1. CHE 1316 Laboratory Measurements & Techniques LECTURE 3 Acids and Bases Titration Curves Strong Acid/Strong Base Titrations Acid/Base Indicators www.baylor.edu/chemistry

2. Acids and Bases Acid = Proton Donor (HA) Base = Proton Acceptor (BOH or B) HA + BOH → A − + B + + H 2 O HA + B → A − + BH + (HCl + NaOH → Cl − + Na + + H 2 O) (HCl + NH 3 → Cl − + NH 4 + ) Section 4.2 & Chapter 15, General Chemistry HA + H 2 OA − + H 3 O + B + H 2 OBH + + OH − (hydronium) (hydroxide) Water: can act as acid or base (amphiprotic)

3. Conjugate Acids and Bases acid 1 + base 2 base 1 + acid 2 H 2 O(l) + NH 3 (aq)OH  (aq) + NH 4 + (aq) CH 3 CO 2 H(aq) + H 2 O(l)CH 3 CO 2  (aq) + H 3 O + (aq) HCl(aq) + H 2 O(l)Cl  (aq) + H 3 O + (aq) H 2 O(l) + H 2 O(l)OH  (aq) + H 3 O + (aq) Acid-base reactions can typically be characterized as two bases competing for a proton.

4. Autoprotolysis and pH In pure water: [H 3 O + ] = 1.0 × 10 −7 M = [OH − ] pH = −log[H 3 O + ] = −log(1.0 × 10 −7 ) = 7.00 (neutral) If acid is added to water: e.g., [H 3 O + ] increases to 1.00 x 10 −3 M. pH = −log(1.00 × 10 −3 ) = 3.00 (acidic) In general: Acidic solutions have [H 3 O + ] > 10 −7 M, so pH < 7 Basic solutions have [H 3 O + ] 7 2 H 2 OH 3 O + + OH −

5. Strong Acids and Bases Strong Acids: dissociate completely into H 3 O + and A −. [H 3 O + ] = C HA ; e.g., 0.1 M HA  [H 3 O + ] = 0.1 M Strong Bases: dissociate completely into M n+ and n OH −. [OH − ] = C MOH × n; e.g., 0.1 M NaOH  [OH − ] = 0.1 M HA + H 2 O A − + H 3 O + M(OH) n M n+ + n OH − MEMORIZE!!

6. Weak Acids and Bases most acids/bases not listed in Table 4.2 Weak Acids: e.g., acetic acid, formic acid, etc. [H 3 O + ] << C HA ; 0.1 M HA  [H 3 O + ] << 0.1 M Weak Bases: e.g,. ammonia, pyridine (typically contain N) [OH − ] << C B ; 0.01 M NH 3  [OH − ] << 0.01 M HA + H 2 O A − + H 3 O + NH 3 + H 2 O NH 4 + + OH −

7. pH and pOH Determine pH and pOH for 0.020 M HCl: K w = [H 3 O + ][OH − ] 1.0 × 10 −14 M 2 = (0.020 M)[OH − ] [OH − ] = 5.0 x 10 −13 M pH = −log[H 3 O + ] = −log(0.020) = 1.70 pOH = −log[OH − ] = −log(5.0 x 10 −13 ) = 12.30 2 H 2 OH 3 O + + OH − K w = 1.0 × 10 −14 M 2 = [H 3 O + ][OH − ]

8. Converting pH and pOH pK w = pH + pOH = 14.00 Determine pH and pOH for 0.050 M NaOH: pOH = −log[OH−] = −log(0.050) = 1.30 pH + pOH = 14.00 pH = 14.00 − pOH = 14.00 − 1.30 pH = 12.70 General Chemistry: Sections 15.1-15.3

9. Strong Acid/Base Titrations Example: Titration of 50.0 mL of 0.100 M HCl (analyte) with 0.100 M NaOH (titrant). Initial soln:0.100 M HCl = 0.100 M H 3 O + ; pH = 1.00 Analytical Reaction: HCl + NaOH → NaCl + H 2 O Each mmol of NaOH added reacts with 1 mmol of HCl. Initial mmol HCl: C HCl × V HCl = 0.100 M × 50.0 mL = 5.00 mmol HCl Addition of 10.0 mL NaOH: C NaOH × V NaOH = 0.100 M × 10.0 mL = 1.00 mmol NaOH

10. Strong Acid/Base Titrations Equivalence Point 50.0 mL NaOH added 1.00 mmol NaOH added per 10.0 mL

11. Strong Acid/Base Titrations Molarity = mmol HCl or NaOH total volume in mL total volume = 50.0 mL + V NaOH

12. Strong Acid/Base Titrations Eqiv. Pt. pH 7.00

13. Indicator Function & Selection Choose indicator so that pK In  equivalence point pH pK In General Chemistry: Section 15.9,15.10

14. Suggested Problems Problem Set 1: p. 114 of lab manual: 1, 3, 4c-g, 5-7, 10-12, 15-17, 19, 21, 23, 25, 26, 28, 30, 32 Problem Set 3. p. 126 of lab manual: 5, 8, 17-19 Additional Problems from CHE 1301 text: General Chemistry: Chapter 15 odd problems pp. 668-671: 21-25, 37-43, 89, 93, 95, 98