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6-4-2011

6-4-2011. Titrations. Definition: Volumetric determination of the amount of an acid or base by addition of a standard acid or base until neutralization has occurred. 2) A burette (buret) is usually used to add the titrant & can be read to ±0.01 mL

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6-4-2011

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  1. 6-4-2011

  2. Titrations Definition: Volumetric determination of the amount of an acid or base by addition of a standard acid or base until neutralization has occurred. 2) A burette (buret) is usually used to add the titrant & can be read to ±0.01 mL 3) a plot of pH vs volume of titrant is called a pH curve or a titration curve.

  3. 4) End Point can be determined by either an indicator or by use of a pH meter. 5) Can easily determine Ka since pH = pKahalf way to the end point for a weak acid, at this point [HA]=[A-], or mmol HA = mmol A-. 6) A quick examination of the pH curve will tell if titrating an acid or base as well as if they are strong or weak.

  4. Overview: Titrations

  5. What will we study? Three types of titrations: • Strong acid with strong base or vice versa • Weak acid with strong base • Weak base with strong acid One of the reactants must be strong. You can not titrate weak acids with weak bases

  6. pH Determination while Titrating a Strong Acid like HCl with a strong base • Before Titration Starts: calculate pH from M of HCl • During Titration & Before End Point: Calculate HCl remaining; then find pH from remaining HCl. • At End Point: Only NaCl left in water; pH = 7 • After End Point: Calculate Excess NaOH; Find pH from excess NaOH.

  7. Assume we have 50 mL of 0.1 M HNO3 to which 10, 30, 50, 70 and 90 mL of 0.1 M NaOH is added:

  8. Assume we have 50 mL of 0.1M HNO3 to which 10, 30, 50, and 70 mL of NaOH is added:

  9. Find the pH of a 25.0 mL solution of 0.100 M HCl after addition of 0, 10, 20, 25, 35, and 45 mL of 0.100 M NaOH. After addition of 0 mL NaOH We have HCl solution only: [H+] = 0.100 M since HCl is a strong acid pH = 1.00 After addition of 10 mL NaOH H+ + OH-g H2O mmol H+ remaining = initial mmol H+ - mmol NaOH added mmol H+ remaining = 0.100*25.0 – 0.100*10 = 1.50 [H+] = {mmol/mL} = 1.5/35 = 0.043 and pH = 1.37

  10. After addition of 20 mL NaOH mmol H+ remaining = initial mmol H+ - mmol NaOH added mmol H+ remaining = 0.100*25.0 – 0.100*20 = 0.50 [H+] = {mmol/mL} = 0.50/45 = 0.011 and pH = 1.95 After addition of 25 mL NaOH mmol H+ remaining = initial mmol H+ - mmol NaOH added mmol H+ remaining = 0.100*25.0 – 0.100*25 = 0 This is the equivalence point [H+] = [OH-] = 10-7 M pH = 7

  11. After addition of 35 mL NaOH mmol OH- excess = mmol NaOH added - mmol H+ mmol OH- excess = = 0.100*35.0 – 0.100*25 = 1.0 [OH-] = {mmol/mL} = 1.0/60 = 0.017 pOH = 1.78, and pH = 14 – 1.78 = 12.22 After addition of 45 mL NaOH mmol OH- excess = mmol NaOH added - mmol H+ mmol OH- excess = = 0.100*45.0 – 0.100*25 = 2.0 [OH-] = {mmol/mL} = 2.0/70 = 0.029 pOH = 1.54, and pH = 14 – 1.54 = 12.46

  12. 25.0 mL of 0.100 M HCl with 0.100 M NaOH (Strong Acid with a Strong Base). Note: Large change in pH near end point and pH at equivalence point = 7.00

  13. pH Determination while Titrating a Weak Acid like HA with a strong base • Before Titration Starts: Initial pH from M of weak acid (HA) • During Titration & Before End Point: Find remaining HA and A- formed; Calculate pH for the buffer containing HA & A-. • At End Point: Only have A-. Calculate pH from reaction of A- with H2O using Kb. • After End Point: Calculate Excess NaOH; then find pH directly from excess OH- .

  14. If one reactant is weak, two steps are needed: • Quantitative step: Find conc. Of remaining acid, the formed conjugate base, and OH- after reaction of weak acid with strong base. • Equilibrium step: use the appropriate equilibrium constant expression to find [H+] or [OH-] and then calculate pH.

  15. Assume we have 50 mL of 0.1M HA (weak) to which 10, 30, 50, 70 and 90 mL of NaOH is added:

  16. Assume we have 50 mL of 0.1M HA (weak) to which 10, 30, 50, 70 and 90 mL of NaOH is added:

  17. Weak Acid-Strong Base Titrations Find the pH of a 25 mL solution of 0.10 M HOAc (Ka = 1.8*20-5), after addition of 0, 5, 15, 25, 30, and 40 mL of 0.10 M NaOH. After addition of 0.0 mL NaOH We have HOAc only, in solution. This is a weak acid which you have studied how to calculate its pH HOAc D H+ + OAc- Initial 0.10 0 0 Change -x +x +x Equilibrium 0.10 – x x x

  18. Ka = x2/(0.10 – x), assume 0.10>>x since Ka is very small. 1.8*10-5 = x2/0.1 X = 1.34*10-3 RE = (1.34*10-3 /0.1)*100 = 1.3%, OK X = [H+] = 1.34*10-3M pH = 2.87

  19. After addition of 5.0 mL NaOH HOAc + OH-g OAC- + H2O mmol HOAc left = initial mmol HOAc – mmol NaoH mmol HOAc left = 0.10*25 – 0.10*5 = 2.0 mmol OAc- formed = 0.5 [HOAc] = 2.0/30 M, and [OAC-] = 0.5/30 M (This is a buffer solution) HOAc D H+ + OAc- Initial 2.0/30 0 0.5/30 Change -x +x +x Equil 2.0/30 – x x 0.5/30 + x

  20. Ka = x(0.5/30 +x)/(2.0/30 – x), assume 0.5/30>>x since Ka is very small (buffer solution). 1.8*10-5 = (0.5/30*x)/(2.0/30) X = 7.2*10-5 Since it is a buffer, no need to calculate RE X = [H+] = 7.2*10-5 M pH = 4.14

  21. After addition of 15.0 mL NaOH HOAc + OH-g OAC- + H2O mmol HOAc left = initial mmol HOAc – mmol NaoH mmol HOAc left = 0.10*25 – 0.10*15 = 1.0 mmol OAc- formed = 1.5 [HOAc] = 1.0/30 M, and [OAC-] = 1.5/30 M (This is a buffer solution) HOAc D H+ + OAc- Initial 1.0/40 0 1.5/40 Change -x +x +x Equil 1.0/40 – x x 1.5/40 + x

  22. Ka = x(1.5/40 +x)/(1.0/40 – x), assume 1.0/40>>x since Ka is very small. 1.8*10-5 = (1.5/40*x)/(1.0/40) X = 1.2*10-5 Since it is a buffer, no need to calculate RE X = [H+] = 1.2*10-5 M pH = 4.92

  23. After addition of 25.0 mL NaOH HOAc + OH-g OAC- + H2O mmol HOAc left = initial mmol HOAc – mmol NaoH mmol HOAc left = 0.10*25 – 0.10*25 = 0.0 This is the equivalence point (all HOAc is consumed) mmol OAc- formed = 2.5 [OAC-] = 2.5/50 = 0.05 M H2O + OAc D OH- + HOAc Initial 0.05 0 0 Change -x +x +x Equil 0.05 – x x x

  24. Kb = x2/(0.05 – x), assume 0.05>>x since Kb is very small. (10-14/1.8*10-5) = x2/0.05 X = 5.3*10-6 RE = (5.3*10-6 /0.05)*100 = 0.01%, OK X = [OH-] = 5.3*10-6 M pOH = 5.28 pH = 14 – pOH = 14 – 5.28 = 8.72 Note that the pH at equivalence point is basic in the titration of weak acids.

  25. After addition of 30.0 mL NaOH HOAc + OH-g OAC- + H2O Initial 2.5 3.0 0 Final 0 0.5 2.5 mmol OH- excess = mmol NaoH added - mmol HOAc Now the solution has [OAc-] = 2.5/55 M and [OH-] = 0.5/55 M.The OH- from the base is much greater than that coming from OAc- since Kb is very small. [OH-] = 9.1*10-3 M, and pOH = 2.04 pH = 14 – 2.04 = 11.96

  26. After addition of 40.0 mL NaOH HOAc + OH-g OAC- + H2O Initial 2.5 4.0 0 Final 0 1.5 2.5 mmol OH- excess = mmol NaoH added - mmol HOAc Now the solution has [OAc-] = 2.5/65 M and [OH-] = 1.5/65 M.The OH- from the base is much greater than that coming from OAc- since Kb is very small. [OH-] = 2.3*10-2 M, and pOH = 1.64 pH = 14 – 1.64 = 12.36

  27. CH3COOH (aq) + OH-(aq) CH3COO-(aq) + H2O (l) CH3COO-(aq) + H2O (l) OH-(aq) + CH3COOH (aq) Summary of Weak Acid-Strong Base Titrations CH3COOH (aq)g CH3COO-(aq) + H+(l) At equivalence point (pH > 7):

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