1. 18 Acid-Base Equilibria 1 18 Additional Aspects of Acid – Base Equilibria Common ion effect solutions containing acids, bases, salts, solvent Buffer solutions solutions containing a weak acid and its salt Indicators (acid-base) colored acids and bases Titration curve chemistry and pH during titration Solutions of salts and polyprotic acids (hydration) reaction of solvent with ions

2. 18 Acid-Base Equilibria 2 Exam paper return The exam papers are piled according to the last digit of your ID number from 0 to 9. All papers are now in front of the Chem123 Lab on the shelf. You may retrieve yours from the hallway in ESC on the first floor.

3. 18 Acid-Base Equilibria 3 K a K b and K w H + + Base = Conjugate_acid of Base + Acid = H + + Conjugate_base of Acid - For example : NH 3 + H 2 O = NH 4 + + OH - K a for NH 4 + = K w / K b for NH 3 HA = H + + A - K b for A - = K w / K a for HA Thus, K a K b = K w for conjugate pairs. A - + H 2 O = HA + OH - [HA] [OH - ] [H + ] K b = ————— ——— [A - ] [H + ] [HA] = ———— [OH - ] [H + ] [A - ] [H + ] 1 = —— K w K a

4. 18 Acid-Base Equilibria 4 K a K b and K w - comparison HA = H + + A - [H + ] [A - ] [OH - ] K a = ———— ——— [HA][OH - ] [A - ] = ————— [H + ] [OH - ] [HA] [OH - ] 1 = —— K w K b A - + H 2 O = HA + OH - [HA] [OH - ] [H + ] K b = ————— ——— [A - ] [H + ] [HA] = ———— [OH - ] [H + ] [A - ] [H + ] 1 = —— K w K a

5. 18 Acid-Base Equilibria 5 K a K b and K w – another way HA + H 2 O (l) = H 3 O + (aq) + A – (aq) K a of HA +) A – + H 2 O (l) = OH – (aq) + HA (aq) K b of A – 2 H 2 O (l) = H 3 O + (aq) + OH – (aq) K w = K a K b Review qn : When you add two equations to get a third, what are the relationship between the K s?

6. 18 Acid-Base Equilibria 6 K a, K b of A - & K w A - + H 2 O = HA + OH - [HA] [OH - ] [H + ] K b = ————— ——— [A - ] [H + ] [HA] = ———— [OH - ] [H + ] [A - ] [H + ] 1 = —— K w K a HA = H + + A - K a A - + H 2 O = HA + OH - K b -- add them together - - H 2 O = H + + OH - K w = K a K b Two ways to show their relationship

7. 18 Acid-Base Equilibria 7 Properties of salt solutions The spectator ions in acid-base reactions form salts. Salts completely ionize in their aqueous solutions. A salt NH 4 A, the ionizes NH 4 A = NH 4 + + A - The ions of salts interact with water (called hydration ), A - + H 2 O = HA + OH - If the anions are stronger base than H 2 O, the solution is basic. NH 4 + + H 2 O = NH 3 + H 3 O + If the cations are stronger acid than H 2 O, the solution is acidic. These competitive reactions make the solution acidic or basic depending on the strength of the acids and bases. Are the solutions of the following salts neutral, basic or acidic NaCl?NaAc?NH 4 Cl? NH 4 Ac? KClO 4 ?KNic?CH 3 NH 3 ClO 4 ? modified

8. 18 Acid-Base Equilibria 8 Hydration problems What is the pH and concentrations of various species of a 0.100 M KA salt solution if K a for HA = 1.0e-5? (K = potassium, HA a general acid ) Solution :KA = K + + A - A – + H 2 O (l) = OH – (aq) + HA (aq) K b = 1e–14 / 1.0e–5 = 1e–9 0.10-xx x x 2 K b = ———— = 1.0e–9 0.10 – x (0.10 – x)  0.10 = [A – ] x =  1e-9*0.1 = 1e–5 = [OH – ] = [HA] pOH = 5 (= pKb / 2 = 10 / 2 ); pH = 14 – 5 = 9 (basic)

9. 18 Acid-Base Equilibria 9 Acidity of Salt Solutions What is the pH of a 1.0-M NaHCO 3 solution? For H 2 CO 3, K a1 = 4.3e-7 and K a2 = 4.8e-11. Solution : In the solution, [Na + ] = 1.0 M and [HCO 3 – ] = 1.0 M. Two competitive reaction takes place: H 2 O + HCO 3 – = H 2 CO 3 + OH – K b = K w / K a1 = 2.33e-8 HCO 3 – = H + + CO 3 2– K a2 = 4.8e-11 Thus, HCO 3 – is a stronger base than an acid. The solution is basic. Discussion : The overall dissociation constant, K overall = K a1 * K a2 = 2.1e-17 Generalization : if K a2 < K b or K a1 * K a2 < K w, the solution of NaHA is basic for the diprotic acid H 2 A.

10. 18 Acid-Base Equilibria 10 Common ion effect Since salts, acids and bases ionize in their solutions, their common ions such as H +, OH –, other cations, anions, will affect the equilibria. Calculate the degree of ionization of HA in a solution containing 0.10 M each of HCl and HA, if K a = 1.0e–5 for HA. Solution: (assume [A-] = x and work out the relations as shown) HCl = H + + Cl – 0.10+x 0.10 (0.10+x) x HA = H + + A – K a = 1.0e–5 = —————— 0.10-x0.10+x x 0.10-x (0.1-x=0.1+x=0.1) x = 1.0e–5 M degree of ionization = 1.0e–5 / 0.1 = 1.0e–4 = 0.01% ReDo if [HA] = 0.01 M and [HCl] = 0.10 M?

11. 18 Acid-Base Equilibria 11 Mass and Charge Balance Equations In a solution of electrolytes, the stoichiometry leads to mass or material balance (mbe) and charge balance equations (cbe). The merit is to consider the various ion species and equilibria in the solution when we write these equations. In a 0.10 M NaHCO 3 solution, mbe: 0.10 = [Na+] = [H 2 CO 3 ] + [HCO 3 – ] + [CO 3 2– ] cbe: [Na + ] + [H + ] = [HCO 3 – ] + 2 [CO 3 2– ] + [OH – ] New See Week 8 Part b problem #1. Consider these equilibrium eqn for mbe and cbe: NaHCO 3 = Na + + HCO 3 - HCO 3 - + H 2 O = H 2 CO 3 + OH - HCO 3 - + H 2 O = CO 3 2- + H 3 O +

12. 18 Acid-Base Equilibria 12 Buffers A buffer contains a weak acid and a salt of the same acid. A buffer contains a weak base and a salt of the same base. Concentrations of various species for a solution containing 0.10 M each of KA and HA, K a = 1.0e–5. KA = K + + A – 0.10 0.10+x x (0.10+x) HA = H + + A – K a = 1.0e–5 = —————— 0.10-x x 0.10+x 0.10-x (0.1-x=0.1+x=0.1) x = 1.0e–5 = [H + ] pH = p K a in this case, because [HA] = [A – ]

13. 18 Acid-Base Equilibria 13 pH of buffers For a weak acid, HA, we have HA = H + + A – [H + ] [A – ] K a = ———— [HA] [A – ] – log K a = – log [H + ] – log ——— [HA] [A – ] [A – ] p K a = pH – log —— pH = p K a + log —— [HA] [HA] [salt] [acid] Adding acid or base only affect the ratio [A–] / [HA], the pH changes little. Henderson- Hasselbach equationn

14. 18 Acid-Base Equilibria 14 A buffer solution A buffer solution is made up using 10.0 mL0.10 M acetic acid (HA, K a = 1.7e-5) and 20.0 0.10 M sodium acetate (NaA). Evaluate its pH. Hint : [A – ] = 20.0*0.1 / (20.0+10.0) = 0.067 M [HA] = 10.0*0.1 / (20.0+10.0) = 0.033 M Henderson- Hasselbach equationn [A – ] pH = p K a + log —— [HA] = – log (1.7e-5) + log ( 2 / 1 ) = 4.77 + 0.30 = 5.07 Find ways to see [A-] / [HA] = 2 / 1 pH of a sol’n by mixing 10.0 mL 0.1 M HCl and 10.0 mL 0.3 M NaA

15. 18 Acid-Base Equilibria 15 Making a buffer of certain pH Find a weak acid with p K a close to the desirable pH Find out the desirable [A-] / [HA] ratio Measure appropriate amount (moles) of acid and its salt Dissolve in appropriate amount of water (volume does not matter) Use a strong acid and a salt or a weak acid and a strong base instead of acid and salt.

16. 18 Acid-Base Equilibria 16 Indicators Indicators are substances whose solutions change color due to changes in pH. HIn = H + + In -, and define the equilibrium constant as K ai, [H + ][In - ] K ai [In - ] K ai = ————— —— = —— [HIn] [H + ] [HIn] The color varies according to the ratio K ai / [H+] ___

17. 18 Acid-Base Equilibria 17 Common indicators NameAcid colorpH range Base color Methyl violet Yellow 0 – 1.6Blue-violet Methy orangeRed3.2-4.4 Yellow LitmusRed5-8Blue PhenolphthaleinColorless8.2 - 10.0Pink ThymolphthaleinColorless9.4-10.6Blue

18. 18 Acid-Base Equilibria 18 Phenolphthalein – a indicator C 20 H 14 O 4 (MW = 318.33) Formal name: 3,3-bis(4-hydroxyphenyl)-1( 3H )-isobenzofuranone, 3,3-bis(4-hydroxyphenyl)phthalide) Colorless in acidic solution Pink in basic solution, C 20 H 12 O 4 2– pH ~ 10

19. 18 Acid-Base Equilibria 19 Titration calculation When a strong acid is titrated with a strong base, consider: The amount of acid present = V a * C a The amount of base NaOH added = V b * C b The amount of acid left = V a * C a - V b * C b The concentration of acid and thus V a * C a - V b * C b [H + ] = ————————— V a + V b Titration of 10.0 mL 1 M HCl using 1 M NaOH Base add [H+] pH 01.00 55/150.48 91/191.28 9.5.5/19.51.59 9.90.1/19.92.30 9.950.05/19.952.60 10NaCl soln7 [OH] 10.50.05 / 20.0511.40 10.100.5 / 20.111.70 111 / 2112.68 155 / 2513.3 2020 / 3013.92 Please plot the curve from the data

20. 18 Acid-Base Equilibria 20 Titration curve Weak HA slide 7 Buffers–slide 10-11 Hydration of salts Slides 6 & 7

21. 18 Acid-Base Equilibria 21 Acid-base equilibria - summary Know names, formula, & properties of some acids & bases: HA, NH 3 Evaluate K a and K b and apply them to calculate [ ]’s of various species using approximation using quadratic equation Theory of polyprotic acids evaluate concentrations of various species Derive and apply the relation K a K b = K w Explain why some salt solutions are acidic or basic Evaluate pH of salts (hydration reacting with water ) Explain theory of buffer evaluate pH of buffer make a buffer solution Explain indicators as weak acids and bases Plot a titration curve (weak acid titrating with strong base)

22. 18 Acid-Base Equilibria 22 Review In solving x in problems involving equilibrium constant, how do we know which method to use. When can we neglect x? Also, when can we use something like Successive Approxiamation? In the example given in class, K was quite small, so as done in other examples, I probably would have neglected x. Then there is Newton's method where we plug in trial values for x. How do we know where a good place is to start? Technically x could be any value!

23. 18 Acid-Base Equilibria 23 Review 1 – pH of weak acid solutions Solutions of a weak acid HA, K a = 1.0e–6 with concentrations of 1.0 M, 0.10 M, 0.00010 M. What are their pH? Solution : HA = H + + A –, Ka = 1.0e-6 C-xx x x 2 K a = ——— C–x C = 1.0 M x =  K a * C = 1e–3 pH = -log0.001 = 3 C = 0.10 M x =  K a * C = 3.2e–4 pH = -log3.2e-4=3.49 C = 0.00010 M x =  K a * C = 1e–5, 10%of C Do not approximate use x 2 + K a x – K a * C = 0 – K a + ( K a 2 + 4 K a* C) x = ————————— = 1.9e–5 2 pH = -log1.9e-5 = 4.72 not 5

24. 18 Acid-Base Equilibria 24 Review 2 – pH of buffer 25.00 mL of 0.10 M HAc (acetic acid, p K a = 4.75) is titrated with 0.20 M NaOH solution. What is the pH when 5.00 mL NaOH has been added? Analysis The solution containing 25.00 mL 0.10 M HAc and 5.00 mL 0.20 M NaOH actually contains 5.00 mL * 0.20 mol/L = 1.0 mmol NaAc, and (25.00*0.10 – 1.0) mol = 1.50 mmol HAc. This solution is a buffer Solution: For a buffer, apply [A - ] 1.0 / total volume pH = p K a + log ------- = 4.75 + log --------- = 4.75 – 0.18 = 4.57 [HA] 1.5/ total volume

25. 18 Acid-Base Equilibria 25 Review 3 – pH of buffer 25.00 mL of 0.10 M HAc (acetic acid, p K a = 4.75) is titrated with 0.20 M NaOH solution. What is the pH when 6.25 mL NaOH has been added? Analysis The solution containing 25.00 mL 0.10 M HAc and 6.25 mL 0.20 M NaOH actually contains 6.25 mL * 0.20 mol/L = 1.25 mmol NaAc, and (25.00*0.10 – 1.25) mol = 1.25 mmol HAc. This solution is a buffer Solution: For a buffer, apply [A - ] 1.25 / total volume pH = p K a + log ------- = 4.75 + log --------- = 4.75 – 0.00 = 4.75 [HA] 1.25 / total volume

26. 18 Acid-Base Equilibria 26 Review 4 – pH of buffer 25.00 mL of 0.10 M HAc (acetic acid, p K a = 4.75) is titrated with 0.20 M NaOH solution. What is the pH when 7.00 mL NaOH has been added? Analysis The solution containing 25.00 mL 0.10 M HAc and 7.00 mL 0.20 M NaOH actually contains 7.00 mL * 0.20 mol/L = 1.4 mmol NaAc, and (25.00*0.10 – 1.4) mol = 1.10 mmol HAc. This solution is a buffer Solution: For a buffer, apply [A - ] 1.4 / total volume pH = p K a + log ------- = 4.75 + log --------- = 4.75 + 0.10 = 4.85 [HA] 1.1/ total volume not 4.57

27. 18 Acid-Base Equilibria 27 Review 5 – pH of salt a solution 25.00 mL of 0.10 M HAc (acetic acid, p K a = 4.75) is titrated with 0.20 M NaOH solution. What is the pH when 12.50 mL NaOH has been added? Analysis The solution containing 25.00 mL 0.10 M HAc and 12.5 mL 0.20 M NaOH actually contains 12.50 mL * 0.20 mol/L = 2.5 mmol NaAc, and (25.00*0.10 – 2.5) mol = 0.00 mmol HAc. [NaA] = 2.5 mmol / (25.0+12.5)mL = 0.067 M Solution: hydration problem A - + H 2 O = HA + OH - 0.067-x x x x 2 K w 10 -14 ------------ = K b = ----- = -------- = 5.62e-10 0.067 – x K a 10 -4.75 x =  (5.62e-10)*0.067 = 1.36e-6 pOH = -log(1.36e-6) = 5.2

28. 18 Acid-Base Equilibria 28 Review 6 – pH of salt and base 25.00 mL of 0.10 M HAc (acetic acid, p K a = 4.75) is titrated with 0.20 M NaOH solution. What is the pH when 15.0 mL NaOH has been added? Analysis The solution containing 25.0 mL * 0.10 mol/L = 2.5 mmol NaAc, [NaAc] = 2.5 mmol /40 mL = 0.0625 M and (15.0 m* 0.20 – 2.5) = 0.5 mmol NaOH [NaOH] = 0.5 mmol / 40 mL = 0.0125 M Solution: The pH is dictated by the NaOH NaOH = Na + + OH - 0.0125 + x A - + H 2 O = HA + OH - 0.0625-x x 0.0125+x The extend of this reaction is small, x << 0.0125, [OH] = 0.0125 M pOH = -log(0.0125) = 1.90 pH = 14.00 – 1.90 = 12.10 Review 5; x = 1.36e-6 M

29. 18 Acid-Base Equilibria 29 Buffer solution sites Suppliers: postapplescientific.com/reagents/buffersoln.html caledonlabs.com/cgi-bin/products.cgi?category=K Buffer solution preparation csudh.edu/oliver/chemdata/buffers.htm biochem.mcw.edu/~simont/java/BufferMaker.html