1. Aqueous Equilibria Unit 19 Acid Base Equilibria: Titrations Dr. Jorge L. Alonso Miami-Dade College – Kendall Campus Miami, FL CHM 1046: General Chemistry and Qualitative Analysis Textbook Reference: Chapter 19 (sec. 5-8) Module 9

2. Aqueous Equilibria Titration A volumetric technique in which one can determine the concentration of a solute in a solution of unknown concentration, by making it react with another solution of known concentration (standard). Standard-of known Conc.(M) Moles B = M x V Moles A = M x V {*TitrationMovie} Soln-Unknown Concentration (M): Acid If: Moles A = Moles B Then: (M x V) A = (M x V) B Known Volume (V) Measure Vol to reach end pt.

3. Aqueous Equilibria Determining the Concentration of Solutions by Titration Example: HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) Neutralization: equivalence point # mol 1(acid) = # mol 2(base) (Standard) x a HA (aq) + x b MOH (aq)  MA (aq) + H 2 O (l) 1 mol 1(acid) = 1 mol 2(base) xaxa xbxb H 2 SO 4(aq) + 2NaOH (aq)  Na 2 SO 4(aq) + H 2 O (l) A known concentration of base (or acid) is slowly added to a solution of acid (or base) until neutralization occurs.

4. Aqueous Equilibria Titration Calculations: Stoichiometry using Molarities Neutralization: 1 moles (acid) = 1 moles (base) Since moles = MV = moles x Liter Liter M A x V A M B x V B = x A HN + x B MOH  MN + HOH = coefficients from balanced equations Where == ηAηA x A x B = ηBηB = * Equation Useful for determining Molarities and Volumes at the Equivalence Point of a Titration *

5. Aqueous Equilibria Solution Stoichiometry Problems: Molarity H 2 SO 4 + 2 NaOH  2HOH + Na 2 SO 4 M A x V A = M B x V B 12 Problem: A volume of 16.3 mL of a 0.30M NaOH solution was used to titrate 25.00 mL H 2 SO 4. What is the concentration of H 2 SO 4 in the solution of unknown concentration? = 0.098 M H 2 SO 4 Titration of a Strong Acid with a Strong Base

6. Aqueous Equilibria Titration Graph: pH vs. Volume Excess acid Excess base Acid = Base Methyl Red Indicator Phenolphthalein Indicator {Titration2} SA SB pH meter mL of NaOH pH 0 1.2 10 1.4 20 1.6 30 1.7 40 1.9 50 7.0 60 12.0 70 12.2 80 12.3 Titration Data:

7. Aqueous Equilibria Titrations: The Strength of Acids & Bases Strong Base with Weak Acid Strong Base with Strong Acid Weak Base with Weak Acid Weak Base with Strong Acid SA SB WA SA WB WA WB Phenolphthalein 8-10

8. Aqueous Equilibria With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle. (2) Titration of a WA with a SB

9. Aqueous Equilibria Selecting Appropriate Indicators an indicator is chosen so that it will change color at a pH just beyond the equivalence point (mid point of the steep vertical portion of the graph). The first point at which the indicator permanently changes color marks the end of the titration and is called the indicator end-point. Dropping a perpendicular from the indicator end-point to the x-axis is a very close estimation of the equivalence point. Phenolph Meth Red ????? Select appropriate indicator for following:

10. Aqueous Equilibria SA SB SA SB (1) Titration of a SA with a SB (or SB with a SA)

11. Aqueous Equilibria Acid-Base Neutralization Equations (1) Strong Acids and Bases are represented in completely dissociated state: as H+ and OH- (2) Weak Acids and Bases are represented in undissociated state: as HA and B (or MOH)

12. Aqueous Equilibria (H 2 CO 3 + K + ) (H 2 CO 3 + Ca 2+ + C 2 H 3 O 2 - ) (H 2 CO 3 + Zn 2+ + SO 4 2 - ) (H 2 C O 3 + Zn 2+ )

13. Aqueous Equilibria HA + OH -  H 2 O + A - (2) Acid Base Neutralization: when not at the end-point or using WA or WB (3) WA or WB Equilibrium problems HA ↔ H + + A - Use: Mole ICEnd TableUse: [ICE] TableI0.0038 η0.0030 η0 η C- 0.0030 η + 0.0030 η End0.0008 η0 η 0.0030 η Mole HA + MOH → MA + H 2 0 (.15M)(.025L) (.10M)(.030L)I0.06100 C -x+x E0.061 - xxx HA ↔ H + A - Equations and Tables used in solving A-B Titration Problems (1) Acid Base Neutralization: when you reach the end-point using SA or SB

14. Aqueous Equilibria Example: 25 mL of 0.15M HCl with 0.10M NaOH. (1) Titration of a SA with a SBI0.0038 η0.0030 η0 C- 0.0030 η + 0.0030 η End0.0008 η00.0030 η (1) What volume of NaOH is required to reach the equivalence point? Mole HX + MOH → MX + H 2 0 (2) What is the pH of the initial strong acid? (strong acid problem) (3) What is the pH prior to the equivalence point? Let’s say after 30. mL of NaOH. (excess SA problem) In strong acid [HA]=[H + ], so pH=-log [H + ] (.15M)(.025L) (.10M)(.030L) pH = -log (1.4 x 10 -2 ) = 1.9 Salt of SA & SB: will not Hydrolyze = -log (0.15) = 0.82 *what is used for neutralization Rx?*

15. Aqueous Equilibria Example: 25 mL of 0.15M HCl with 0.1M NaOH. (4) What is the pH at the equivalence point? (5) What is the pH after the equivalence point? Lets say after 40. mL of NaOH. (excess SB problem) I0.0038 η0.0040 η0 η C- 0.0038 η + 0.0038 η End0 η 0.0002 η 0.0038 η mole HX + MOH → MX + H 2 0 (.15M)(.025L) (.10M)(.040L) Salt of SA & SB: will not Hydrolyze Only salt + water present and salt will not hydrolyze water since it is derived from SA & SB. So pH = 7 pOH = -log(3.1x10 -3 ) = 2.5 pH + pOH = pKw pH = pKw - pOH pH = 14 – 2.5 = 11.5 (1) Titration of a SA with a SB *what is used for neutralization Rx?*

16. Aqueous Equilibria The pH >7 at the equivalence point. Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. (2) Titration of a WA with a SB HA ↔ H + + A - SB (OH - ) HA + OH -  H 2 O + A - *what is used for equilibrium Rx?* *what is used for neutralization Rx?*

17. Aqueous Equilibria Example: 25 mL of 0.15M HC 2 H 3 O 2 (K a = 1.8 X10 -5 ) with 0.10M NaOH. (2) What is the initial pH of the acetic acid? (Before titration, WA Equilibrium problem)I0.1500 C-x+x E0.15 - xxx HA ↔ H + A - (1) What volume of NaOH is required to reach the equivalence point? (2) Titration of a WA with a SB

18. Aqueous Equilibria Example: 25 mL of 0.15M HC 2 H 3 O 2 (K a = 1.8 X10 -5 ) with 0.10M NaOH. (3) What is the pH prior to the equivalence point? Let’s say after 30. mL of NaOH. (WA Buffer problem) I0.0038 η0.0030 η0 η C- 0.0030 η + 0.0030 η End0.0008 η0 η 0.0030 η mole HA + MOH → MA + H 2 0 (.15M)(.025L) (.10M)(.030L) Salt of WA & SB: WILL Hydrolyze H 2 O (2) Titration of a WA with a SB

19. Aqueous Equilibria Example: 25 mL of 0.15M HC 2 H 3 O 2 (K a = 1.8 X10 -5 ) with 0.1M NaOH. (4) What is the pH at the equivalence point? This happens @ 37.5 mL (Hydrolysis of Salt derived from a WA & SB) I0.0038 η 0 η C- 0.0038 η + 0.0038 η End0 η 0.0038 η mole HA + MOH → MA + H 2 0 (.15M)(.025L) (.10M)(.038L) Salt of WA & : will Hydrolyze water Salt is NaC 2 H 3 O 2 Na + = derived from SB (NaOH), will not hydrolyze. C 2 H 3 O 2 - = derived from WA (acetic acid) it WILL hydrolyze water. (2) Titration of a WA with a SB

20. Aqueous Equilibria Example: 25 mL of 0.15M HC 2 H 3 O 2 (K a = 1.8 X10 -5 ) with 0.1M NaOH. (4) What is the pH at the equivalence point? This happens @ 37.5 mL (Hydrolysis of Salt derived from a WA & SB) I0.06100 C -x+x E0.061 - xxx C 2 H 3 O 2 - + H 2 O↔ HC 2 H 3 O 2 + OH - [C 2 H 3 O 2 - ] =.I0.0038 η 0 η C- 0.0038 η + 0.0038 η End0 η 0.0038 η mole HA + MOH → MA + H 2 0 (.15M)(.025L) (.10M)(.038L) pH + pOH = pKw pH = pKw - pOH pH = 14 – 5.2 = 8.8 Salt is NaC 2 H 3 O 2 (2) Titration of a WA with a SB

21. Aqueous Equilibria 2005B Q1

22. Aqueous Equilibria Mole ICEnd Table:

23. Aqueous Equilibria The pH at the equivalence point in these titrations is < 7. Methyl red is the indicator of choice. MOH ↔ M + + OH - SA (H 3 O + ) H 3 O + + MOH  2H 2 O + M + (3) Titration of a WB with a SA

24. Aqueous Equilibria Calculation of pH B + H +  H 2 O + A - (1) Acid Base Neutralization(2) WB Equilibrium problem B+ H 2 O ↔ BH + + OH - Use: Mole ICEnd TableUse: [ICE] TableI0.0038 η0.0030 η0 η C- 0.0030 η + 0.0030 η End0.0008 η0 η 0.0030 η Mole B + H+ → A-+ H 2 0 (.15M)(.025L) (.10M)(.030L)I0.06100 C -x+x E0.061 - xxx B + H 2 O ↔ BH + OH - (3) Titration of a WB with a SA: Weak base and strong acid

25. Aqueous Equilibria 2007A Q1 Titration: weak acid with strong base

26. Aqueous Equilibria Use: Mole ICEnd Table Mole HA + OH - → A - + H 2 0 (.40 M)(.025L) (.40M)(.015L)I0.010 η0.0060 η0 η C- 0.0060 η + 0.0060 η End0.0040 η0 η 0.0060 η = 0.15 M F - (e) What’s pH? = 0.10 M HF For F-: For HF:

27. Aqueous Equilibria Use: [ICE] Table HA H + ↔ A - I0.1000.15 C -x+x E0.10 - xx0.15 + x 0.15 M F - 0.10 M HF

28. Aqueous Equilibria Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation. K a1 K a2 K a3

29. Aqueous Equilibria 2005B Q1 Titration: weak acid strong base

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34. Aqueous Equilibria 2000 QA Titration: weak base strong acid

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36. Aqueous Equilibria 2001 Q3 Titration: weak acid strong base

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38. Aqueous Equilibria 2002A Q1 Titration: weak acid strong base

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40. Aqueous Equilibria 2003A Q1Titration: weak base strong acid

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43. Aqueous Equilibria 2006B Q1Titration: weak acid strong base

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46. Aqueous Equilibria Expressing Concentrations of Solutions: Molarity (& Normality*) * For MDC students only!

47. Aqueous Equilibria moles of solute Liters of solution = Molarity (M) = coefficients for the acid (A) and the base (B) from the balanced neutralization equations Where M A x V A = M B x V B (mol/L) A x L A (mol/L B ) x L B mol A B = = x A HN + x B MOH  MN + HOH

48. Aqueous Equilibria M A x V A M B x V B (mol/L) A x L A (mol/L B ) x L B mol A B = = = moles A M B x V B =

49. Aqueous Equilibria For titrations: M B x V B = Since x A HN + x B MOH  MN + HOH

50. Aqueous Equilibria mol of solute L of solution M = Molarity (M) vs. Normality (N) equiv of solute L of solution N = A/B = # H + or #OH - Redox = #e - involved in balanced redox equation. Where: M = N When n = 1 That is when using HCl, KHP NaOH But not when using H 2 SO 4, Ca(OH) 2 Lesson for MDC students only:

51. Aqueous Equilibria Acidg-MM (g/  ) + H 2 0 to HCl36 g + 1L = H 2 SO 4 98 g + 1L = H 3 PO 4 98 g + 1L = Molarity (M) vs. Normality (N) Molarity (  /L) Normality (eq/L) 1M 1N 2N 3N ====== g-EW 36/1 =36 98/2 =49 98/3 =32.7 Eq(g/gEW) 36/36 98/49 98/33

52. Aqueous Equilibria 2M H 3 PO 4 3M Ca(OH) 2 Using Molarity 1N H 3 PO 4 1N Ca(OH) 2 Using Normality 2H 3 PO 4 + 3 Ca(OH) 2  6 HOH + Na 3 PO 4 N = M or M = N N A x V A = N B x V B Using Normality for titrations:

53. Aqueous Equilibria Titration of a WA with a SB The pH at the equivalence point will be >7. Phenolphthalein is commonly used as an indicator in these titrations. Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.

54. Aqueous Equilibria Titration: measuring the Equivalence Point (H + = OH - ) A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base. The end- point of a titration is when indicator changes color. Methyl red in base (range R4-6Y) Phenolphthalein in base (range C 8-10 F)

55. Aqueous Equilibria Titration of a SA with a SB Excess acid Excess base H + = OH - x A HN + x B MOH  MN (aq) + HOH 1. From the start of the titration the pH goes up slowly. Just before the equivalence point, the pH increases rapidly. 2. At the equivalence point, moles H + = moles OH -, and the solution contains only water and the salt from the cation of the base and the anion of the acid. 3. Just after the equivalence point, the pH increases rapidly. As more base is added, the increase in pH again levels off.

56. Aqueous Equilibria Titration Solution of unknown concentration (M 1 x V 1 = #mol 1 ) Neutralization: 1 mol a = 1 mol b (equivalence point) {*TitrationMovie} Solution of known concentration (M 2 x V 2 = #mol 2 ) xaxa xbxb

57. Aqueous Equilibria Stoichiometric/Volumetric Calculations ACID BASE M A x V A M B x V B = xAxA xBxB ACID BASE x A HN + x B MOH  MN + HOH ==

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61. Aqueous Equilibria 2007 (A)

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