Titration Curves (Part II): Weak Acid by Strong Base

1. Titration Curves (Part II):Weak Acid by Strong Base Let’s look at this type of Titration Curve: http://www.kentchemistry.com/links/AcidsBases/flash/titrateWeakAcidStrongBase.swf

2. Several Important differences: --initially most of the weak acid is whole molecules NOT ions --the pH at equivalence point is NOT 7

3. Important Feature--- UNCHANGED: For equal volumes of acid solutions of the SAMEmolarity, the volume of base required to titrate to the equivalence point is independent of whether the acid is strong or weak.

4. Calculating Titration Curve: Weak acid by a Strong Base What is the pH at each of the following points in the titration of 25 ml of 0.1 M H C-2H3O2 by 0.1 M NaOH --Before ANY NaOH added(initial pH) --After 10 ml 0.1 M NaOH added(before) --After 25 ml 0.1 M NaOH added(neutralization) --After 26ml 0.1 M NaOH added(after)

5. Initial pH H C2H3O2 <-----> H+ + C2H3O2 - 0.1M-x x x Ka = 1.8 x 10-5 = x2/ .1-x As2 x = 1.34 x 10-3 M = [H+ ] pH = 2.88

6. After 10 ml of 0.1M NaOH VNaOH MNaOH = 10 ml(0.1 M) = = 1.0 mmol OH - H C2H3O2 + OH -<-----> H2O + C2H3O2- init 2.5 mmol 1.0 mmol react -1.0 mmol -1.0 mmol +1.0 mmol after 1.5 mmol 1.0 mmol

7. Calculation- H C2H3O2 <-----> H+ + C2H3O2 - 1.5 mmol -x X 1.5 mmol +x(0.04 M) 1st calc. MHCl : 1.5 mmol/12.5 +10ml = 0.04 29M 2ndcalc. MNaOH: 1.0 mmol/12.5 +10 ml = 0.0286 M Ka = 1.8 x 10-5 = x (0.0286 M)/ 0.0429 M [H+]=x=2.7 x 10-5 pH = 4.58

8. Want to try Henderson -Hasselbach? pH = pKa + Log [C2H3O2 -] / [H C2H3O2 ] pH = 4.76 + Log (0.0286 M / 0.0429 M) pH = 4.76 + (-.17) pH = 4.58

9. After 25 ml NaOH added(Neutralization) VNaOH MNaOH = 25 ml(0.1 M) = = 2.5 mmol OH - H C2H3O2 + OH -<-----> H2O + C2H3O2- init 2.5 mmol 2.5 mmol react -2.5mmol -2.5mmol +2.5 mmol after 0 mmol 0 mmol 2.5 mmol

10. Calculations: ONLY C2H3O2- ION LEFT Acetate ion hydrolyzes: C2H3O2- + H2O<-----> H C2H3O2 + OH - 2.5 mmol-x x x Kh or b = Kw/Ka= 5.7 x 10-10 5.7 x 10-10 = x2 / 2.5 mmol / 50 ml(0.05 M) [OH - ] = X = 5.7 x 10-6 pOH = 5.28 pH = 8.75

11. After 26 ml 0.1 M NaOH VNaOH MNaOH = 26 ml(0.1 M) = = 2.6 mmol OH - H C2H3O2 + OH -<-----> H2O + C2H3O2- init 2.5 mmol 2.6 mmol react -2.5mmol -2.5mmol +2.5 mmol after 0 mmol 0.1 mmol 2.5 mmol

12. Calculation That 0.1 mmol of Strong base controls the pH [OH - ] = 0.1 mmol / 25 ml + 26 ml [OH - ] = 1.96 x 10-3 M pOH = 2.71 pH = 11.29 WHEW !!

13. Principal Features: Weak Acid:Strong Base Titration Curve 1. Initial pH higher than in strong acid titration. 2. Sharp increase at start of titration 3. Long gradual change in pH before equivalence point. 4. At 1/2 neutralization point, pH = pKa 5. pH at equivalence point > 7.

14. 6. Beyond equivalence point, titration curve identical to strong acid:strong base titration curve. 7. Steep portion at equivalence point- short.

15. Practice Problem: WA/WB Starting with 15 ml of 0.1 M H C2H3O2 What is pH ? After 5 ml 0.1 M NaOH added initially-- After 5 ml 0.1 M NaOH added After 10 ml 0.1 M NaOH added After 15 ml 0.1 M NaOH added After 25 ml 0.1 M NaOH added

16. Summary of Titration Calculations: Titration of Strong Acid by a Strong Base: Molarity of STRONG Acid= M1 Volume of STRONG Acid= V1 Molarity of STRONG Base= M2 Volume of STRONG Base= V2

17. Before Equivalence Point: [H +] = M1V1 - M2V2 / V1 + V2 At Equivalence Point pH = 7

18. After Equivalence Point: [OH - ] = M2V2 - M1V1 / V1 + V2

19. Titration of Weak Acid by a Strong Base: Before Equivalence Point: [H +] = Ka X (M1V 1- M2V2 / V1 + V2) At Equivalence Point: [OH - ] = ( Kw / Ki) X (M1V1 / V1 + V2)

20. After equivalence Point: [OH - ] = M2V2 - M1V1 / V1 + V2

21. Acid -Base Indicators

22. An acid-base indicator is a weak acid for which the non-ionized species( HIn) has one colorand the conjugate base, the anion(In -) has a different color.

23. There are LOTS of Indicators

24. It’s equilibrium looks like HIn + H2O <---> H3O + + In - color(1)color(2) We can write an equation relating pH, pKa and concentrations of indicator molecules and anions in exactly the same way as we formulated buffer equations, i.e. the Henderson-Hasselbach equation.

25. Henderson-Hasselbach for Indicators: pH = pK a + log [In - ] / [HIn] In general- if 90% or more of the indicator is in the form of HIn, the solution would be one color(1). In general- if 90% or more of the indicator is in the form of In -, the solution would be one color(2).

26. The complete change in color takes place over 2 pH units. Acid Color[In - ] / [HIn]<0.1 pH < pKa -1 Intermediate Color[In - ] / [HIn] ≈ 1 pH = pKa Base Color[In - ] / [HIn] > 10 pH > pKa +1

27. For Bromthymol blue, pKa = 7.1 Acid Color [In - ] / [HIn]<0.1 pH < 6.1(YELLOW) Intermediate Color [In - ] / [HIn] ≈ 1 pH =7.1(GREEN) Base Color [In - ] / [HIn] > 10 pH > 8.1(BLUE)

28. What indicator should we use for a titration of 50 ml of HC2H3O2 with 0.1 M NaOH? What is the pH at the equivalence point? Do you remember how to find the equivalence point of a weak acid/strong base titration?

29. At equivalence pt. 50 ml NaOH and 50 ml HC2H3O2 VNaOH MNaOH = 50 ml(0.1 M) = = 5.0 mmol OH - H C2H3O2 + OH -<-----> H2O + C2H3O2- init 5.0 mmol 5.0 mmol react -5.0mmol -5.0mmol +5.0 mmol after 0 mmol 0 mmol 5.0 mmol

30. Calculations: ONLY C2H3O2- ION LEFT Acetate ion hydrolyzes: C2H3O2- + H2O<-----> H C2H3O2 + OH - 5.0 mmol-x x x Kh or b = Kw/Ki = 5.7 x 10-10 5.7 x 10-10 = x2 / 5.0 mmol / 100 ml(0.05 M) [OH - ] = X = 5.7 x 10-6 pOH = 5.28 pH = 8.75

31. Now- We need an indicator which changes near pH 8.72. At this point we need a Table of indicators.

32. pH and Color Changes for some common Acid-Base Indicators: 14 12 10 8 6 4 2 pH Scale alizarin yellow bromthymol blue methyl violet Phenolphthalein methyl orange

33. At a pH of 8.72, Phenolphthalein would be a good choice.