1. Acid/Base Titrations

2. Titrations Titration Curve – always calculate equivalent point first Strong Acid/Strong Base Regions that require “different” calculations B/F any base is added Half-way point region At the equivalence point After the equivalence point

3. Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr First -find Volume at equivalence M 1 V 1 = M 2 V 2 (0.050 L)(0.02000M) = 0.1000 V V = 10.0 mL

4. Strong Acid/Strong Base 50.00 mL of 0.02000 M KOH Titrated with 0.1000 M HBr Second – find initial pH pH = - logA H ~ -log [H + ] pOH = -logA OH ~ -log [OH - ] pH = 12.30

5. Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr Third– find pH at mid-way volume KOH (aq) + HBr (aq) -> H 2 O (l) + KBr(aq) Before After 0.001000 mol0.0006000 mol 0.000400 mol0 mol Limiting Reactant 0.0006000 mol pH = 11.8 (~6 ml)

6. Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr Fourth – find pH at equivalence point KOH (aq) + HBr (aq) -> H 2 O (l) + KBr(aq) Before After 0.001000 mol0.0010000 mol 0 mol 0.0010000 mol pH = 7.0

7. Strong Acid/Strong Base 50 mL of 0.02000 M KOH Titrated with 0.1000 M HBr Finally – find pH after equivalence point KOH (aq) + HBr (aq) -> H 2 O (l) + KBr(aq) Before After 0.001000 mol0.001200 mol 0 mol0.0002000 mol 0.0010000 mol pH = 2.5 12 ml Limiting Reactant

10. Titration of WEAK acid with a strong base

12. Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid K a = 1.8 x 10 -5 Titrant = 0.100 M NaOH First, calculate the volume at the equivalence-point M 1 V 1 = M 2 V 2 (0.0250 L) 0.1000 M = 0.1000 M (V 2 ) V 2 = 0.0250 L or 25.0 mL

13. Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid K a = 1.8 x 10 -5 Titrant = 0.100 M NaOH Second, Calculate the initial pH of the acetic acid solution

14. Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid K a = 1.8 x 10 -5 Titrant = 0.100 M NaOH Third, Calculate the pH at some intermediate volume

15. Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid K a = 1.8 x 10 -5 Titrant = 0.100 M NaOH Fourth, Calculate the pH at equivalence

16. Titration of a weak acid solution with a strong base. 25.0 mL of 0.1000M acetic acid K a = 1.8 x 10 -5 Titrant = 0.100 M NaOH Finally calculate the pH after the addition 26.0 mL of NaOH

18. Initial pH Buffer Region Equivalence point M 1 V 1 =M 2 V 2 pH @ equivalence? pH after equivalence?

19. Initial pH Buffer Region Equivalence point M 1 V 1 =M 2 V 2 pH @ equivalence pH after equivalence?

20. Initial pH Buffer Region Equivalence point M 1 V 1 =M 2 V 2 pH @ equivalence pH after equivalence Dominated by remaining [OH - ]

22. Weak Base titrated with strong acid Consider a 100 ml of a 0.0100 M base with 0.0500 M HCl K b = 1 x 10 -5

23. Initial pH Buffer Region pH @ equivalence pH after equivalence Dominated by remaining [H + ]