 Download Download Presentation Unit 19 Acid Base Equilibria: Titrations

# Unit 19 Acid Base Equilibria: Titrations

Download Presentation ## Unit 19 Acid Base Equilibria: Titrations

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

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

3. Determining the Concentration of Solutions by Titration A known concentration of base (or acid) is slowly added to a solution of acid (or base) until neutralization occurs. (Standard) Example: HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l) xaHA (aq) + xbMOH (aq) MA (aq) + H2O (l) H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + H2O(l) Neutralization: equivalence point # mol1(acid) = # mol2(base) 1 mol1(acid) = 1 mol2(base) xa xb

4. Titration Calculations: Stoichiometry using Molarities xA HN + xB MOH  MN + HOH Neutralization: 1 moles(acid)= 1 moles(base) xAxB Since moles = MV = moles x Liter Liter ηA ηB MA x VAMB x VB = = = = = = coefficients from balanced equations Where * Equation Useful for determining Molarities and Volumesat the Equivalence Point of a Titration *

5. Solution Stoichiometry Problems: Molarity Problem: A volume of 16.3 mL of a 0.30M NaOH solution was used to titrate 25.00 mL H2SO4. What is the concentration of H2SO4 in the solution of unknown concentration? H2SO4 + 2 NaOH 2HOH + Na2SO4 MA x VA = MB x VB 1 2 = 0.098 M H2SO4 Titration of a Strong Acid with a Strong Base

6. Titration Graph: pH vs. Volume Titration Data: Excess base SB Phenolphthalein Indicator Acid = Base Methyl Red Indicator pH meter Excess acid SA {Titration2}

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

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

9. Selecting Appropriate Indicators Select appropriate indicator for following: Phenolph Meth Red Meth Red Phenolph ????? • 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.

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

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

12. (H2CO3 + K+) (H2CO3 + Ca2+ + C2H3O2- ) (H2C O3 + Zn2+) (H2CO3 + Zn2+ + SO42-)

13. Equations and Tables used in solving A-B Titration Problems (1) Acid Base Neutralization: when you reach the end-pointusing SA or SB (2) Acid Base Neutralization: when not at the end-point or using WA or WB (3) WA or WB Equilibrium problems HA ↔ H+ + A- HA + OH- H2O + A- Use: MoleICEnd Table Use: [ICE] Table (.15M)(.025L) (.10M)(.030L)

14. (1) Titration of a SA with a SB (1) What volume of NaOH is required to reach the equivalence point? Example: 25 mL of 0.15MHCl with 0.10MNaOH. (2) What is the pH of the initial strong acid? (strong acid problem) = -log (0.15) = 0.82 In strong acid [HA]=[H+], so pH=-log [H+] (3) What is the pH prior to the equivalence point? Let’s say after 30. mL of NaOH. (excess SA problem) *what is used for neutralization Rx?* (.15M)(.025L) (.10M)(.030L) pH = -log (1.4 x 10-2) = 1.9 Salt of SA & SB: will not Hydrolyze

15. (1) Titration of a SA with a SB (4) What is the pH at the equivalence point? Only salt + water present and salt will not hydrolyze water since it is derived from SA & SB. So pH = 7 Example: 25 mL of 0.15MHCl with 0.1MNaOH. (5) What is the pH after the equivalence point? Lets say after 40. mL of NaOH. (excess SB problem) *what is used for neutralization Rx?* (.15M)(.025L) (.10M)(.040L) Salt of SA & SB: will not Hydrolyze pOH = -log(3.1x10-3) = 2.5 pH + pOH = pKw pH = pKw - pOH pH = 14 – 2.5 = 11.5

16. (2) Titration of a WA with a SB SB (OH-) HA ↔ H+ + A- HA + OH- H2O + A- *what is used for neutralization Rx?* • Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. • The pH >7 at the equivalence point. *what is used for equilibrium Rx?*

17. (2) Titration of a WAwith a SB (1) What volume of NaOH is required to reach the equivalence point? Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.10M NaOH. (2) What is the initial pH of the acetic acid? (Before titration, WA Equilibrium problem)

18. (2) Titration of a WA with a SB (3) What is the pH prior to the equivalence point? Let’s say after 30. mL of NaOH. (WA Buffer problem) Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.10M NaOH. (.15M)(.025L) (.10M)(.030L) Salt of WA & SB: WILL Hydrolyze H2O

19. (2) Titration of a WA with a SB Example: 25 mL of 0.15M HC2H3O2 (Ka= 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) (.15M)(.025L) (.10M)(.038L) Salt of WA & : will Hydrolyze water Salt is NaC2H3O2 Na+ = derived from SB (NaOH), will not hydrolyze. C2H3O2- = derived from WA (acetic acid) it WILL hydrolyze water.

20. (2) Titration of a WA with a SB (4) What is the pH at the equivalence point? This happens @ 37.5 mL (Hydrolysis of Salt derived from a WA & SB) (.15M)(.025L) (.10M)(.038L) [C2H3O2-] =. Salt is NaC2H3O2 pH + pOH = pKw pH = pKw - pOH Example: 25 mL of 0.15M HC2H3O2 (Ka= 1.8 X10-5) with 0.1M NaOH. pH = 14 – 5.2 = 8.8

21. 2005B Q1

22. MoleICEnd Table:

23. (3) Titration of a WB with a SA SA (H3O+) • The pH at the equivalence point in these titrations is < 7. • Methyl red is the indicator of choice. MOH↔ M+ + OH- H3O+ + MOH  2H2O + M+

24. (3) Titration of a WB with a SA: Calculation of pH (1) Acid Base Neutralization (2) WB Equilibrium problem B+ H2O ↔ BH+ + OH- B + H+ H2O + A- Weak base and strong acid Use: MoleICEnd Table Use: [ICE] Table (.15M)(.025L) (.10M)(.030L)

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

26. Use: MoleICEnd Table (.40 M)(.025L) (.40M)(.015L) (e) What’s pH? For F-: = 0.15 M F- For HF: = 0.10 M HF

27. 0.15 M F- 0.10 M HF Use: [ICE] Table

28. Titrations of Polyprotic Acids Ka3 In these cases there is an equivalence point for each dissociation. Ka2 Ka1

29. Titration: weak acid strong base 2005B Q1

30. 2005B Q1

32. Titration: weak acid strong base 2002A Q1

33. 2003A Q1 Titration: weak base strong acid

34. 2006B Q1 Titration: weak acid strong base

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

36. molesof solute Litersof solution = Molarity (M) xA HN + xB MOH  MN + HOH mol mol A = B (mol/L)A x LA (mol/LB) x LB = MA x VA = MB x VB = coefficients for the acid (A) and the base (B) from the balanced neutralization equations Where

37. mol mol A = B (mol/L)A x LA (mol/LB) x LB = MA x VAMB x VB = molesAMB x VB =

38. xA HN + xB MOH  MN + HOH For titrations: Since MB x VB =

39. Lesson for MDC students only: equiv of solute L of solution mol of solute L of solution M = N = Molarity (M) vs. Normality (N) Where: M = N When n = 1 That is when using HCl, KHP NaOH But not when using H2SO4, Ca(OH)2 A/B = # H+ or #OH- Redox= #e- involved in balanced redox equation.