1. 1 6-4-2011

2. 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. 3 4) End Point can be determined by either an indicator or by use of a pH meter. 5) Can easily determine K a since pH = pK a half 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. 4 Overview: Titrations

5. 5 What will we study? Three types of titrations: 1.Strong acid with strong base or vice versa 2.Weak acid with strong base 3.Weak base with strong acid One of the reactants must be strong. You can not titrate weak acids with weak bases

6. 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. 7 Assume we have 50 mL of 0.1 M HNO 3 to which 10, 30, 50, 70 and 90 mL of 0.1 M NaOH is added: HNO 3 + NaOH  NaCl + H 2 O 0.1*10 0.1*50 10 mL 15Initial 04Final [H + ] = 4/60 M, pH = 1.18 04/60 MConc 0.1*300.1*5030 mL 35Initial 02Final [H + ] = 2/80 M, pH = 1.60 2/80 MConc

8. 8 Assume we have 50 mL of 0.1M HNO 3 to which 10, 30, 50, and 70 mL of NaOH is added: HNO 3 + NaOH  NaCl + H 2 O 0.1*50 50 mL 55Initial 00Final Equivalence point, pH = 7Conc 0.1*700.1*5070 mL 75Initial 20Final [OH - ] = 2/120 M, pOH = 1.78, pH = 12.22 2/120 M0Conc

9. 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 -  H 2 O 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. 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. 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. 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. 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 H 2 O using K b.  After End Point: Calculate Excess NaOH; then find pH directly from excess OH -.

14. 14 If one reactant is weak, two steps are needed: 1.Quantitative step: Find conc. Of remaining acid, the formed conjugate base, and OH - after reaction of weak acid with strong base. 2.Equilibrium step: use the appropriate equilibrium constant expression to find [H + ] or [OH - ] and then calculate pH.

15. 15 Assume we have 50 mL of 0.1M HA (weak) to which 10, 30, 50, 70 and 90 mL of NaOH is added: HA + OH -  A - + H 2 O 00.1*10 0.1*50 10 mL 015Initial 104Final Buffer1/6004/60 MConc 00.1*300.1*5030 mL 035Initial 302Final Buffer3/8002/80 MConc

16. 16 Assume we have 50 mL of 0.1M HA (weak) to which 10, 30, 50, 70 and 90 mL of NaOH is added: HA + OH -  A - + H 2 O 00.1*50 50 mL 055Initial 500Final Salt, use K b 5/100 M00Conc 00.1*700.1*5070 mL 075Initial 520Final pOH from base, OH - 5/120 M2/120 M0Conc

17. 17 Weak Acid-Strong Base Titrations Find the pH of a 25 mL solution of 0.10 M HOAc (K a = 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  H + + OAc - Initial 0.10 0 0 Change -x +x +x Equilibrium 0.10 – x x x

18. 18 K a = x 2 /(0.10 – x), assume 0.10>>x since K a is very small. 1.8*10 -5 = x 2 /0.1 X = 1.34*10 -3 RE = (1.34*10 -3 /0.1)*100 = 1.3%, OK X = [H + ] = 1.34*10 -3 M pH = 2.87

19. 19 After addition of 5.0 mL NaOH HOAc + OH -  OAC - + H 2 O 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  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. 20 K a = x(0.5/30 +x)/(2.0/30 – x), assume 0.5/30>>x since K a 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. 21 After addition of 15.0 mL NaOH HOAc + OH -  OAC - + H 2 O 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  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. 22 K a = x(1.5/40 +x)/(1.0/40 – x), assume 1.0/40>>x since K a 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. 23 After addition of 25.0 mL NaOH HOAc + OH -  OAC - + H 2 O 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 H 2 O + OAc  OH - + HOAc Initial 0.05 0 0 Change -x +x +x Equil 0.05 – x x x

24. 24 K b = x 2 /(0.05 – x), assume 0.05>>x since K b is very small. (10 -14 /1.8*10 -5 ) = x 2 /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. 25 After addition of 30.0 mL NaOH HOAc + OH -  OAC - + H 2 O Initial2.5 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 K b is very small. [OH - ] = 9.1*10 -3 M, and pOH = 2.04 pH = 14 – 2.04 = 11.96

26. 26 After addition of 40.0 mL NaOH HOAc + OH -  OAC - + H 2 O Initial2.5 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 K b is very small. [OH - ] = 2.3*10 -2 M, and pOH = 1.64 pH = 14 – 1.64 = 12.36

27. 27 Summary of Weak Acid-Strong Base Titrations CH 3 COOH (aq)  CH 3 COO - (aq) + H + (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) At equivalence point (pH > 7):