1. 8 - 1 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

2. 8 - 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. 8 - 3 Another useful equation is:  pH + pOH = 14.00 It follows that:  [H + ] = 10 -pH and [OH - ] = 10 -pOH

4. 8 - 4 Acidic, basic, or neutral solutions can be distinguished as shown below: SolutionpH[H + ]pOH[OH - ] Acidic< 7.00> 10 -7 > 7.00< 10 -7 Neutral= 7.00= 10 -7 = 7.00= 10 -7 Basic> 7.00< 10 -7 < 7.00> 10 -7

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

6. 8 - 6 Strong Acids There are six strong acids.  HCl, HNO 3, HClO 4, HI, HBr, and H 2 SO 4  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 H 2 SO 4 is not intuitively obvious to the most casual observer.

7. 8 - 7  All the strong acids are assumed to have one ionizable hydrogen including H 2 SO 4.  The ionization of H 2 SO 4 must be shown as: H 2 SO 4 (aq) → H + (aq) + HSO 4 - (aq)  The single arrow as always indicates that the ionization is 100% and there are no H 2 SO 4 molecules remaining in solution.

8. 8 - 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. 8 - 9 Another Strong Acid Problem Determine the pH and pOH of a 10 -10 M HNO 3 solution. [HNO 3 ] = 10 -10 M HNO 3 (aq) → H + (aq) + NO 3 - (aq) pH = -log[H + ] = -log(10 -10 ) = 10 What??

10. 8 - 10 How can a strong acid have a pH corresponding to a base? The answer lies with the autoionization of water. H 2 O(l) + H 2 O(l) H 3 O + (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. 8 - 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. 8 - 12 Weak Acids Equilibrium Constant H 2 O(l) + H 2 O(l) H 3 O + (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 K eq = [H + ] [OH - ] K w = K eq = [H + ] [OH - ] = 1.00 × 10 -14

13. 8 - 13 For the reaction H 2 O(l) + H 2 O(l) H 3 O + (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 K eq.

14. 8 - 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. 8 - 15  The effect of temperature on the K eq depends on which reaction is endothermic.  The inverse relationship between H + and OH - in any solution is given by K w.  Remember the numerical value for K w is both reaction and temperature dependent.

16. 8 - 16 A Weak Acid Problem Acetic acid, HC 2 H 3 O 2, has a K a = 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. K a = 1.8 x 10 -5 n = 2.0 mol HC 2 H 3 O 2 V = 1.0 L [HC 2 H 3 O 2 ] = n V 2.0 mol HC 2 H 3 O 2 1.0 L = 2.0 M

17. 8 - 17 HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 - (aq) [ ] I 2.0 0 0 [ ] c -x +x +x [ ] e 2.0 - x x x KaKa = [H + ] [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] 1.8 x 10 -5 = 2.0 - x x x ×

18. 8 - 18 At this point in the problem an approximation will be made called the 5% rule.  You will note that K a = 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. 8 - 19 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: 1.8 x 10 -5 = 2.0 - x x x × ≈ x2x2 2.0 %ion = [H + ] [HC 2 H 3 O 2 ] × 100%

20. 8 - 20 %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. 6.0 x 10 -3 M 2.0 M × 100%= 0.30%

21. 8 - 21 To learn more about the Solver, visit http://www2.ohlone.edu/people2/joconnell/ti/s olver8384.pdf Another useful quantity that is used is the pK a of an acid and is defined as pK a = -log K a. So for the current problem, acetic acid, K a = 1.8 x 10 -5 and pK a = 4.74 pK a ’s are more convenient to work with than Ka’s.

22. 8 - 22 Combining Acidic Solutions What is the pH of a solution obtained by mixing 235 mL of 0.0245 M HNO 3 with 554 mL of 0.438 M HClO 4 ? [HNO 3 ] = 0.0245 M[HClO 4 ] = 0.438 M V 1 = 235 mLV 2 = 554 mL HNO 3 (aq) → H + (aq) + NO 3 - (aq) HClO 4 (aq) → H + (aq) + ClO 4 - (aq)

23. 8 - 23 [HNO 3 ] = n/V 1 mol HNO 3 n1n1 = 0.0245 mol HNO 3 1.00 L × 235 mL × 1 L 10 3 mL × × 1 mol H + = 5.76 × 10 -3 mol H + 1 mol HNO 3 n2n2 = 0.438 mol HClO 4 1.00 L × 554 mL × 1 L 10 3 mL × × 1 mol H + = 2.42 × 10 -1 mol H + 1 mol HClO 4

24. 8 - 24. [H + ] = nTnT VTVT = 2.48 × 10 -1 mol H + 789 mL × 1 L 10 3 mL = 0.314 M

25. 8 - 25 Polyprotic Acids Polyprotic acids are those acids, H 2 CO 3 and H 2 SO 4, which have more than one ionizable hydrogen. The K a value for each hydrogen is different and K a vastly decreases with each hydrogen. The first hydrogen is the easiest to remove because the original molecule or ion is the strongest acid.

26. 8 - 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.  H 2 CO 3 (aq) H + (aq) + HCO 3 - (aq)  HCO 3 - (aq) H + (aq) + CO 3 2- (aq)  The greater attraction between the HCO 3 - and the H + will make it more difficult to ionize.

27. 8 - 27 The pH of a diprotic or a triprotic acid is determined almost completely by the first ionization.

28. 8 - 28 Diprotic Acid Problem Calculate the pH of a 0.0020 M solution of carbonic acid. [H 2 CO 3 ] = 0.0020 M K a1 = 4.4 x 10 -7 K a2 = 4.7 x 10 -11

29. 8 - 29 H 2 CO 3 (aq) H + (aq) + HCO 3 - (aq) [ ] I 0.0020 0 0 [ ] c -x +x +x [ ] e 0.0020 - x x x 4.4 x 10 -7 = 0.0020 - x x x × K a1 = [H + ] [HCO 3 - ] [H 2 CO 3 ] 4.4 x 10 -7 = 0.0020 - x x x ×

30. 8 - 30 We will make the assumption that 0.0020 – x ≈ 0.0020 and verify the assumption after the calculation is made. x = [H + ] = [HCO 3 - ] = 3.0 x 10 -5 M 4.4 x 10 -7 = 0.0020 x2x2 %ion = [H + ] [HCO 3 - ] × 100% %ion = 3.0 x 10 -5 M 2.0 x 10 -3 M × 100% =1.5%

31. 8 - 31 The %ion = 1.5% is well within the 5%, therefore the assumption is valid. HCO 3 - (aq) H + (aq) + CO 3 2- (aq) [ ] I 3.0 x 10 -5 3.0 x 10 -5 0 [ ] c -x +x +x [ ] e 3.0 x 10 -5 -x 3.0 x 10 -5 +x x You must use the concentrations of H + and HCO 3 - resulting from the first hydrogen ionizing.

32. 8 - 32 B K a2 = [H + ] [CO 3 2- ] [HCO 3 - ] 4.7 x 10 -11 = (3.0 x 10 -5 + x) x × (3.0 x 10 -5 - x) 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. 8 - 33 x = [H + ] = [CO 3 2- ] = 4.7 x 10 -11 M %ion = [H + ] [HCO 3 - ] × 100% %ion = 4.7 x 10 -11 M 3.0 x 10 -5 M × 100% =1.6 x 10 -4 % The %ion = 1.6 x 10 -4 % is well within the 5%, therefore the assumption is valid.

34. 8 - 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. 8 - 35 Common Ion Equilibria Calculate the pH of a solution containing 0.075 M nitrous acid and 0.20 M sodium nitrite. K a = 4.5 x 10 -4 [NaNO 2 ] = 0.20 M [HNO 2 ] = 0.075 M NaNO 2 (aq) → Na + (aq) + NO 2 - (aq)

36. 8 - 36 HNO 2 (aq) H + (aq) + NO 2 - (aq) [ ] I 0.075 0 0.20 [ ] c -x +x +x [ ] e 0.075 - x x x KaKa = [H + ] [NO 2 - ] [HNO 2 ] 4.5 x 10 -4 = 0.075 - x x (0.20 + x) × ≈ 0.075 0.20x

37. 8 - 37 [H + ] = 1.7 x 10 -4 M %ion = [H + ] [HNO 2 ] × 100% %ion = 1.7 x 10 -4 M × 100% 0.075 M =0.23% The 5% rule applies so the previous two assumptions are valid. pH = -log[H + ] = -log(1.7 x 10 -4 ) = 3.77