate Acid Base Pairs • Conjugate Base - The species remaining after an acid has transferred its proton. • Conjugate Acid - The species produced after base has accepted a proton. • HA & :A- - conjugate acid/base pair • :A- - conjugate base of acid HA • :B & HB+ - conjugate acid/base pair • HB+ - conjugate acid of base :B

  9. Examples of Bronsted-Lowry Acid Base Systems • Note: Water can act as acid or base • Acid Base Conjugate AcidConjugate Base • HCl + H2O  H3O+ + Cl- • H2PO4- + H2O  H3O+ +HPO42- • NH4+ + H2O  H3O+ + NH3 • Base Acid Conjugate AcidConjugate Base :NH3 + H2O  NH4++ OH- • PO43- + H2O  HPO42- + OH-

  10. G.N. Lewis Definition • Lewis • Acid - an electron pair acceptor • Base - an electron pair donor

  11. The pH Scale pH [H3O+ ] [OH- ] pOH

  12. pH and acidity • Acidity or Acid Strength depends on Hydronium Ion Concentration [H3O+] • The pH system is a logarithmic representation of the Hydrogen Ion concentration (or OH-) as a means of avoiding using large numbers and powers. pH = - log [H3O+] = log(1 / [H3O+]) pOH = - log [OH-] = log(1 / OH-]) • In pure water [H3O+] = 1 x 10-7 mol / L (at 20oC)  pH = - log(1 x 10-7) = - (0 - 7) = 7 • pH range of solutions: 0 - 14 pH < 7 (Acidic) [H3O+] > 1 x 10-7 m / L pH > 7 (Basic) [H3O+] < 1 x 10-7 m / L

  13. pH and acidity Kw = [H3O+] [OH-] = 1.0 x10-14 In pure water [H3O+] = [OH-]= 1.0 x10-7 pH + pOH = 14

  14. Calculating the pH pH = - log [H3O+] Example 1: If [H3O+] = 1 X 10-10pH = - log 1 X 10-10 pH = - (- 10) pH = 10 Example 2: If [H3O+] = 1.8 X 10-5pH = - log 1.8 X 10-5 pH = - (- 4.74) pH = 4.74

  15. Indicators

  16. pH and acidity The pH values of several common substances are shown at the right. Many common foods are weak acids Some medicines and many household cleaners are bases.

  17. Neutralization An acid will neutralize a base, giving a salt and water as products Examples Acid Base Salt water HCl + NaOH  NaCl + H2O H2SO4 + 2 NaOH  Na2SO4 + 2 H2O H3PO4 + 3 KOH  K3PO4 + 3 H2O 2 HCl + Ca(OH) 2 CaCl2 + 2 H2O A salt is an ionic compound that is formed from thepositive ion (cation)of thebaseand thenegative ion (anion) of the acid

  18. Neutralization Calculations If the concentration of acid or base is expressed in Molarity or mol dm-3 then: --The volume in dm3 multiplied by the concentration yields moles (mol) . -- If the volume is expressed in cm3 the same product yields millimoles (mmol) mol dm-3 x dm3 = mole mol dm-3 x cm3 = (0.001) x mole = mmol

  19. Neutralization Problems • The volume of solution in dm3 multiplied by concentration in moles dm-3 will yield moles. • If an acid and a base combine in a 1 to 1 ratio, the moles of acid will equal the moles of base. • Therefore the volume of the acid multiplied by the concentration of the acid is equal to the volume of the base multiplied by the concentration of the base. Vacid C acid = V base C base If any three of the variables are known, it is possible to determine the fourth.

  20. Neutralization Problems Example 1:Hydrochloric acid reacts with potassium hydroxide according to the following reaction: HCl + KOH  KCl + H2O If 15.00 cm3 of 0.500 M HCl exactly neutralizes 24.00 cm3 of KOH solution, what is the concentration of the KOH solution? Solution: Vacid Cacid = Vbase Cbase (15.00 cm3 )(0.500 M) = (24.00 cm3 ) Cbase Cbase = (15.00 cm3 )(0.500 M) (24.00 cm3 ) Cbase = 0.313 M

  21. Neutralization Problems Whenever an acid and a base do not combine in a 1 to 1 ratio, a mole factor must be added to the neutralization equation n Vacid C acid = V base C base The mole factor (n) is the number of times the moles the acid side of the above equation must be multiplied so as to equal the base side. (or vice versa) Example H2SO4 + 2 NaOH  Na2SO4 + 2 H2O The mole factor is 2 and goes on the acid side of the equation. The number of moles of H2SO4 is one half that of NaOH. Therefore the moles of H2SO4 are multiplied by 2 to equal the moles of NaOH.

  22. Neutralization Problems Example 2:Sulfuric acid reacts with sodium hydroxide according to the following reaction: H2SO4 + 2 NaOH  Na2SO4 + 2 H2O If 20.00 cm3 of 0.400 M H2SO4 exactly neutralizes 32.00 cm3 of NaOH solution, what is the concentration of the NaOH solution? Solution: In this case the mole factor is 2 and it goes on the acid side, since the mole ratio of acid to base is 1 to 2. Therefore 2 Vacid Cacid = Vbase Cbase 2 (20.00 cm3 )(0.400 M) = (32.00 cm3 ) Cbase Cbase = (2) (20.00 cm3 )(0.400 M) (32.00 cm3 ) Cbase = 0.500 M

  23. Neutralization Problems Example 3:Phosphoric acid reacts with potassium hydroxide according to the following reaction: H3PO4 + 3 KOH  K3PO4 + 3 H2O If 30.00 cm3 of 0.300 M KOH exactly neutralizes 15.00 cm3 of H3PO4 solution, what is the concentration of the H3PO4 solution? Solution: In this case the mole factor is 3 and it goes on the acid side, since the mole ratio of acid to base is 1 to 2. Therefore 3 Vacid Cacid = Vbase Cbase 3 (15.00 cm3 )(Cacid) = (30.00 cm3 ) (0.300 M) Cacid = (30.00 cm3 )(0.300 M) (3) (15.00 cm3 ) Cacid = 0.200 M

  24. Neutralization Problems Example 4:Hydrochloric acid reacts with calcium hydroxide according to the following reaction: 2 HCl + Ca(OH)2 CaCl2 + 2 H2O If 25.00 cm3 of 0.400 M HCl exactly neutralizes 20.00 cm3 of Ca(OH)2 solution, what is the concentration of the Ca(OH)2 solution? Solution: In this case the mole factor is 2 and it goes on the base side, since the mole ratio of acid to base is 2 to 1. Therefore Vacid Cacid = 2 Vbase Cbase (25.00 cm3) (0.400) = (2) (20.00 cm3) (Cbase) Cbase = (25.00 cm3 ) (0.400 M) (2) (20.00 cm3 ) Cbase = 0.250 M

  25. Error Propagation Neutralization Problems Va Ca = VbCb so Ca = (Vb Cb)/Va If Va = 20.22 cm3+ 0.03 Vb = 19.30 cm3+ 0.03 and Cb = 0.517 M+ 0.0011 Then Ca = (Vb Cb)/Va = (19.30 cm3)(0.517 M)/(20.22 cm3) Ca = 0.493 M Error Propagation add up the relative errors 0.03+0.03+0.0011 20.22 19.30 0.517 = 0.00148 + 0.00155 + 0.00213 = 0.00516 The uncertainty in the answer is therefore (0.493M)(0.00516) = + 0.00254 Hence the final answer should be reported as 0.493 M+ 0.003 M

  26. Acid Base Dissociation • Acid-base reactions are equilibrium processes. • The ratio of the concentrations of the reactants and products is constant for a given temperature at equilibrium • It is known as the Acid or Base Dissociation Constant. • The stronger the acid or base, the larger the value of the dissociation constant.

  27. Acid Strength • Strong Acid - Transfers all of its protons to water; - Completely ionized; - Strong electrolyte; - The conjugate base is weaker and has a negligible tendency to be protonated. • Weak Acid - Transfers only a fraction of its protons to water; • - Partly ionized; - Weak electrolyte; - The conjugate base is stronger, readily accepting protons from water • As acid strength decreases, base strength increases. • The stronger the acid, the weaker its conjugate base • The weaker the acid, the stronger its conjugate base

  28. Acid Dissociation Constants Dissociation constants for some weak acids

  29. Base Strength • Strong Base - all molecules accept a proton; - completely ionizes; - strong electrolyte; - conjugate acid is very weak, negligible tendency to donate protons. • Weak Base - fraction of molecules accept proton; - partly ionized; - weak electrolyte; - the conjugate acid is stronger. It more readily donates protons. • As base strength decreases, acid strength increases. • The stronger the base, the weaker its conjugate acid. • The weaker the base the stronger its conjugate acid.

  30. Common Strong Acids/Bases Strong Acids Hydrochloric Acid Nitric Acid Sulfuric Acid Perchloric Acid Strong Bases Sodium Hydroxide Potassium Hydroxide *Barium Hydroxide *Calcium Hydroxide *While strong bases they are not very soluble

  31. Water as an Equilibrium System • Water has the ability to act as either a Bronsted- Lowry acid or base. • Autoionization – spontaneous formation of low concentrations of [H+] and OH-] ions by proton transfer from one molecule to another. • Equilibrium Constant for Water

  32. Weak Acid Equilibria A weak acid is only partially ionized. Both the ion form and the unionized form exist at equilibrium HA + H2O  H3O+ + A- The acid equilibrium constant is Ka = [H3O+ ] [A-] [HA] Ka values are relatively small for most weak acids. The greatest part of the weak acid is in the unionized form

  33. Weak Acid Equilibrium Constants Sample problem .A certain weak acid dissociates in water as follows: HA + H2O  H3O+ + A- If the initial concentration of HA is 1.5 M and the equilibrium concentration of H3O+ is 0.0014 M. Calculate Ka for this acid Solution Ka = [H3O+ ] [A-] [HA] I C E Substituting [HA] 1.5 -x 1.5-x Ka = (0.0014)2 = 1.31 x 10-6 [A-] 0 +x x 1.4986 [H3O+ ] 0 +x x x = 0.0014 1.5-x = 1.4986

  34. Weak Acid Equilibria Concentration Problems Problem 1.A certain weak acid dissociates in water as follows: HA + H2O  H3O+ + A- The Ka for this acid is 2.0 x 10-6. Calculate the [HA] [A-], [H3O+] and pH of a 2.0 M solution Solution Ka = [H3O+ ] [A-] = 2.0 x 10-6 [HA] I C E Substituting [HA] 2.0 -x 2.0-x Ka = x2 = 2.0 x 10-6 [A-] 0 +x x 2.0-x [H3O+ ] 0 +x x If x <<< 2.0 it can be dropped from the denominator The x2 = (2.0 x10-6)(2.0) = 4.0 x10-6 x = 2.0 x 10-3 [A-] = [H3O+ ] = 2.0 x10-3 [HA] = 2.0 - 0.002= 1.998 pH = - log [H3O+ ] =-log (2.0 x 10-3) = 2.7

  35. Weak Acid Equilibria Concentration Problems Problem 2. Acetic acid is a weak acid that dissociates in water as follows: CH3COOH + H2O  H3O+ + CH3COO- The Ka for this acid is 1.8 x 10-5. Calculate the [CH3COOH],[CH3COO-] [H3O+] and pH of a 0.100 M solution Solution Ka = [H3O+ ] [CH3COO-]= 1.8 x 10-5 [CH3COOH] I C E Substituting [CH3COOH] 0.100 -x 0.100-x Ka = x2 = 1.8 x 10-5 [CH3COO- ] 0 +x x 0.100-x [H3O+] 0 +x x If x <<< 0.100 it can be dropped from the denominator The x2 = (1.8 x10-5)(0.100) = 1.8 x10-6 x = 1.3 x 10-3 [CH3COO--] = [H3O+ ] = 1.3 x10-3 [CH3COOH] = 0.100 - 0.0013 = 0.0987 pH = - log [H3O+ ] =-log (1.3 x10-3) = 2.88

  36. Weak Base Equilibria Weak bases, like weak acids, are partially ionized. The degree to which ionization occurs depends on the value of the base dissociation constant General form: B + H2O  BH+ + OH- Kb = [BH+][OH-] [B] Example NH3 + H2O  NH4+ + OH- Kb = [NH4+][ OH-] [NH3]

  37. Weak Base Equilibrium Constants Sample problem .A certain weak base dissociates in water as follows: B + H2O  BH+ + OH- If the initial concentration of B is 1.2 M and the equilibrium concentration of OH- is 0.0011 M. Calculate Kb for this base Solution Kb = [BH+ ] [OH-] [B] I C E Substituting [B] 1.2 -x 1.2-x Kb = (0.0011)2 = 1.01 x 10-6 [OH-] 0 +x x 1.1989 [BH+ ] 0 +x x x = 0.0011 1.2-x = 1.1989

  38. Weak Base Equilibria Example1. Ammonia dissociates in water according to the following equilibrium NH3 + H2O  NH4+ + OH- Kb = [NH4+][ OH-] = 1.8 x 10-5 [NH3] Calculate the concentration of [NH4+][ OH-] [NH3 ]and the pH of a 2.0M solution. I C E Substituting [NH3] 2.0 -x 2.0-x Kb = x2 = 1.8x 10-5 [OH-] 0 +x x 2.0-x [NH4+] 0 +x x If x <<< 2.0 it can be dropped from the denominator The x2 = (1.8 x10-5)(2.0) = 3.6 x10-5 x = 6.0 x 10-3 [OH-] = [NH4+] = 6.0 x10-3 [NH3] = 2.0- 0.006= 1.994 pOH = - log [OH-] =-log (6.0 x10-3) = 2.22 pH = 14-pOH = 14-2.22 = 11.78

  39. Amphoteric Solutions • A chemical compound able to react with both an acid or a base is amphoteric.    • Water is amphoteric. The two acid-base couples of water are H3O+/H2O and H2O/OH-It behaves sometimes like an acid, for example • And sometimes like a base : • Hydrogen carbonate ion HCO3- is also amphoteric, it belongs to the two acid-base couples H2CO3/HCO3- and HCO3-/CO32-

  40. Common Ion Effect The common ion effect is a consequence of Le Chatelier’s Principle When the salt with the anion (i.e. the conjugate base) of a weak acid is added to that acid, • It reverses the dissociation of the acid. • Lowers the percent dissociation of the acid. A similar process happens when the salt with the cation (i.e, conjugate acid) is added to a weak base. These solutions are known as Buffer Solutions.

  41. Buffer Solutions - Characteristics • A solution that resists a change in pH. It is pH stable. • A weak acid and its conjugate base form an acid buffer. • A weak base and its conjugate acid form a base buffer. • We can make a buffer of any pH by varying the concentrations of the acid/base and its conjugate.

  42. Buffer Solution Calculations Calculate the pH of a solution that is 0.50 M CH3COOH and 0.25 M NaCH3COO. CH3COOH + H2O  H3O+ + CH3COO-(Ka = 1.8 x 10-5) Solution Ka = [H3O+ ] [CH3COO-]= 1.8 x 10-5 [CH3COOH] I C E. Substituting [CH3COOH] 0.50 -x 0.50-x Ka = x (0.25+x) = 1.8 x 10-5 [CH3COO-] 0.25 +x 0.25+x (0.50-x) [H3O+] 0 +x xIf x <<< 0.25 it can be dropped from both expressions in ( ) since adding or subtracting a small amount will not significantly change the value of the ratio Then the expression becomes x(0.25)/(0.50) = 1.8 x 10-5 x = 3.6 x 10-5 = [H3O+] pH = - log [H3O+] =-log(3.6 x 10-5 ) = 4.44

  43. Buffer Solutions - Equations 1. Ka = [H3O+] [A-] [HA] • [H3O+] = Ka [HA] [A-] The [H3O+] depends on the ratio [HA]/[A-] Taking the negative log of both sides of equation 2 above • pH = -log(Ka [HA]/[A-]) • pH = -log(Ka) - log([HA]/[A-]) • pH = pKa + log([A-]/[HA])

  44. Henderson Hasselbach Equation • pH = pKa + log([A-]/[HA]) • pH = pKa + log(Base/Acid) • This expression is known as the Henderson-Hasselbach equation. It provides a shortcut from using the I.C.E. model for buffer solutions where the concentration of both [A-] and [HA] are significantly greater than zero.

  45. Using the Henderson -Hasselbach Equation pH = pKa + log([A-]/[HA]) Example Calculate the pH of the following of a mixture that contains 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4) HC3H5O3 + H2O  H3O+ + C3H5O3- Solution Using the Henderson-Hasselbach equation pH = - log (1.4 x 10-4) + log ( 0.25/0.75 ) = 3.85 + (-0.477) = 3.37

  46. Henderson-Hasselbach Equation and Base Buffers For a base a similar expression can be written pOH = pKb + log ([BH+] / [B]) pOH = pKb + log ([Acid] / [Base]) Example: Calculate the pH of a solution that contains 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5) Solution pOH = - log(1.8 x 10-5) + log (0.40/0.25) = 4.74 + 0.204 = 4.94 pH = 14 - pOH = 14 - 4.94 = 9.06

  47. Henderson-Hasselbach Equation & Base Buffers Methyl amine is a weak base with a Kb or 4.38 x 10-4 CH3NH2 + H2O  CH3NH3+ + OH- Calculate the pH of a solution that is 0.10 M in methyl amine and 0.20 M in methylamine hydrochloride. pOH = pKb + log ([BH+] / [B]) Solution pOH = -log (4.38 x 10-4) + log (0.20 / 0.10) = 3.36 + 0.30 = 3.66 pH = 14- 3.66 = 10.34

  48. Additional Buffer Problems How many grams of sodium formate, NaCHOO, would have to be dissolved in 1.0 dm3 of 0.12 M formic acid, CHOOH, to make the solution a buffer of pH 3.80?Ka= 1.78 x 10-4 pH = pKa + Log ([A-]/[HA]) Solution 3.80 = -log (1.78 x 10-4) + Log [A-] - Log [0.12] 3.80 = 3.75 + Log [A-] - (-0.92) Log [A-]= 3.80 - 3.75 - 0.92 = - 0.87 [A-] = 10-0.87 = 0.135 mol dm-3 The molar mass of NaCHOO = 23+12+1+2(16) = 58.0 gmol-1 So (0.135 mol dm-3)(58.0 gmol-1) = 7.8 grams per dm-3

  49. Relationship of Ka, Kb & Kw • HA weak acid. Its acid ionization is • A- is the conjugate base Its base ionization is • Multiplying Ka and Kb and canceling like terms

  50. Titration Curves • A graph showing pH vs volume of acid or base added • The pH shows a sudden change near the equivalence point • The Equivalence point is the point at which the moles of OH- are equal to the moles of H3O+

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