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Weak Acids & Bases: A weak acid is not completely dissociated:- HA  H + + A -

Weak Acids & Bases: A weak acid is not completely dissociated:- HA  H + + A - Similarly, weak bases: B + H 2 O  BH + + OH - pKa = -log 10 Ka pKb = -log 10 Kb The smaller the pKa value - the STRONGER the acid e.g.

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Weak Acids & Bases: A weak acid is not completely dissociated:- HA  H + + A -

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  1. Weak Acids & Bases: A weak acid is not completely dissociated:- HA  H+ + A- Similarly, weak bases: B + H2O  BH+ + OH- pKa = -log10 Ka pKb = -log10 Kb The smaller the pKa value - the STRONGER the acid e.g. weak acid; pKa = 4 (Ka = 10-4) weaker acid; pKa = 6 (Ka = 10-6) even weaker acid; pKa = 8 (Ka = 10-8) Similar for bases (smaller pKb - STRONGER weak base) Ka Kb “stronger weak acids” Dr. David Johnson: CH0004 Lecture Notes - Term 2

  2. Conjugate Pairs: The acid HA and its corresponding base (A-) are said to be a conjugate acid-base pair. Similarly for B and BH+. The conjugate base of a strong acid is a strong base. (The conjugate acid of a strong base is a strong acid) Similarly, The conjugate base of a weak acid is a weak base. (The conjugate acid of a weak base is a weak acid) For a weak conjugate acid-base pair: Ka . Kb = Kw (pKa + pKb = 14) so, As the acid becomes weaker, the base becomes “stronger” (but never a strong base) - and vice versa.... Dr. David Johnson: CH0004 Lecture Notes - Term 2

  3. Conjugate Acid Base Pairs: pKb 3.36 3.226 Dr. David Johnson: CH0004 Lecture Notes - Term 2

  4. Calculation of the pH of Weak Acids & Bases: Calculate the pH of 0.02 molar benzoic acid (pKa 4.202) C6H5COOH  C6H5COO- + H+ Initial conc M 0 0 Eqm conc (M-x) x x Ka = x2 / (M-x)  x2 + Ka x - KaM = 0 which is a quadratic equation of the form: ax2 + bx + c =0 (a = 1; b = Ka; c = -KaM) Solving the quadratic for x: x = 1.09 x 10-3 i.e. [H+] = 1.09 x 10-3 M pH = 2.96 Note:- Solution to a quadratic equation: Dr. David Johnson: CH0004 Lecture Notes - Term 2

  5. Alternative (to the quadratic): [H+] = (KaM)½ = (6.28 x 10-5 x 0.02) ½ = 1.12 x 10-3pH = 2.95 Note: the above is only an approximation and doesn’t work with very dilute or weak solutions, but usually gives a very close answer. Dr. David Johnson: CH0004 Lecture Notes - Term 2

  6. In general: To find the pH of a weak acid, use: [H+] = (KaM)½ To find the pH of a weak base, use: [OH-] = (KbM)½ and pH = (14 - pOH) e.g. Calculate the pH of 0.1 M ammonia (pKb = 4.757) [OH-] = (KbM)½ = (0.10 x 1.75 x 10-5)½ = 1.32 x 10-3 So pH = 11.12 Kb Dr. David Johnson: CH0004 Lecture Notes - Term 2

  7. Fractional Compositions: or “” values (very important with EDTA later!!) Consider the weak acid HA of molarity M HA  H+ + A- M = [HA] + [A-] Substitute (2) into (1) and re-arrange: Ka (1) (2) = HA Dr. David Johnson: CH0004 Lecture Notes - Term 2

  8. Fractional Compositions: Similarly: Note: HA + A- = 1 HA = A- = H+ Dr. David Johnson: CH0004 Lecture Notes - Term 2

  9. Fractional Compositions - Summary: HA  H+ + A- HA H+ A- HA + A- = 1 HA = [H+] / ([H+] + Ka) H+ = Ka / ([H+] + Ka) Alternative Nomenclature: HA is sometimes called 0 A- is sometimes called 1 Additional Equation: Ka = H+2 M(Ka = 2 C) equal Dr. David Johnson: CH0004 Lecture Notes - Term 2

  10. Fractional Compositions - Example: The acid dissociation constant of a weak acid (at 0.100 mol dm-3) is 1.0 x 10-6. Calculate: (i) the pH (ii) the fractional dissociation and (i) [H+] = (Ka M) =  1.0 x 10-7 = 3.162 x 10-4  pH = 3.5 (ii) H+ = Ka / ([H+] + Ka) = (1.0 x 10-6) / (3.162 x 10-4 + 1.0 x 10-6) = 3.154 x 10-3 Check: Ka = 2 C = (3.154 x 10-3)2 x 0.1 = 1.0 x 10-6 Dr. David Johnson: CH0004 Lecture Notes - Term 2

  11. Fractional Compositions - Example: Calculate the fractional composition of a weak acid (pKa = 5) at pH = 2, 3, 4, 5, 6, 7, 8 and 9 pH = 2  [H+] = 1 x 10-2 Ka = 1 x 10-5 HA = [H+] / ([H+] + Ka) = 0.999 A- = (1 - HA) = 0.001 pH = 3 HA = 0.99 A- = 0.01 pH = 4 HA = 0.909 A- = 0.091 pH = 5 HA = 0.5 A- = 0.5 pH = 6 HA = 0.091 A- = 0.909 pH = 7 HA = 0.01 A- = 0.99 pH = 8 HA = 0.001 A- = 0.999 pH = 9 HA = 0.0001 A- = 0.9999 Dr. David Johnson: CH0004 Lecture Notes - Term 2

  12. Fractional Composition Diagram: - based on previous sheet data  HA A- or  H+ Dr. David Johnson: CH0004 Lecture Notes - Term 2

  13. Buffers: A buffer is a solution which resists changes in pH when small amounts of acid or base are added - or when dilution occurs. The buffers consists of a mixture of an acid and its conjugate base. The Henderson Hasselbach Equation: for a weak acid: Ka = [H+] [A-] / [HA] take logs and rearrange...... pH = pKa + log ([A-] / [HA]) similarly, for a weak base: pH = pKa + log ([B] / [BH+]) i.e. for buffers: pH = pKa + log ([salt] / [acid]) Dr. David Johnson: CH0004 Lecture Notes - Term 2

  14. The Henderson Hasselbach Equation: for buffers: pH = pKa + log ([salt] / [acid]) “de-protonated form” “protonated form” Dr. David Johnson: CH0004 Lecture Notes - Term 2

  15. pH Change in Buffers: [salt] / [acid]pH 100:1 pKa + 2 10:1 pKa + 1 1:1 pKa 1:10 pKa - 1 1:100 pKa - 2 Example calculation: Calculate the pH of a buffer prepared by dissolving 2.53 g of oxoacetic acid (CHOCOOH, pKa 3.46, RMM 74.04) and 5.13 g of potassium oxoacetate (CHOCOOK, RMM 112.13) in 200 cm3 water. [salt] = 5.13 / 112.13 [acid] = 2.53 / 74.04 note - volume term missed out as it would cancel pH = 3.46 + log 1.339 = 3.59 Dr. David Johnson: CH0004 Lecture Notes - Term 2

  16. How / Why do Buffers Work? Consider a buffer prepared by dissolving “tris” (12.43 g, RMM 121.14, pKa 8.075) and “tris hydrochloride” (4.67 g, RMM 157.60) in 1 litre of water. pH = 8.075 + log ( [12.43 / 121.14] ) [4.67 / 157.60] = 8.61 If strong acid is added, some of the buffer base will be converted into its conjugate acid, and the value ([salt] / [acid]) changes - BUT NOT BY MUCH!!!! Similarly, if strong base is added, the buffer acid is converted to its conjugate base, and the value ([salt] / [acid]) changes the other way - AGAIN NOT BY MUCH!!! 0.1026 0.0296 Dr. David Johnson: CH0004 Lecture Notes - Term 2

  17. How / Why do Buffers Work? Example calculation: If 12.0 cm3 of 1.00 molar HCl is added to the tris buffer from the previous sheet: B + H+  BH+ (“tris”) (tris) (HCl) Initial moles 0.1026 negligible 0.0120 0.0296 Final moles 0.0906 0.0416 (= 0.1026 - 0.0120) (= 0.0296 + 0.0120) new pH = 8.075 + log (0.0906 / 0.0416) = 8.41 Dr. David Johnson: CH0004 Lecture Notes - Term 2

  18. new pH = 8.075 + log (0.0906 / 0.0416) = 8.41 i.e. the pH of the buffer has not changed very much by adding a limited amount of molar strong acid!!! Adding 12 cm3 changed pH from 8.61 to 8.41 if 12 cm3 had been added to an unbuffered solution (of either the salt or the acid) the pH would have decreased to 1.93!!!! Why??? Because the strong acid (or base) is consumed by one of the components of the buffer without unduly affecting the ratio (which is in a “log” term anyway). i.e. If you add HCl, B is converted to BH+, if you add NaOH, BH+ is converted to B. As long as you don’t add too much HCl or NaOH (thus consuming all of B or BH+), the log term in Henderson Hasselbach doesn’t change by much The buffer has its maximum ability to resist pH change at pH = pKa. (BUFFER CAPACITY). Dr. David Johnson: CH0004 Lecture Notes - Term 2

  19. Buffer Capacity: - how well a buffer solution resists pH change when an acid or base is added. Usually demonstrated as follows; A buffer is prepared such that the overall molarity is 1.00 (i.e. [HA] + [A-] = 1 mole). Then, 0.01 moles of either strong acid or strong base is added - and the new pH calculated. Results show the best result at: pH = pKa - see diagram on next sheet…... Dr. David Johnson: CH0004 Lecture Notes - Term 2

  20. Buffer Capacity: ±0.017 Maximum buffer capacity at pH = pKa Dr. David Johnson: CH0004 Lecture Notes - Term 2

  21. Buffer capacity - Examples: 4 points on previous diagram: Buffer prepared such that [HA] + [A-] = 1. Then 0.01 moles of either [H+] or [OH-] are added at a particular pH and the resultant CHANGE in pH is calculated. Examples from notes (remember pKa = 5) Method: (a) calculate initial moles of each species at the pH required (b) set up equilibrium table for changes as acid/base added (c) re-calculate using Henderson-Hasselbach Example 1: pH = 5 and 0.01 moles of acid added: (a) H/H pH = pKa + log ([A-] / [HA]) 5 = 5 + log ([A-] / [HA])  [A-] = [HA] egn X moles [HA] + [A-] = 1 mole eqn Y Combine X and Y to give: [HA] = 0.5 and [A-] = 0.5 (b) A- H+ HA Initial 0.5 0.01 0.5 Equilibrium 0.49 0.51 (c) New pH = 5 + log (0.49 / 0.51) = 4.983 Change in pH = -0.017 Example 2: pH = 5 and 0.01 moles of base added: Similar to above - except reverse: New pH = 5 + log (0.51 / 0.49) Change in pH = +0.017 Dr. David Johnson: CH0004 Lecture Notes - Term 2

  22. Buffer capacity - Examples: 4 points on previous diagram - continued: Example 3: pH = 3.5 and 0.01 moles of acid added: (a) H/H pH = pKa + log ([A-] / [HA]) 3.6 = 5 + log ([A-] / [HA])  [A-] = 0.0398 [HA] egn X moles [HA] + [A-] = 1 mole eqn Y Combine X and Y to give: [HA] = 0.9617 and [A-] = 0.0383 (b) A- H+ HA Initial 0.0383 0.01 0.9617 Equilibrium 0.0283 0.9717 (c) New pH = 3.6 + log (0.0283 / 0.9717) = 3.464 Change in pH = -0.136 Example 4: pH = 3.6 and 0.01 moles of base added: (a) same initial situation to above (b) HA OH- A- + H2O Initial 0.9617 0.01 0.0383 Equilibrium 0.9517 0.0483 (c) New pH = 3.6 + log (0.0483 / 0.9517) = 3.705 Change in pH = +0.105 Dr. David Johnson: CH0004 Lecture Notes - Term 2

  23. Commonly Used Buffers: Dr. David Johnson: CH0004 Lecture Notes - Term 2

  24. Preparing Buffers: Buffers are usually prepared by starting with a measured amount of either a weak acid (HA) or a weak base (B). Then OH- is added to make a mixture of HA / A- (a buffer) or H+ is added to make a mixture of B / BH+ (a buffer). Example calculation: How many millilitres (cm3) of 0.500 M NaOH should be added to 10.00 g of tris hydrochloride (BH+, from previous example) in order to give a buffer of pH 7.60 in a volume of 250 cm3? ____________________________________ no moles tris hydrochloride: = 10 / 157.60 [RMM] = 0.0635 moles Dr. David Johnson: CH0004 Lecture Notes - Term 2

  25. BH+ OH- B initial moles: 0.0635 x 0 final moles: (0.0635 - x) 0 x Use Henderson Hasselbach 7.60 = 8.075 + log ([salt] / [acid]) -0.475 = log { x / (0.0635 - x) } 0.335 = x / (0.0635 - x) x = 0.0159 mol so, what volume of 0.5 M NaOH gives 0.0159 mol 0.5 V = 0.0159 V = 0.0318 L = 31.8 cm3 i.e. mix 31.8 cm3 of 0.5 M NaOH with 10.00 g tris in 250 cm3 (note: volume not important in buffers really!!!) to give a pH of 7.60. Dr. David Johnson: CH0004 Lecture Notes - Term 2

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