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18 Additional Aspects of Acid – Base Equilibria

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

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18 Additional Aspects of Acid – Base Equilibria

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  1. 18 Additional Aspects of Acid – BaseEquilibria 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 18 Acid-BaseEquilibria

  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. 18 Acid-BaseEquilibria

  3. KaKb and Kw A- + H2O = HA + OH- [HA] [OH-] [H+]Kb = ————— ——— [A-] [H+] [HA] = ———— [OH-][H+][A-][H+] 1 = —— KwKa H+ + Base = Conjugate_acid of Base+Acid = H+ + Conjugate_base of Acid- For example: NH3 + H2O = NH4+ + OH-Ka for NH4+ = Kw / Kb for NH3 HA = H+ + A-Kb for A- = Kw / Ka for HA Thus, KaKb = Kw for conjugate pairs. 18 Acid-BaseEquilibria

  4. KaKb and Kw - comparison HA = H+ + A- [H+] [A-] [OH-]Ka = ———— ——— [HA] [OH-] [A-] = ————— [H+][OH-][HA][OH-] 1 = —— KwKb A- + H2O = HA + OH- [HA] [OH-] [H+]Kb = ————— ——— [A-] [H+] [HA] = ———— [OH-][H+][A-][H+] 1 = —— KwKa 18 Acid-BaseEquilibria

  5. KaKb and Kw – another way HA + H2O (l) = H3O+ (aq) + A–(aq) KaofHA+)A– + H2O (l) = OH–(aq) + HA (aq) KbofA– 2 H2O (l) = H3O+ (aq) + OH– (aq) Kw = KaKb Review qn:When you add two equations to get a third, what are the relationship between the Ks? 18 Acid-BaseEquilibria

  6. Ka, Kb of A- & Kw Two ways to show their relationship A- + H2O = HA + OH- [HA] [OH-] [H+]Kb = ————— ——— [A-] [H+] [HA] = ———— [OH-][H+][A-][H+] 1 = —— KwKa • HA = H+ + A-Ka • A- + H2O = HA + OH-Kb • - add them together - - • H2O = H+ + OH-Kw = KaKb 18 Acid-BaseEquilibria

  7. Properties of salt solutions The spectator ions in acid-base reactions form salts. Salts completely ionize in their aqueous solutions. A salt NH4A, the ionizes NH4A = NH4+ + A- The ions of salts interact with water (called hydration), A- + H2O = HA + OH-If the anions are stronger base than H2O, the solution is basic. NH4+ + H2O = NH3 + H3O+If the cations are stronger acid than H2O, 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 acidicNaCl? NaAc? NH4Cl? NH4Ac? KClO4? KNic? CH3NH3ClO4? 18 Acid-BaseEquilibria modified

  8. Hydration problems What is the pH and concentrations of various species of a 0.100 M KA salt solution if Ka for HA = 1.0e-5? (K = potassium, HA a general acid) Solution: KA = K+ + A-A– + H2O (l) = OH–(aq) + HA (aq) Kb= 1e–14/1.0e–5 = 1e–9 0.10-x x x x2Kb = ———— = 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) 18 Acid-BaseEquilibria

  9. Acidity of Salt Solutions What is the pH of a 1.0-M NaHCO3 solution? For H2CO3, Ka1 = 4.3e-7 and Ka2 = 4.8e-11. Solution: In the solution, [Na+] = 1.0 M and [HCO3–] = 1.0 M. Two competitive reaction takes place: H2O + HCO3– = H2CO3 + OH–Kb = Kw / Ka1 = 2.33e-8 HCO3– = H+ + CO32–Ka2 = 4.8e-11 Thus, HCO3 – is a stronger base than an acid. The solution is basic. Discussion: The overall dissociation constant, Koverall = Ka1* Ka2 = 2.1e-17 Generalization: if Ka2 < Kb or Ka1*Ka2 < Kw, the solution of NaHA is basic for the diprotic acid H2A. 18 Acid-BaseEquilibria

  10. Since salts, acids and bases ionize in their solutions, their common ions such as H+, OH–, other cations, anions, will affect the equilibria. Common ion effect Calculate the degree of ionization of HA in a solution containing 0.10 M each of HCl and HA, if Ka = 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–Ka = 1.0e–5 = —————— 0.10-x 0.10+xx 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? 18 Acid-BaseEquilibria

  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 NaHCO3 solution, mbe: 0.10 = [Na+] = [H2CO3] + [HCO3 –] + [CO32–] cbe: [Na+] + [H+] = [HCO3 –] + 2 [CO32–] + [OH–] Consider these equilibrium eqn for mbe and cbe: NaHCO3 = Na+ + HCO3-HCO3- + H2O = H2CO3 + OH-HCO3- + H2O = CO32- + H3O+ New See Week 8 Part b problem #1. 18 Acid-BaseEquilibria

  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, Ka = 1.0e–5. KA = K+ + A– 0.10 0.10+x x (0.10+x) HA = H+ + A–Ka = 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 = pKa in this case, because [HA] = [A–] 18 Acid-BaseEquilibria

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

  14. A buffer solution A buffer solution is made up using 10.0 mL0.10 M acetic acid (HA, Ka = 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 Find ways to see [A-] / [HA] = 2 / 1 [A–]pH = pKa + log —— [HA] = – log (1.7e-5) + log (2/1) = 4.77 + 0.30 = 5.07 Henderson-Hasselbach equationn pH of a sol’n by mixing 10.0 mL 0.1 M HCl and 10.0 mL 0.3 M NaA 18 Acid-BaseEquilibria

  15. Making a buffer of certain pH Find a weak acid with pKa 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. 18 Acid-BaseEquilibria

  16. Indicators Indicators are substanceswhose solutions change color due to changes in pH. HIn = H+ + In-, and define the equilibrium constant as Kai, [H+][In-]Kai[In-]Kai = ——————— = ——[HIn] [H+] [HIn] The color varies according to the ratioKai / [H+] ___ 18 Acid-BaseEquilibria

  17. Common indicators 18 Acid-BaseEquilibria

  18. Phenolphthalein – a indicator C20H14O4 (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, C20H12O42– pH ~ 10 18 Acid-BaseEquilibria

  19. Titration calculation Titration of 10.0 mL 1 M HCl using 1 M NaOH Base add [H+] pH 0 1.0 05 5/15 0.489 1/19 1.289.5 .5/19.5 1.599.9 0.1/19.9 2.309.95 0.05/19.95 2.60 When a strong acid is titrated with a strong base, consider: The amount of acid present = Va * Ca The amount of base NaOH added = Vb * Cb The amount of acid left = Va * Ca - Vb * Cb The concentration of acid and thus Va * Ca - Vb * Cb [H+] = —————————Va + Vb 10 NaCl soln 7 [OH]10.5 0.05 / 20.05 11.4010.10 0.5 / 20.1 11.7011 1 / 21 12.6815 5 / 25 13.320 20 / 30 13.92 Please plot the curve from the data 18 Acid-BaseEquilibria

  20. Titration curve Buffers–slide 10-11 Hydration of saltsSlides 6 & 7 Weak HAslide 7 18 Acid-BaseEquilibria

  21. Acid-base equilibria - summary Know names, formula, & properties of some acids & bases: HA, NH3Evaluate Ka and Kb and apply them to calculate [ ]’s of various species using approximation using quadratic equationTheory of polyprotic acids evaluate concentrations of various speciesDerive and apply the relation KaKb = KwExplain why some salt solutions are acidic or basicEvaluate pH of salts (hydration reacting with water)Explain theory of buffer evaluate pH of buffer make a buffer solutionExplain indicators as weak acids and basesPlot a titration curve (weak acid titrating with strong base) 18 Acid-BaseEquilibria

  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! 18 Acid-BaseEquilibria

  23. Review 1 – pH of weak acid solutions Solutions of a weak acid HA, Ka = 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-x x x C = 1.0 Mx = Ka*C = 1e–3pH = -log0.001 = 3 x2 Ka = ——— C–x C = 0.10 Mx = Ka*C = 3.2e–4pH = -log3.2e-4=3.49 C = 0.00010 Mx = Ka*C = 1e–5, 10%of C Do not approximate usex2 + Ka x – Ka*C = 0 – Ka + (Ka2 + 4 Ka*C)x = ————————— = 1.9e–5 2pH = -log1.9e-5 = 4.72 not 5 18 Acid-BaseEquilibria

  24. Review 2 – pH of buffer 25.00 mL of 0.10 M HAc (acetic acid, pKa = 4.75) is titrated with 0.20 M NaOH solution. What is the pH when 5.00 mL NaOH has been added? AnalysisThe 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 volumepH = pKa + log ------- = 4.75 + log --------- = 4.75 – 0.18 = 4.57 [HA] 1.5/ total volume 18 Acid-BaseEquilibria

  25. Review 3 – pH of buffer 25.00 mL of 0.10 M HAc (acetic acid, pKa = 4.75) is titrated with 0.20 M NaOH solution. What is the pH when 6.25 mL NaOH has been added? AnalysisThe solution containing 25.00 mL 0.10 M HAc and 6.25 mL 0.20 M NaOH actually contains 6.25mL * 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 volumepH = pKa + log ------- = 4.75 + log --------- = 4.75 – 0.00 = 4.75 [HA] 1.25 / total volume 18 Acid-BaseEquilibria

  26. Review 4 – pH of buffer 25.00 mL of 0.10 M HAc (acetic acid, pKa = 4.75) is titrated with 0.20 M NaOH solution. What is the pH when 7.00 mL NaOH has been added? AnalysisThe 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 volumepH = pKa + log ------- = 4.75 + log --------- = 4.75 + 0.10 = 4.85 [HA] 1.1/ total volumenot 4.57 18 Acid-BaseEquilibria

  27. Review 5 – pH of salt a solution 25.00 mL of 0.10 M HAc (acetic acid, pKa = 4.75) is titrated with 0.20 M NaOH solution. What is the pH when 12.50 mL NaOH has been added? AnalysisThe 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- + H2O = HA + OH-0.067-x x x x2Kw 10-14------------ = Kb = ----- = -------- = 5.62e-10 0.067 – x Ka 10-4.75 x = (5.62e-10)*0.067 = 1.36e-6 pOH = -log(1.36e-6) = 5.2 18 Acid-BaseEquilibria

  28. Review 6 – pH of salt and base 25.00 mL of 0.10 M HAc (acetic acid, pKa = 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 Mand (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- + H2O = HA + OH-0.0625-x x 0.0125+xThe extend of this reaction is small, x << 0.0125, [OH] = 0.0125 M Review 5; x = 1.36e-6 M pOH = -log(0.0125) = 1.90pH = 14.00 – 1.90 = 12.10 18 Acid-BaseEquilibria

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

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