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# CM4106 Review of Lesson 3 (Part 1)

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Buffer Solutions/ pH Calculation / Maximum Buffering Capacity

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### CM4106 Review of Lesson 3 (Part 1)

1. 1. CM4106 Chemical Equilibria & ThermodynamicsLesson 3 (Part 1)Additional Aspects of Acid-Base Equilibria(Topic 3.1 – 3.5)A Chemistry Education Blog by Mr Tanhttp://chemistry-mr-tan-yong-yao.blogspot.sg/
2. 2. Fundamentals:Important relations p = -log10 At 25ºC: pH + pOH = pKw = 14 pKa + pKb = 14 [H+][OH–] = 10-14 Ka x Kb = Kw = 10-14
3. 3. (I) Calculations for acids OR bases pH, pKa, [H+], Ka pOH, pKb, [OH–], Kb Step 1: Determine what is present in the solution. (A) Acid : Strong Acid vs Weak Acid Monoprotic Acid / Diprotic Acid / Triprotic Acid Concentration of Acid (B) Base: Strong Base vs Weak Base Monoprotic base / Diprotic base / Triprotic base Concentration of Bas Take note of concentration of acid / bases (For dilute solutions, we need to take into consideration of [H+] / [OH-] from auto-ionization from water) Step 2: Use the appropriate equations for the respective species.
4. 4. (I) pH calculations of Acid OR Base x2 Ka = Strong Acid Weak Acid ([HA] – x)Strong acids dissociate HA(aq) ⇌ H+(aq) + A-(aq)completely into ions inaqueous solution. I [HA] 0 0 C -x +x +x[H+] = [HA] E [HA] - x +x +x x2 Kb = Strong Base Weak Base ([B] – x)Strong bases dissociate B + H2O ⇌ BH+ + OH-completely into ions in I [B] - 0 0aqueous solution. C -x - +x +x[OH-] = [B] E [B] - x - x x
5. 5. (I) pH calculations of Acid OR Base1. Determine if acid/ base is strong or weak (more common)2. For weak acids, if asked to determine Ka, pH or [H+], you can save time by using the formula: [H+] = Ka × c [OH–] = Kb × c Assumption: x is negligibleExample:Calculate the pH of 0.50M HF solution at 25ºC where the Ka = 7.1 x 10-4 [H+][F-] x2 Ka = = = 7.1 x 10-4 [HF] (0.50 – x) Assumption: For weak acids, x must be very small  0.50 – x ≈ 0.5 x2 = 7.1 x 10-4 x = [H+] = 0.0188 M Assumption is valid; 0.50 x < 5% of [HF]initial pH = 1.73 (to 2 d.p.)
6. 6. Calculate the pH of the following solutions at 298K0.10 mol dm−3 CH3COOH (pKa = 4.75) Weak acid solution (monoprotic acid) CH3COOH(aq) + H2O (l) ⇌ CH3COO(aq) + H3O+(aq) Initial (M) 0.10 - 0.00 0.00 Change (M) -x - +x +x Eqm (M) 0.10 - x - +x +x [CH3COO][H3O+] x2 Ka = = = 10 4.75 = 1.79 x 10-5 [CH3COOH] (0.10 – x) Assumption: For weak acids, x must be very small  0.10 – x ≈ 0.10 x2 Remember to validate Assumption = 1.79 x 10-5 Assumption is justified, 0.10 x < 5% of [CH3COOH]initial x = [H3O+] = 1.338 x 10-3 M Concentration: 2 s.f. pH = 2.87 ( 2 d.p.) pH: 2 d.p.
7. 7. (II) Calculations for mixture of acids AND bases pH, pKa, [H+], α pOH, pKb, [OH–] Step 1: Determine what is present in the solution. (A) acid Stoichiometric amounts of acid and base – base salt solution (B) buffer Non-stoichiometric amounts of acid and base – likely to be a buffer (C) salt Step 2: Use the appropriate equations for the respective species.
8. 8. (II) (a) or (c) pH calculation of salt solutions COVERED IN GREATER DETAILS IN TOPIC 3.8Salt solutions can be (i) neutral (ii) weak acids or (iii) weak bases Basic SaltCH3COONa (aq) → CH3COO- (aq) + Na+ (aq)CH3COO- (aq) + H2O (l) ⇌ CH3COOH (aq) + OH- (aq)conjugate base pH > 7 Acidic Salt NH4Cl (aq) → NH4+ (aq) + Cl- (aq) NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O+ (aq)conjugate acid pH < 7
9. 9. (II) (b) pH of Buffer Solutions Acid Buffer weak acid + conjugate base weak acid Assume negligible dissociation of acid due to common ion effect CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq) CH3COONa(aq) → CH3COO-(aq) + Na+(aq) conjugate base Assume full dissociation of salt [salt] pH = pKa + lg (strong electrolyte) [acid] Basic Buffer weak base + conjugate acid weak base Assume negligible dissociation of base due to common ion effect NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq) NH4Cl(aq) → NH4+(aq) + Cl-(aq) conjugate acidAssume full dissociation of salt(strong electrolyte) pOH = pKb + lg [salt] [base]
10. 10. (II) (b) pH of Buffer Solutions1. Determine if buffer is acidic/ basic2. When using H-H equation, you can save time by just calculating no. of moles of salt and acid/base since total volume is the same and cancels out. Useful for MCQs! nsalt/V nsalt/V pH = pKa + lg pOH = pKb + lg nacid/V nbase/V3. Need to be sensitive to the condition (maximum buffering capacity)where [salt] = [acid] which simply means pH = pKaSimilarly for basic buffer, [salt] = [base]  pOH = pKb
11. 11. [salt] pH  pK a  lg Sample Calculation [acid]• Find the pH of a buffer made from adding 3.28 g of sodium ethanoate, CH3COO-Na+ to 1.00 dm3 of 0.0100 M ethanoic acid, CH3COOH . (Ka = 1.80 x 10-5) Useful Information Molar Mass of CH3COO-Na+ = 82.03 g/mol Amount of CH3COO-Na+ present in 3.28 g = (3.28 g / 82.03) = 0.0400 mol [CH3COO-Na+] = 0.0400 M [CH3COOH] = 0.0100 M [salt] pH  pK a  lg [acid] 5  0.0400    lg(1.80  10 )  lg   pH = 5.3466  0.0100  = 5.347 (3 d.p.) Page 46
12. 12. An alternative approach – Sample Calculation ICE Table • Find the pH of a buffer made from adding 3.28 g of sodium ethanoate, CH3COO-Na+ to 1.00 dm3 of 0.0100 M ethanoic acid, CH3COOH. (Ka = 1.80 x 10-5) n(CH3COO-Na+) present in 3.28 g = (3.28 g / 82.03)Useful Information = 0.0400 molMolar Mass of CH3COO -Na+ = 82.03 g/mol [CH3COO-Na+] = 0.0400 M [CH3COOH] = 0.0100 M CH3COOH(aq) + H2O (l) ⇌ H3O+ (aq) + CH3COO (aq) Initial [ ] (M) 0.0100 M - 0 0.0400 M Change [ ] (M) -x - +x +x Equilibrium [ ] (M) 0.0100 - x - x 0.0400 + x Ka = [H3O+][CH3COO-] x (0.0400 + x) = = 1.80 x 10-5 [CH3COOH] (0.0100 – x) Solve for x, pH = - lg [H3O+] = - lg (4.497 x 10-6) x = [H3O+] = 4.497 x 10-6 M = 5.3470 = 5.347 (3 d.p.)
13. 13. Calculating pH change after small amounts of acid / base is added to buffer solutions1) Determine new [salt]new and [acid]new / [base]new How? a) Determine n(acid / base) added to buffer solution b) Determine the change to n(salt)buffer and n(acid)buffer / n(base)buffer by considering the neutralization action of a buffer c) Determine new total volume of the buffer solution [salt]new pH new  pK a  lg [acid] new
14. 14. CM4106Chemical Equilibria &Thermodynamics Review of Pre-Quiz 3 Topic 3 (3.1 – 3.5) Additional Aspects of Acid-Base EquilibriaBuffer Solutions, pH calculation, Maximum Buffer Capacity
15. 15. Question 1 (a)• Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching agent. It dissociates in water as represented by the equation below. HOCl(aq) + H2O(l) ⇌ OCl-(aq) + H3O+(aq) Ka = 3.2 x10-8a) Write the equilibrium-constant expression for the dissociation of HOCl in water. [OC l  ][H 3O  ] Ka  [HOC l ]
16. 16. Question 1(b) b) Calculate the molar concentration of H3O+ in a 0.14 M solution of HOCl. HOCl(aq) + H2O(l) ⇌ OCl-(aq) + H3O+(aq) Ka = 3.2 x10-8 HOCl (aq) + H2O (l) ⇌ H3O+ (aq) + OCl- (aq) Initial [ ] (M) 0.14 - 0 0 Change [ ] (M) -y - +y +y Equilibrium [ ] (M) 0.14 – x - x y y2 Solve for y, Ka  (0.14  y) y = 6.7 x 10-5 y23.2  10 8  [H3O+] = 6.7 x 10-5 M (2 s.f.) (0.14  y ) Note: units must be present
17. 17. Question 1 (c)c) A mixture of HOCl and sodium hypochlorite (NaOCl) can be used as a buffer. Write two equations to show how this buffer solution can control pH.• Addition of H3O+: OCl(aq) + H3O+(aq)  HOCl(aq) + H2O(l) Note: state symbols must be present• Addition ofOH : HOCl(aq) + OH(aq)  OCl (aq) + H2O(l)
18. 18. Question 1(d)HOCl reacts with NaOH according to the reaction representedbelow: HOCl(aq) + OH-(aq)  OCl-(aq) + H2O(l)Bob, a budding young chemist, decides to make a buffer byadding a volume of 10.0 mL of 0.56 M NaOH to 50.0 mL of 0.14 MHOCl solution. Assume that the volumes are additive. HOCl(aq) + H2O(l) ⇌ OCl-(aq) + H3O+(aq) Ka = 3.2 x10-8Preliminary Considerations:Reaction of NaOH with HOCl will form NaOCl (salt containing conjugatebase OCl-).Hence, if HOCl is in excess, the final mixture will contain some excessHOCl that is unreacted and some NaOCl formed as a result of theneutralization reaction.  BUFFER SOLUTION (Mixture of acid andconjugagte base (salt)
19. 19. Question 1(d)(i)• Calculate the pH of the buffer solution Henderson-Hasselbalch Equation to [salt] pH  pK a  log calculate pH of buffer equation [acid] Need to determine [salt] / [acid] HOCl (aq) + NaOH (aq)  NaOCl (aq) + H2O(l)n HOCl = (50.0/1000)(0.14) = 7.0 x 10-3 mol (excess reagent)n NaOH = (10.0/1000)(0.56) = 5.6 x 10-3 molTotal Volume = 50.0 + 10.0 = 60.0 cm3 = 0.0600 Ln HOCl remaining n NaOCl formed = 5.6 x10-3 mol = 7.0 x 10-3 - 5.6 x 10-3 = 1.4 x 10-3 mol [OCl] = 5.6 x 10-3 mol / 0.0600 L[HOCl] = 1.4 x 10-3 mol / 0.0600 L = 9.33 x 10-2 M = 2.33 x 10-2 M
20. 20. Question 1(d)(i) • Calculate the pH of the buffer solution.[HOCl] = 1.4 x 10-3 mol / 0.0600 L [OCl] = 5.6 x 10-3 mol / 0.0600 L = 2.33 x 10-2 M = 9.33 x 10-2 M [salt] pH  pK a  log [acid] 8 (9.33  10 -2 ) pH   lg(3.2 10 )  lg (2.33  10 -2 ) pH  8.097
21. 21. Question 1(d)(ii) • Bob would like to prepare a new buffer solution. How many grams of solid NaOH must he add to a 50.0 mL of 0.20 M HOCl to obtain a buffer that has a pH of 6.0? • Assume that the addition of the solid NaOH results in a negligible change in volume. (Mr NaOH = 40)Let the no of mole of NaOH required be y mol:n HOCl = (50/1000)(0.20) = 0.01 mol n HOCl remaining = (0.01 - y) moln NaOH = y mol n NaOCl formed = y mol NaOH (s) + HOCl (aq)  NaOCl (aq) + H2O (l)
22. 22. Question 1(d)(ii) • Bob would like to prepare a new buffer solution. How many grams of solid NaOH must he add to a 50.0 mL of 0.20 M HOCl to obtain a buffer that has a pH of 6.0? • Assume that the addition of the solid NaOH results in a negligible change in volume. (Mr NaOH = 40) n HOCl remaining = (0.01 - y) mol Total Volume = 50.0 mL = 0.0500 L n NaOCl formed = y mol  y    [salt] 6.0  - log (3.2 x 10 )  lg  0.0500  ypH  pK a  log -8  1.4949  lg [acid]  0.01  y  0.01  y    0.0500  Solve for y,Mass of NaOH (s) required y= 3.101 x 10-4 x 40 y = n (NaOCl) formed 10 1.4949  = n (NaOH) required 0.01  y= 0.1240 g = 3.101x 10-4 mol