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Introduction to Acids and Bases

Introduction to Acids and Bases. Pure or distilled water undergoes a very slight ionization as shown below. H 2 O(l) H + (aq) + OH - (aq) The equilibrium constant for water is given by: K w = [H + ][OH - ] = 1.00 x 10 -14. Note that the product of [H + ] and [OH - ] is a

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Introduction to Acids and Bases

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  1. Introduction to Acids and Bases • Pure or distilled water undergoes a very • slight ionization as shown below. • H2O(l) H+(aq) + OH-(aq) The equilibrium constant for water is given by: • Kw = [H+][OH-] = 1.00 x 10-14

  2. Note that the product of [H+] and [OH-] is a • constant at 25°C and that [H+] and [OH-] are • inversely proportional to each other. • The pH scale has been devised to determine • the molarity of the hydrogen ion, [H+], in an • aqueous solution and is given by: • pH = -log[H+] Similarly, pOH is defined as: • pOH = -log[OH-]

  3. Another useful equation is: • pH + pOH = 14.00 It follows that: • [H+] = 10-pH and [OH-] = 10-pOH

  4. Acidic, basic, or neutral solutions can be • distinguished as shown below:

  5. An alternative approach to the relationship • between pH and pOH is shown below. Increasing Acidity Decreasing Acidity pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Neutral Increasing Acidity Decreasing Acidity pOH 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

  6. Strong Acids • There are six strong acids. • HCl, HNO3, HClO4, HI, HBr, and H2SO4 • When given any of the above acids always assume 100% ionization. • 100% ionization is shown by using a single arrow in the ionization equation. • The ionization equation for H2SO4 is not intuitively obvious to the most casual observer.

  7. All the strong acids are assumed to have one ionizable hydrogen including H2SO4. • The ionization of H2SO4 must be shown as: H2SO4(aq) → H+(aq) + HSO4-(aq) • The single arrow as always indicates that the ionization is 100% and there are no H2SO4 molecules remaining in solution.

  8. Strong Acid Problem • Determine the pH and pOH of a 0.20 M HCl • solution. • [HCl] = 0.20 M • HCl(aq) → H+(aq) + Cl-(aq) • pH = -log[H+] = -log(0.20) = 0.70 • pH + pOH = 14.00 • pOH = 14.00 – 0.70 = 13.30

  9. Another Strong Acid Problem • Determine the pH and pOH of a 10-10 M HNO3 • solution. • [HNO3] = 10-10 M • HNO3(aq) → H+(aq) + NO3-(aq) • pH = -log[H+] = -log(10-10) = 10 • What??

  10. How can a strong acid have a pH • corresponding to a base? • The answer lies with the autoionization of • water. • H2O(l) + H2O(l) H3O+(aq) + OH-(aq) • [ ]I 0 0 • [ ]c +1.0 x 10-7 +1.0 x 10-7 • [ ]e 1.0 x 10-7 1.0 x 10-7

  11. Although water ionizes only to a slight extent, • there is a dynamic equilibrium which remains • intact. • Distilled or pure water has a pH = 7 because • the [H+] = [OH-] = 1.0 x 10-7 M. • [H+]T = [H+]water + [H+]nitric acid • [H+]T = 1.0 x 10-7 M + 10-10 M ≈ 1.0 x 10-7 M • pH = -log[H+] = -log(1.0 x 10-7) = 7.00

  12. Weak Acids Equilibrium Constant • H2O(l) + H2O(l) H3O+(aq) + OH-(aq) • [ ]I 0 0 • [ ]c +1.0 x 10-7 +1.0 x 10-7 • [ ]e 1.0 x 10-7 1.0 x 10-7 • Keq = [H+] [OH-] • Kw = Keq = [H+] [OH-] = 1.00 × 10-14

  13. For the reaction • H2O(l) + H2O(l) H3O+(aq) + OH-(aq) • remember the following: • This is an example of a heterogeneous equilibria because more than one phase (state) is present. • When the reactants/products are solids or liquids, their concentration are nearly constant and incorporated into the Keq.

  14. Pure solids and liquids are not incorporated into the equilibrium expression. • The position of the equilibrium is independent of the amount of solid or liquid present. • The solvent for the reaction does not appear in the equilibrium expression. • Ions or molecules appear as their molar concentrations.

  15. The effect of temperature on the Keq depends on which reaction is endothermic. • The inverse relationship between H+ and OH- in any solution is given by Kw. • Remember the numerical value for Kw is both reaction and temperature dependent.

  16. A Weak Acid Problem • Acetic acid, HC2H3O2, has a Ka = 1.8 x 10-5. • Calculate the hydrogen ion concentration in a • solution prepared by adding 2.0 moles of • acetic acid to form one liter of solution. • Ka = 1.8 x 10-5 n = 2.0 mol HC2H3O2 V = 1.0 L • [HC2H3O2] = • [HC2H3O2] = n V 2.0 mol HC2H3O2 2.0 M = 1.0 L

  17. [H+] [C2H3O2-] Ka = [HC2H3O2] x x × 1.8 x 10-5 = 2.0 - x • HC2H3O2(aq) H+(aq) + C2H3O2-(aq) • [ ]I 2.00 0 • [ ]c -x+x+x • [ ]e 2.0 - xxx

  18. At this point in the problem an approximation • will be made called the 5% rule. • You will note that Ka = 1.8 x 10-5 is a very small number and that x may be very small compared to the initial concentration of the acid. • This assumption will not always be true and should always be tested after solving for x which is the [H+].

  19. x x x2 × [H+] ≈ 1.8 x 10-5 = 100% = × %ion 2.0 - x 2.0 [HC2H3O2] • The approximation made will be 2.0 – x ≈ 2.0 • which avoids using the quadratic formula. • x = [H+] = 6.0 x 10-3 M • The 5% rule is given by:

  20. 6.0 x 10-3 M 100% = 0.30% × 2.0 M • %ion = • The 0.30% is well within the tolerance of 5% • so the assumption is valid. • If the % ionization was more than 5%, then the • assumption would not be valid and you would • have to use the quadratic formula. • There are quadratic formula programs for the • TI calculator and also the Solver function on • the TI.

  21. To learn more about the Solver, visit • http://www2.ohlone.edu/people2/joconnell/ti/solver8384.pdf • Another useful quantity that is used is the pKa • of an acid and is defined as pKa = -log Ka. • So for the current problem, acetic acid, • Ka = 1.8 x 10-5 and pKa = 4.74 • pKa’s are more convenient to work with than • Ka’s.

  22. Combining Acidic Solutions • What is the pH of a solution obtained by • mixing 235 mL of 0.0245 M HNO3 with 554 mL • of 0.438 M HClO4? • [HNO3] = 0.0245 M [HClO4] = 0.438 M • V1 = 235 mL V2 = 554 mL • HNO3(aq) → H+(aq) + NO3-(aq) • HClO4(aq) → H+(aq) + ClO4-(aq)

  23. 0.0245 mol HNO3 1 L n1 235 mL × × × = 1.00 L 103 mL 1 mol H+ 5.76 × 10-3 mol H+ = × 1 mol HNO3 • [HNO3]= n/V 1 mol HNO3 0.438 mol HClO4 1 L n2 554 mL × × × = 1.00 L 103 mL 1 mol H+ 2.42 × 10-1 mol H+ = × 1 mol HClO4

  24. nT [H+] = VT • . 2.48 × 10-1 mol H+ [H+] = = 0.314 M 1 L 789 mL × 103 mL

  25. Polyprotic Acids • Polyprotic acids are those acids, H2CO3 and • H2SO4, which have more than one ionizable • hydrogen. • The Ka value for each hydrogen is different • and Ka vastly decreases with each hydrogen. • The first hydrogen is the easiest to remove • because the original molecule or ion is the • strongest acid.

  26. The reason for the increased difficultly is • because each successive hydrogen is trying • to be removed from a more negative species. • Compare the two ionization equations for • carbonic acid. • H2CO3(aq) H+(aq) + HCO3-(aq) • HCO3-(aq) H+(aq) + CO32-(aq) • The greater attraction between the HCO3- and the H+ will make it more difficult to ionize.

  27. The pH of a diprotic or a triprotic acid is • determined almost completely by the first • ionization.

  28. Diprotic Acid Problem • Calculate the pH of a 0.0020 M solution of • carbonic acid. • [H2CO3] = 0.0020 M • Ka1 = 4.4 x 10-7 Ka2 = 4.7 x 10-11

  29. [H+] [HCO3-] Ka1 = [H2CO3] x x × 4.4 x 10-7 = 0.0020 - x • H2CO3(aq) H+(aq) + HCO3-(aq) • [ ]I 0.00200 0 • [ ]c -x+x+x • [ ]e 0.0020 - xxx x x × 4.4 x 10-7 = 0.0020 - x

  30. x2 4.4 x 10-7 = 0.0020 [H+] 100% = × %ion [HCO3-] 3.0 x 10-5 M 100% = = 1.5% × %ion 2.0 x 10-3 M • We will make the assumption that 0.0020 – x ≈ • 0.0020 and verify the assumption after the • calculation is made. • x = [H+] = [HCO3-] = 3.0 x 10-5 M

  31. The %ion = 1.5% is well within the 5%, • therefore the assumption is valid. • HCO3-(aq) H+(aq) + CO32-(aq) • [ ]I 3.0 x 10-53.0 x 10-5 0 • [ ]c -x+x+x • [ ]e 3.0 x 10-5-x3.0 x 10-5+xx • You must use the concentrations of H+ and HCO3- resulting from the first hydrogen ionizing.

  32. x (3.0 x 10-5 + x) × 4.7 x 10-11 = (3.0 x 10-5 - x) [H+] [CO32-] Ka2 = [HCO3-] • B The extremely small value of the exponent, 10-11, allows for the assumption 3.0 x 10-5 ± x ≈ 3.0 x 10-5 which simplifies the above expression to

  33. [H+] 100% = × %ion • x = [H+] = [CO32-] = 4.7 x 10-11 M [HCO3-] 4.7 x 10-11 M 100% = = 1.6 x 10-4% × %ion 3.0 x 10-5 M The %ion = 1.6 x 10-4 % is well within the 5%, therefore the assumption is valid.

  34. What was the point of doing this rather long • and tedious problem? • These steps can be applied to any diprotic or • triprotic acid and what you verify is that the • [H+] contribution from the second or third • hydrogen can be ignored. • So, what is the pH of a 0.0020 M solution of • carbonic acid? • pH = -log[H+] = -log(3.0 x 10-5) = 4.52

  35. Common Ion Equilibria • Calculate the pH of a solution containing • 0.075 M nitrous acid and 0.20 M sodium • nitrite. • Ka = 4.5 x 10-4 • [NaNO2] = 0.20 M [HNO2] = 0.075 M • NaNO2(aq) → Na+(aq) + NO2-(aq)

  36. [H+] [NO2-] • HNO2(aq) H+(aq) + NO2-(aq) • [ ]I 0.0750 0.20 • [ ]c -x+x+x • [ ]e 0.075 - xxx Ka = [HNO2] 0.20x x (0.20 + x) × 4.5 x 10-4 ≈ = 0.075 0.075 - x

  37. [H+] 100% = × %ion [HNO2] • [H+] = 1.7 x 10-4 M 1.7 x 10-4 M 100% = × = 0.23% %ion 0.075 M The 5% rule applies so the previous two assumptions are valid. pH = -log[H+] = -log(1.7 x 10-4) = 3.77

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