1. BUFFER – A solution of about equal amounts of a weak acid and its weak conjugate base A buffer is resistant to changes in pH because it can neutralize any strong acid or base added to it 2A-1 (of 16) SOLUTION EQUILIBRIA

2. A solution contains both HF and F - Weak acid: Weak conjugate base: HF F-F- 1 HCl molecule If a strong acid is added to the buffer: 2A-2 (of 16)

3. The STRONG ACID (H 3 O + ) is reacted away completely by the WEAK CONJUGATE BASE (F - ) A solution contains both HF and F - Weak acid: Weak conjugate base: HF F - 1 HCl molecule  The original [H 3 O + ] is virtually unchanged If a strong acid is added to the buffer: 2A-3 (of 16) H 3 O + + F - → HF + H 2 O

4. A solution contains both HF and F - Weak acid: Weak conjugate base: HF F - If a strong base is added to the buffer: OH - + HF → H 2 O + F - The STRONG BASE (OH - ) is reacted away completely by the WEAK ACID (HF)  The original [H 3 O + ] is virtually unchanged 1 OH - ion 2A-4 (of 16)

5. The higher the concentrations of the weak acid and weak conjugate base in the buffer, the more strong acid or strong base it can neutralize This buffer can neutralize 2 H 3 O + ’s and 2 OH - ’s 2A-5 (of 16)

6. Calculate the pH of a solution 0.25 M in hydrofluoric acid (K a = 7.2 x 10 -4 ) and 0.50 M in sodium fluoride x0.50 + x HF (aq) + H 2 O (l) ⇆ H 3 O + (aq) + F - (aq) Initial M’s Change in M’s Equilibrium M’s 0.25 ~00.50 - x + x 0.25 - x K a = [H 3 O + ][F - ] _____________ [HF] 7.2 x 10 -4 = x(0.50 + x) ______________ (0.25 – x) 7.2 x 10 -4 = x(0.50) _________ 0.25 3.6 x 10 -4 = x 2A-6 (of 16) 3.44= pH = [H 3 O + ] A buffer: contains a weak acid (HF) and its weak conjugate base (F - ) CALCULATING THE pH OF A BUFFER SOLUTION

7. [H 3 O + ] x [F - ] = K a ______ [HF] - log [H 3 O + ] - log [F - ] ______ [HF] Calculate the pH of a solution 0.25 M in hydrofluoric acid (K a = 7.2 x 10 -4 ) and 0.50 M in sodium fluoride 2A-7 (of 16) CALCULATING THE pH OF A BUFFER SOLUTION [H 3 O + ] x [F - ] = K a ______ [HF] pH - log [F - ] = pK a ______ [HF] = - log K a [H 3 O + ][F - ] = K a _____________ [HF] pH = pK a + log [F - ] ______ [HF] log [H 3 O + ] + log [F - ] ______ [HF] = log K a log

8. pH = pK a + log [A - ] _____ [HA] HENDERSON-HASSELBALCH EQUATION Calculate the pH of a solution 0.25 M in hydrofluoric acid (K a = 7.2 x 10 -4 ) and 0.50 M in sodium fluoride CALCULATING THE pH OF A BUFFER SOLUTION 2A-8 (of 16)

9. pH = pK a + log [A - ] _____ [HA] DAVID HASSELHOFF EQUATION This can be used when a solution contains both a weak acid and its weak conjugate base Calculate the pH of a solution 0.25 M in hydrofluoric acid (K a = 7.2 x 10 -4 ) and 0.50 M in sodium fluoride CALCULATING THE pH OF A BUFFER SOLUTION 2A-9 (of 16)

10. Calculate the pH of a solution 0.25 M in hydrofluoric acid (K a = 7.2 x 10 -4 ) and 0.50 M in sodium fluoride CALCULATING THE pH OF A BUFFER SOLUTION pK a = -log K a pH = pK a + log [F - ] ______ [HF] = -log (7.2 x 10 -4 )= 3.143 = 3.143 + log (0.50 M) ___________ (0.25 M) = 3.44 Determine the pK a of the weak acid Use the David Hasselholf Equation 2A-10 (of 16)

11. pH = pK a + log [weak conj base] ______________________ [weak acid] pH = pK a + log (n weak conj base ) _________________ (n weak acid ) CALCULATING THE pH OF A BUFFER SOLUTION Useful when buffers are produced from mixing different solutions together pH = pK a + log (n weak conj base ) / vol ________________________ (n weak ccid ) / vol 2A-10 (of 16)

12. Find the pH of a buffer solution that is prepared by mixing 300. mL of 2.50 M acetic acid (K a = 1.8 x 10 -5 ) and 200. mL of 2.50 M sodium acetate. 2A-12 (of 16) HC 2 H 3 O 2 : Weak AcidNaC 2 H 3 O 2 : Salt w/ Weak Conj. Base Solutions are mixed  calculate the moles of all species present x 0.300 L solution= 0.750 mol HC 2 H 3 O 2 2.50 mol HC 2 H 3 O 2 _______________________ L solution 2.50 M NaC 2 H 3 O 2  2.50 M Na + and 2.50 M C 2 H 3 O 2 - x 0.200 L solution= 0.500 mol C 2 H 3 O 2 - 2.50 mol C 2 H 3 O 2 - _______________________ L solution

13. pK a = -log K a pH = pK a + log n C 2 H 3 O 2 - _________ n HC 2 H 3 O 2 = -log (1.8 x 10 -5 )= 4.745 = 4.745 + log (0.500 mol) ______________ (0.750 mol) = 4.57 Determine the pK a of the weak acid Use the David Hasselholf Equation Find the pH of a buffer solution that is prepared by mixing 300. mL of 2.50 M acetic acid (K a = 1.8 x 10 -5 ) and 200. mL of 2.50 M sodium acetate. HC 2 H 3 O 2 : Weak AcidNaC 2 H 3 O 2 : Salt w/ Weak Conj. Base 2A-13 (of 16)

14. Find the pH of a solution that is prepared by mixing 150. mL of 0.300 M nitrous acid (K a = 4.0 x 10 -4 ) and 100. mL of 0.200 M sodium nitrite. 2A-14 (of 16) HNO 2 : Weak AcidNaNO 2 : Salt w/ Weak Conj. Base Solutions are mixed  calculate the moles of all species present x 0.150 L solution= 0.0450 mol HNO 2 0.300 mol HNO 2 _____________________ L solution 0.200 M NaNO 2  0.200 M Na + and 0.200 M NO 2 - x 0.100 L solution= 0.0200 mol NO 2 - 0.200 mol NO 2 - ____________________ L solution

15. pK a = -log K a pH = pK a + log n NO 2 - _______ n HNO 2 = -log (4.0 x 10 -4 )= 3.398 = 3.398 + log (0.0200 mol) ________________ (0.0450 mol) = 3.05 Determine the pK a of the weak acid Use the David Hasselholf Equation Find the pH of a solution that is prepared by mixing 150. mL of 0.300 M nitrous acid (K a = 4.0 x 10 -4 ) and 100. mL of 0.200 M sodium nitrite. HNO 2 : Weak AcidNaNO 2 : Salt w/ Weak Conj. Base 2A-15 (of 16)

16. What must be the [NO 2 - ]/[HNO 2 ] to make a buffer with a pH = 3.00? 2A-16 (of 16) 3.00 = 3.398+ log [NO 2 - ] _________ [HNO 2 ] -0.398 = log [NO 2 - ] _________ [HNO 2 ] 0.40 = [NO 2 - ] _________ [HNO 2 ] HNO 2 : Weak AcidNO 2 - : Weak Conjugate Base When the pK a of the weak acid is within 1 pH unit of the desired pH, then the weak acid / weak conjugate base combination is a good choice for preparing the buffer = 0.40 ______ 1.0

18. Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and 1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added. 2B-1 (of 18) HC 2 H 3 O 2 : Weak AcidNaC 2 H 3 O 2 : Salt w/ Weak Conj. Base NaOH: Strong Base Look up the K a of the weak acid and determine its pK a pK a = -log K a = -log (1.8 x 10 -5 )= 4.745

19. Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and 1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added. 2B-2 (of 18) HC 2 H 3 O 2 : Weak AcidNaC 2 H 3 O 2 : Salt w/ Weak Conj. Base NaOH: Strong Base Calculate the moles of all species present x 0.500 L solution= 0.750 mol HC 2 H 3 O 2 1.50 mol HC 2 H 3 O 2 _______________________ L solution 1.00 M NaC 2 H 3 O 2  1.00 M Na + and 1.00 M C 2 H 3 O 2 - x 0.500 L solution= 0.500 mol C 2 H 3 O 2 - 1.00 mol C 2 H 3 O 2 - _______________________ L solution 0.100 mol NaOH  0.100 mol Na + and 0.100 mol OH -

20. Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and 1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added. 2B-3 (of 18) Strong bases react completely with acids HC 2 H 3 O 2 : Weak AcidNaC 2 H 3 O 2 : Salt w/ Weak Conj. Base NaOH: Strong Base the weak acid is reacted away by the strong base the weak conjugate base is produced OH - (aq) + HC 2 H 3 O 2 (aq) → H 2 O (l) + C 2 H 3 O 2 - (aq)

21. Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and 1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added. 2B-4 (of 18) Strong bases react completely with acids HC 2 H 3 O 2 : Weak AcidNaC 2 H 3 O 2 : Salt w/ Weak Conj. Base NaOH: Strong Base OH - (aq) + HC 2 H 3 O 2 (aq) → Initial moles Reacting moles Final moles 0.1000.500 00.600 – 0.100 + 0.100 0.750 0.650 – 0.100 H 2 O (l) + C 2 H 3 O 2 - (aq)

22. Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and 1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added. NaOH: Strong Base HC 2 H 3 O 2 : Weak AcidNaC 2 H 3 O 2 : Salt w/ Weak Conj. Base 2B-5 (of 18) David Hasselholf Equation pH = pK a + log n C 2 H 3 O 2 - _________ n HC 2 H 3 O 2 = 4.745 + log 0.600 mol _____________ 0.650 mol = 4.71

23. Find the pH of 500. mL of a buffer solution that is 1.50 M in acetic acid and 1.00 M in sodium acetate if 0.100 moles of sodium hydroxide are added. NaOH: Strong Base HC 2 H 3 O 2 : Weak AcidNaC 2 H 3 O 2 : Salt w/ Weak Conj. Base 2B-6 (of 18) Amend the David Hasselholf Equation pH = pK a + log n C 2 H 3 O 2 - _________ n HC 2 H 3 O 2 + n OH - __________ – n OH - = 4.745 + log (0.500 mol + 0.100 mol) ______________________________ (0.750 mol – 0.100 mol) = 4.71

24. 200. mL of a solution 0.40 M in ammonia and 0.30 M in ammonium chloride are mixed with 100. mL of 0.45 M hydrochloric acid. Find the pH. 2B-7 (of 18) NH 3 : Weak BaseNH 4 Cl: Salt w/ Weak Conj. Acid HCl: Strong Acid Look up the K a of the weak acid and determine its pK a pK a = -log K a = -log (5.6 x 10 -10 )= 9.252

25. 200. mL of a solution 0.40 M in ammonia and 0.30 M in ammonium chloride are mixed with 100. mL of 0.45 M hydrochloric acid. Find the pH. 2B-8 (of 18) NH 3 : Weak BaseNH 4 Cl: Salt w/ Weak Conj. Acid HCl: Strong Acid Calculate the moles of all species present x 0.200 L solution= 0.080 mol NH 3 0.40 mol NH 3 _________________ L solution x 0.200 L solution= 0.060 mol NH 4 + 0.30 mol NH 4 + __________________ L solution x 0.100 L solution= 0.045 mol H 3 O + 0.45 mol H 3 O + __________________ L solution

26. 200. mL of a solution 0.40 M in ammonia and 0.30 M in ammonium chloride are mixed with 100. mL of 0.45 M hydrochloric acid. Find the pH. 2B-9 (of 18) Strong acids react completely with bases NH 3 : Weak BaseNH 4 Cl: Salt w/ Weak Conj. Acid HCl: Strong Acid

27. 200. mL of a solution 0.40 M in ammonia and 0.30 M in ammonium chloride are mixed with 100. mL of 0.45 M hydrochloric acid. Find the pH. 2B-10 (of 18) NH 3 : Weak BaseNH 4 Cl: Salt w/ Weak Conj. Acid HCl: Strong Acid Amend the David Hasselholf Equation pH = pK a + log n NH 3 ______ n NH 4 + – n H 3 O + ___________ + n H 3 O + = 9.255 + log (0.080 mol – 0.045 mol) _______________________________ (0.060 mol + 0.045 mol) = 8.78

28. A buffer can be made by mixing solutions of: (1)a weak acid and a salt containing its weak conjugate base (2)a weak acid and a lesser amount of a strong base (3)a weak base and a lesser amount of a strong acid 2B-11 (of 18)

29. (1)If 600. mL of 0.50 M nitrous acid and 400. mL of 0.50 M sodium nitrite are mixed, a buffer is produced. Find the pH. 2B-12 (of 18) HNO 2 : Weak AcidNaNO 2 : Salt containing the Conjugate Base  the solution is a buffer pK a = -log K a = -log (4.0 x 10 -4 )= 3.398 Look up the K a of the weak acid and determine its pK a

30. (1)If 600. mL of 0.50 M nitrous acid and 400. mL of 0.50 M sodium nitrite are mixed, a buffer is produced. Find the pH. 2B-13 (of 18) HNO 2 : Weak AcidNaNO 2 : Salt containing the Conjugate Base  the solution is a buffer Calculate the moles of all species present x 0.600 L solution= 0.30 mol HNO 2 0.50 mol HNO 2 ____________________ L solution x 0.400 L solution= 0.20 mol NO 2 - 0.50 mol NO 2 - __________________ L solution

31. (1)If 600. mL of 0.50 M nitrous acid and 400. mL of 0.50 M sodium nitrite are mixed, a buffer is produced. Find the pH. 2B-14 (of 18) HNO 2 : Weak AcidNaNO 2 : Salt containing the Conjugate Base pH = pK a + log n NO 2 - _______ n HNO 2 = 3.398 + log (0.20 mol) _____________ (0.30 mol) = 3.22 Use the David Hasselholf Equation  the solution is a buffer

32. (2)If 600. mL of 0.50 M nitrous acid and 400. mL of 0.50 M sodium hydroxide are mixed, a buffer is produced. Find the pH. 2B-15 (of 18) HNO 2 : Weak AcidNaOH: Strong Base Calculate the moles of all species x 0.600 L solution= 0.30 mol HNO 2 0.50 mol HNO 2 ____________________ L solution x 0.400 L solution= 0.20 mol OH - 0.50 mol OH - _________________ L solution

33. (2)If 600. mL of 0.50 M nitrous acid and 400. mL of 0.50 M sodium hydroxide are mixed, a buffer is produced. Find the pH. 2B-16 (of 18) HNO 2 : Weak AcidNaOH: Strong Base Use the Amended David Hasselholf Equation pH = pK a + log n NO 2 - _________ n HNO 2 + n OH - __________ – n OH - = 3.398 + log (0 mol + 0.20 mol) ___________________________ (0.30 mol – 0.20 mol) = 3.70

34. (3)If 500. mL of 0.60 M sodium nitrite and 250. mL of 0.80 M hydrochloric acid are mixed, a buffer is produced. Find the pH. 2B-17 (of 18) Calculate the moles of all species NaNO 2 : Salt containing a Weak BaseHCl: Strong Acid x 0.500 L solution= 0.30 mol NO 2 - 0.60 mol NO 2 - ___________________ L solution x 0.250 L solution= 0.20 mol H 3 O + 0.80 mol H 3 O + _________________ L solution

35. (3)If 500. mL of 0.60 M sodium nitrite and 250. mL of 0.80 M hydrochloric acid are mixed, a buffer is produced. Find the pH. 2B-18 (of 18) NaNO 2 : Salt containing a Weak BaseHCl: Strong Acid Use the Amended David Hasselholf Equation pH = pK a + log n NO 2 - _________ n HNO 2 – n H 3 O + ___________ + n H 3 O + = 3.398 + log (0.30 mol – 0.20 mol) _____________________________ (0 mol + 0.20 mol) = 3.10

37. ACID-BASE TITRATIONS 2C-1 (of 18) Acid-base reactions with either a strong base or strong acid run to completion Determinations of volumes or molarities are stoichiometric calculations

38. x 1 ____________ 0.0100 L If a 10.0 mL sample of an acetic acid solution is titrated with 24.5 mL of a 0.250 M lithium hydroxide solution, find the molarity of the acetic acid HC 2 H 3 O 2 + LiOH →H(OH) + LiC 2 H 3 O 2 x M 10.0 mL 0.250 M 24.5 mL 1 mol x 1 mol HC 2 H 3 O 2 ___________________ 1 mol LiOH 0.250 mol LiOH x 0.0245 L __________________ L = 0.613 M HC 2 H 3 O 2 2C-2 (of 18) This is a stoichiometric calculation

39. x L _____________________ 0.400 mol NaOH 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (a)Calculate the volume of NaOH solution needed to neutralize the HZ solution. NaOH + HZ →NaZ + H(OH) 0.400 M x mL 1.00 M 20.0 mL 1 mol x 1 mol NaOH ________________ 1 mol HZ 1.00 mol HZ x 0.0200 L __________________ L = 0.0500 L NaOH solution= 50.0 mL NaOH solution 2C-3 (of 18)

40. 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (b)Calculate the pH of the HZ solution before the titration if the K a of HZ is 1.0 x 10 -5. xx HZ (aq) + H 2 O (l) ⇆ H 3 O + (aq) + Z - (aq) Initial M’s Change in M’s Equilibrium M’s 1.00 00 - x + x 1.00 - x K a = [H 3 O + ][Z - ] ____________ [HZ] 1.00 x 10 -5 = x 2 ____________ (1.00 – x) x = 3.162 x 10 -3 M pH = -log (3.162 x 10 -3 ) = 2.50 2C-4 (of 18) The pH of a weak acid is determined from an ICE table for the ionization reaction of the weak acid

41. 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (c)Calculate the pH of the solution after 10.0 mL of the NaOH was added. This solution will contain both a weak acid and its conjugate base Calculate moles of all species 2C-5 (of 18) x 0.0200 L solution= 0.0200 mol HZ 1.00 mol HZ ________________ L solution x 0.0100 L solution= 0.00400 mol OH - 0.400 mol OH - __________________ L solution

42. 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (c)Calculate the pH of the solution after 10.0 mL of the NaOH was added. 2C-6 (of 18) Use the Amended David Hasselholf Equation pH = pK a + log n Z - _________ n HZ + n OH - __________ – n OH - = 5.00 + log (0 mol + 0.00400 mol) ___________________________________ (0.0200 mol – 0.00400 mol) = 4.40

43. 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (d)Calculate the pH of the solution after 25.0 mL of the NaOH was added. = 0.0100 mol OH - 2C-7 (of 18) Calculate moles of all species x 0.0200 L solution= 0.0200 mol HZ 1.00 mol HZ ________________ L solution x 0.0250 L solution= 0.0100 mol NaOH0.400 mol OH - __________________ L solution

44. 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (d)Calculate the pH of the solution after 25.0 mL of the NaOH was added. 2C-8 (of 18) Use the Amended David Hasselholf Equation pH = pK a + log n Z - _________ n HZ + n OH - __________ – n OH - = 5.00 + log (0 mol + 0.0100 mol) _________________________________ (0.0200 mol – 0.0100 mol) = 5.00 At the half equivalence point, pH = pK a

45. 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (e)Calculate the pH of the solution at the equivalence point. (remembering 50.0 mL of base needed) Calculate moles of all species 2C-9 (of 18) = 0.0200 mol OH - x 0.0200 L solution= 0.0200 mol HZ 1.00 mol HZ ________________ L solution x 0.0500 L solution= 0.0200 mol NaOH0.400 mol OH - __________________ L solution

46. 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (e)Calculate the pH of the solution at the equivalence point. At the equivalence point, the solution contains only the conjugate base, Z - 2C-10 (of 18) Use the Amended David Hasselholf Equation pH = pK a + log n Z - _______ n HZ = 5.00 + log (0 mol + 0.0200 mol) _________________________________ (0.0200 mol – 0.0200 mol) = ??? + n OH - __________ – n OH -

47. 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (e)Calculate the pH of the solution at the equivalence point. Calculate the molarity of the Z - in the solution at the equivalence point K b = K w ____ K a = 1.0 x 10 -9 = 1.00 x 10 -14 ______________ 1.0 x 10 -5 2C-11 (of 18) = 0.2857 M mol Z - 0.0200 mol Z - __________________ 0.0700 L Calculate the K b of the weak base Z - The pH of a weak base is determined from an ICE table for the ionization reaction of the weak base Z - (aq) + H 2 O (l) ⇆ HZ (aq) + OH - (aq)

48. Initial M’s Change in M’s Equilibrium M’s 0.2857 x x 00 - x+ x 0.2857 - x 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (e)Calculate the pH of the solution at the equivalence point. K b = [HZ][OH - ] _____________ [Z - ] 1.0 x 10 -9 = x 2 _______________ (0.2857 – x) 1.0 x 10 -9 = x 2 __________ (0.2857) 1.69 x 10 -5 = x 4.77 = pOH = [OH - ] 9.23 = pH 2C-12 (of 18)

49. 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (f) Calculate the pH of the solution after 60.0 mL of the NaOH was added. 2C-13 (of 18) Calculate moles of all species = 0.0240 mol OH - x 0.0200 L solution= 0.0200 mol HZ 1.00 mol HZ ________________ L solution x 0.0600 L solution= 0.0240 mol NaOH0.400 mol OH - __________________ L solution

50. Impossible – the strong base did not all react away There are 0.0040 moles of excess OH - in the solution Use the Amended David Hasselholf Equation pH = pK a + log n Z - ______ n HZ = 5.00 + log (0 mol + 0.0200 mol) _________________________________ (0.0200 mol – 0.0240 mol) = ??? + n OH - __________ – n OH - 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (f) Calculate the pH of the solution after 60.0 mL of the NaOH was added. 2C-14 (of 18)

51. 20.0 mL of a 1.00 M HZ solution is titrated with a 0.400 M NaOH solution. (f) Calculate the pH of the solution after 60.0 mL of the NaOH was added. 2C-15 (of 18) Calculate the molarity of the OH - in the solution = 0.050 M mol OH - 0.0040 mol OH - ____________________ 0.0800 L pOH= 1.30 pH=12.70

52. At the half-neutralization point, half HZ, half Z -  pH = pK a At the start of the titration : pure HZ At the equivalence point : pure Z - A weak acid-strong base titration produces an equivalence point pH > 7 because at the equivalence point the solution contains a weak base (Z - ) Weak Acid-Strong Base Titration (HZ and NaOH) 2C-16 (of 18)

53. INDICATORS Phenolphthalein is a weak organic acid (pK a = 9.75) colorless pink HInd (aq) + H 2 O (l) ⇆ H 3 O + (aq) + Ind - (aq) Color change occurs when [Ind - ] = [Hind] colorless if more HInd, pink if more Ind - 2C-17 (of 18) Phenolphthalein will change color at pH = 9.75  good for titrations in which the equivalence point pH is slightly basic K a = [H 3 O + ], so when K a = [H 3 O + ][Ind - ] _______________ [Hind] pK a = pHor

54. A strong acid-strong base titration produces an equivalence point pH = 7 because at the equivalence point the solution contains a non base (A - ) Strong Acid-Strong Base Titration (HA and NaOH) An indicator with a pK a = 7 should be used for this titration 2C-18 (of 18)