[H O ][CH COO ] [CH COOH] 0.2 x

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Evan Fox
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1 CEM1612 Answers to Problem Sheet 6 1. (a) 0.2 M acetic acid As acetic acid is a weak acid, [ 3 O + ] must be calculated: C 3 2 O 3 O + C 3 initial 0.2 large 0 0 change - negligible + + final 0.2 large The equilibrium constant K a is given by: K a [ O ][C ] = = [C ] As pk a = 4.76 = log 10 K a so K a = As K a is very small, 0.2 ~ 0.2 and hence: 2 = or = M = [ 3 O + ] ence, the p is given by: p = log 10 [ 3 O + ] = log 10 [0.0019] = 2.7 (b) 0.2 M sodium acetate As C 3 is a weak base, [O ] must be calculated in a similar way: C 3 2 O O C 3 initial 0.2 large 0 0 change - negligible + + final 0.2 large The equilibrium constant K b is given by: K - [O ][C 2 3] b = = - [C ] For an acid and its conjugate base: so pk a + pk b = pk b = = 9.24

2 As pk b = 9.24, K b = Again, K b is very small, 0.2 ~ 0.2 and hence: 2 = or = M = [O ] ence, the po is given by: po = log 10 [O ] = log 10 [ ] = 5.0 Finally, p + po = 14 so p = = 9.0 (c) A buffer that is 0.2 M in acetic acid and 0.2 M in sodium acetate This solution contains an acid and its conjugate base so the enderson- asselbalch equation can be used: p = pk a + log10 As [acetic acid] = [sodium acetate], log10 = log 10(1) = 0 and so p = pk a = (a) As the α- group has the lowest pk a value, it is the most acidic. At p = 1.81, the α- group is in equilibrium with its conjugate base. The pk a values of the imidazole - and the α- + 3 groups are higher than the p value, both eist predominately in the protonated form at p = acid conjugate base (b) As p = 6.05 is higher than its pk a value, the α- group will eist predominately in its conjugate base form. The pk a value for the α- 3 + group is higher than the p value so it will eist predominately in its protonated form. The ring - group will be in equilibrium with its conjugate base form.

3 3 3 acid conjugate base (c) As p = 9.15 is higher than their pk a values, the α- and imidazole - groups will eist predominately in their conjugate base forms. The α- 3 group is in equilibrium with its conjugate base form The equilibrium of interest is: acid conjugate base 2 PO 4 - (aq) + 2 O(l) PO 4 2- (aq) + 3 O + (l) where 2 PO 4 _ is the acid and PO 4 2- is the base. The enderson-asselbalch equation can be used to calculate the required concentrations of each: p = pk a + log10 or log10 = p pk a log 10 = ( ) = 0.20 ence, = Both the acid and base have initial concentrations of 0.10 M so the ratio of the volume of each required is: V base = V acid = 1.58

4 When the two solutions are added together, their total volume = V base + V acid = 1.0 L. ence, V acid = V base V base = 1.0 V base 1.58 so V base = 0.61 L and V acid = 1.0 V base = 0.39 L 4. The concentration of the acid that has dissociated at equilibrium can be calculated from the p: A 2 O 3 O + A initial 0.6 large 0 0 change - negligible + + final 0.6 large At equilibrium, [ 3 O + ] = [ 3 O + ] from A + [ 3 O + ] from water. As [ 3 O + ] from water ~ 10-7, it can be assumed that the p is due to the [ 3 O + ] from A. Therefore: so p = log 10 [ 3 O + ] ~ log[ 3 O + ] from A = log = 4.0 = 10 4 M The initial concentration of A is 0.6 M. The percentage dissociation is: + from A 4 4 [ O 3 ] = 100% = 100% = 0.017% [ A] + [ A ] (a) The titration is a 1:1 reaction between a weak acid and a strong base. (i) Before addition of any base, the p of the M acetic solution follows the same method used in Q1(a). Following the same steps (with a M solution in place of a 0.2 M solution) gives the equilibrium constant K a is given by: K a [ O ][C ] = = [C ] With the approimation that ~ 0.100, 2 = or = M = [ 3 O + ] ence, the p is given by: p = log 10 [ 3 O + ] = log 10 [ ] = 2.88 (ii) Addition of the strong base O - (aq) leads to the neutralization reaction:

5 C 3 (aq) + O - (aq) à C 3 - (aq) + 2 O(l) The number of moles of C 3 - is equal to the number of moles of O - (aq) which have been added. The number of moles of C 3 remaining is equal to the initial number of moles of C 3 minus the number of moles of O - (aq) which have been added. Initially, there is 50.0 ml of M C 3 so n(c 3 ) initial = concentration volume = M L = mol 25.0 ml of M ao contains n(o - ) added = concentration volume = M L = mol Thus, after addition and the neutralization reaction, there are: n(c 3 ) = n(c 3 ) initial - n(o - ) added = ( ) mol = mol n(c 3 - ) = n(o - ) added = mol These quantities are now in a solution with total volume ( ) ml = 75.0 ml, so their concentrations are: c(c 3 ) = n / V = mol / L = M c(c 3 - ) = n / V = mol / L = M Using the enderson-asselbalch equation gives p = pk a + log M = log10 = M At the ½ equivalence point, when the number of moles of base added is equal to ½ the number of moles of acid originally present, the p is equal to the pk a. (iii) 45.0 ml of M ao corresponds to n(o - ) added = concentration volume = M L = mol Thus, after addition and the neutralization reaction, there are: n(c 3 ) = n(c 3 ) initial - n(o - ) added = ( ) mol = mol

6 n(c 3 - ) = n(o - ) added = mol These quantities are now in a solution with total volume ( ) ml = 95.0 ml, so their concentrations are: c(c 3 ) = n / V = mol / L = M c(c 3 - ) = n / V = mol / L = M Using the enderson-asselbalch equation gives M p = log10 = M 5.71 (ote: In fact one could have skipped the conversion of moles into concentration step, because the volume of solution is the same for C 3 and C3 - and it would just cancel out) (iv) The addition of 50.0 ml of M ao corresponds to the equivalence point as the number of moles of added base is equal to the number of moles of acid initially present. owever, the p is not equal to 7 for this point of the titration of a weak acid with a strong base ml of M ao corresponds mol so whilst this leads to n(c 3 ) = 0.00 mol, it also leads to n(c 3 - ) = mol. A solution of a weak base has p > 7. As the total volume is ( ) ml = ml, the concentration of C 3 - is: c(c 3 - ) = n / V = mol / L = M The calculation of the p now follows that in Q1(b) with the concentration of M: K - [O ][C 2 3] b - [C3 ] = = = As pk b = 9.24, K b = Again, K b is very small, ~ and hence: 2 = or = M = [O ] ence, the po is given by: po = log 10 [O ] = log 10 [ ] = 5.27 Finally, p + po = 14 so

7 p = = 8.73 (v) Beyond the equivalence point, ecess base is being added to a solution of a weak base and the p is controlled by the ecess amount of O - (aq). The calculation of the p is then eactly the same as in Q7(a)(v) from Problem Sheet ml of M ao corresponds to mol. Of this, mol is used to react with the acid. The total volume is ( ) L = L. ence ( mol) ( mol) [O (aq)] = = M (0.105 L) ence, po = -log 10 ( ) = 2,32 and p = = (vi) 75.0 ml of M ao corresponds to mol Of this, mol is used to react with the acid and the total volume is ( ) L = L. ence ( mol) ( mol) [O (aq)] = = M (0.125 L) ence, po = -log 10 (0.0200) = 1.70 and p = = 12.3 (b) Using these values, the p curve for the titration can be constructed and is shown below (in pink) p volume of ao added / ml (c) The figure also includes the data from Q7 in Problem Sheet 5 for a strong acid / strong base titration (in blue). For a weak acid / strong base titration: (i) The initial p is higher. For a strong acid and a weak acid with the same concentration, the p of the strong acid is lower.

8 (ii) (iii) (iv) The p at the ½ equivalence point is equal to the pk a of the weak acid. At the equivalence point, the solution contains weak base and so the p > 7. After the equivalence point, the p is determined only the concentration of ecess strong base and is thus the same for the two titrations.

Amino acids have the general structure: All amino acids have at least two acidic protons. One is the proton

Amino acids have the general structure: All amino acids have at least two acidic protons. One is the proton Worksheet rganic Acids and Bases II Amino Acids Acids are proton donors in aqueous solutions: A + 2 3 + + A - acid base conjugate conjugate base acid The strength of an acid is measured by the [ 3 + ]