1. Updates Assignment 06 is due Mon., March 12 (in class) Midterm 2 is Thurs., March 15 and will cover Chapters 16 & 17 –Huggins 10, 7-8pm –For conflicts: ELL 221, 6-7pm (must arrange at least one week in advance)

2. Acid-Base Equilibria and Solubility Equilibria Chapter 17

3. Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol CH 3 CO 2 H and 0.300 mol CH 3 CO 2 Na to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

4. Calculating pH Changes in Buffers Before the reaction, since mol CH 3 CO 2 H = mol CH 3 CO 2 − pH = pK a = −log (1.8  10 −5 ) = 4.74

5. Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: CH 3 CO 2 H (aq) + OH − (aq)  CH 3 CO 2 − (aq) + H 2 O (l) CH 3 CO 2 HCH 3 CO 2 − OH − Before reaction0.300 mol 0.020 mol After reaction0.280 mol0.320 mol0.000 mol New concentrations: 0.280 mol/ 1 L = 0.280 M CH 3 CO 2 H 0.320 mol/ 1 L = 0.320 M CH 3 CO 2 -

6. Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320) (0. 280) pH = 4.74 + 0.06 pH = 4.80

7. Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) KF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3 - is it conjugate acid buffer solution 17.3

8. = 9.20 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH 4 + (aq) H + (aq) + NH 3 (aq) pH = pK a + log [NH 3 ] [NH 4 + ] pK a = 9.25 pH = 9.25 + log [0.30] [0.36] = 9.17 NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (moles) end (moles) 0.0290.001 0.024 0.0280.00.025 pH = 9.25 + log [0.25] [0.28] [NH 4 + ] = 0.028 0.10 final volume = 80.0 mL + 20.0 mL = 100 mL [NH 3 ] = 0.025 0.10 17.3

9. Titration In a titration a solution of accurately known concentration is added gradually to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

10. Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

11. Titrations Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink) 17.3 Change in indicator color occurs at the point when all of the acid or base being titrated has reacted. At this point any additional titrant reacts with the indicator.

12. Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.

13. Just before and after the equivalence point, the pH increases rapidly. Titration of a Strong Acid with a Strong Base

14. At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid. Titration of a Strong Acid with a Strong Base

15. As more base is added, the increase in pH again levels off. Titration of a Strong Acid with a Strong Base

16. Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. The pH at the equivalence point will be >7. Phenolphthalein is commonly used as an indicator in these titrations. Titration of a Strong Acid with a Strong Base Weak acid/strong base or strong acid/weak base

17. Strong Acid-Strong Base Titrations NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) OH - (aq) + H + (aq) H 2 O (l) 17.4

18. Weak Acid-Strong Base Titrations CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (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): 17.4

19. Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq) At equivalence point (pH < 7): 17.4 H + (aq) + NH 3 (aq) NH 4 Cl (aq)

20. Exactly 100 mL of 0.10 M HNO 2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O (l) start (moles) end (moles) 0.01 0.0 0.01 NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) 0.050.00 -x-x+x+x 0.05 - x 0.00 +x+x xx [NO 2 - ] = 0.01 0.200 = 0.05 M Final volume = 200 mL K b = [OH - ][HNO 2 ] [NO 2 - ] = x2x2 0.05-x = 2.2 x 10 -11 0.05 – x  0.05x  1.05 x 10 -6 = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02

21. Acid-Base Indicators HIn (aq) H + (aq) + In - (aq)  10 [HIn] [In - ] Color of acid (HIn) predominates  10 [HIn] [In - ] Color of conjugate base (In - ) predominates 17.5

22. The titration curve of a strong acid with a strong base. 17.5

23. Which indicator(s) would you use for a titration of HNO 2 with KOH ? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein 17.5