Titration of Mixtures of Acids Using a Strong Base For two acids to titrate separately, the ka of the stronger one should be at least 104 times greater than the other. If this condition is not satisfied, an end point will be observed. Always, the stronger acid will be titrated first since it suppresses the dissociation of the weaker one.

Example Find the pH of a 50 mL solution containing 0. 10 M HCl and 0 Example Find the pH of a 50 mL solution containing 0.10 M HCl and 0.10 M HOAc (ka = 1.75x10-5) after addition of 0, 25, 50, 75, 100, and 150 mL of 0.10 M NaOH. Solution 1. After addition of 0 mL NaOH The solution contains a mixture of HCl and HOAc. We have seen earlier that HOAc dissociation will be suppressed in presence HCl and the pH can be calculated as follows: HOAc D H+ + OAc-

Ka = (0.10 + x) x/(0.10 – x) Assume 0.10 >> x since ka is small 1.75x10-5 = 0.10 x/0.10 x = 1.75x10-5 Relative error = (1.75x10-5/0.10) x100 = 1.8x10-2% Therefore [H+] = 0.10 + 1.8x10-5 = 0.1 pH = 1.0 It is clear that all H+ comes from the strong acid.

2. After addition of 25 mL NaOH initial mmol HCl = 0. 10 x 50 = 5 2. After addition of 25 mL NaOH initial mmol HCl = 0.10 x 50 = 5.0 mmol NaOH added = 0.10 x 25= 2.5 mmol HCl left = 5.0 – 2.5 = 2.5 [H+]HCl = 2.5/75 [HOAc] = 0.10 x 50 / 75 = 5.0/75 HOAc D H+ + OAc-

Ka = (2. 5/75 + x) x/((5. 0/75) – x) Assume 2 Ka = (2.5/75 + x) x/((5.0/75) – x) Assume 2.5/75 >> x since ka is small 1.75x10-5 = 2.5/75x/(5.0/75) x = 3.5 x10-5 Relative error = (3.5x10-5/(2.5/75)) x100 = 0.11% Therefore [H+] = 2.5/75 + 5.4x10-5 = 2.5/75 pH = 1.48 It is clear that all H+ still comes from the strong acid since dissociation of the weak acid is limited and in presence of strong acid the dissociation of the weak acid is further suppressed.

3. After addition of 50 mL NaOH Initial mmol HCl = 0. 10 x 50 = 5 3. After addition of 50 mL NaOH Initial mmol HCl = 0.10 x 50 = 5.0 mmol NaOH added = 0.10 x 50 = 5.0 mmol HCl left = 5.0 – 5.0 = ?? This is the first equivalence point. At this point the HCl is over and the solution contains HOAc only and we should calculate the pH of the solution from the h+ produced from dissociation of the HOAc. [HOAc] = 5.0/100 = 0.05 M HOAc D H+ + OAc-

Ka = [H+][OAc-]/[HOAc] Ka = x * x / (0.05 – x) Ka is very small. Assume 0.05 >> x 1.75*10-5 = x2/0.05 x = 9.4x10-4 Relative error = (9.4x10-4/0.05) x 100 = 1.9% The assumption is valid and the [H+] = 9.4x10-4 M pH = 3.03 pOH = 14 – 3.03 = 10.97

4. After addition of 75 mL NaOH Remember that 50 mL of NaOH were consumed in the reaction with HCl and thus only 25 mL NaOH will actually react with HOAc. Initial mmol HOAc = 0.10 x 50 = 5.0 Mmol NaOH added = 0.10 x 25 = 2.5 mL Mmpl HOAC left = 5.0 – 2.5 = 2.5 [HOAc] = 2.5/125 mmol OAc- formed = 2.5 [OAc-] = 2.5/125 Therefore, a buffer is formed and the pH can be calculated as follows: HOAc D H+ + OAc-

Ka = x (2. 5/125 – x)/ (2. 5/125 – x) Assume 2. 5/125 >> x 1 Ka = x (2.5/125 – x)/ (2.5/125 – x) Assume 2.5/125 >> x 1.75x10-5 = x (2.5/125)/2.5/125 x = 1.75x10-5 Relative error = {1.75x10-5/(2.5/125)} x 100 = 0.088% The assumption is valid [H+] = 1.75x10-5 pH = 4.76

5. After addition of 100 mL NaOH Remember that 50 mL of NaOH were consumed in the reaction with HCl and thus only 50 mL NaOH will actually react with HOAc. Initial mmol HOAc = 0.10 x 50 = 5.0 Mmol NaOH added = 0.10 x 50 = 5.0 mL Mmol HOAC left = 5.0 – 5.0 = ?? This is the second equivalence point mmol OAc- formed = 5.0 [OAc-] = 5.0/150 Therefore, we only have the salt where the pH can be calculated as follows: OAc- + H2O D HOAc + OH-

Kb = kw/ka Kb = 10-14/1.75x10-5 = 5.7x10-10 Kb = [HOAc][OH-]/[OAc-] Kb = x * x/(5.0/150 – x) , assume that 5.0/150 >>x 5.7x10-10 = x2/(5.0/150), x = 4.4x 10-6 Relative error = (4.4x10-6/(5.0/150)) x100 = 0.013% [OH-] = 4.4x10-6 M [H+] = 2.3x10-9 M = [OH-]water

Therefore, the [OH-]acetate >> [OH-]water pOH = 5.36 pH = 14 – 5.36= 8.64 6. After addition of 150 mL NaOH Remember that 50 mL NaOH were consumed in reaction with HCl and 50 mL were consumed in reaction with HOAc. Therefore, we have added 50 mL NaOH in excess: mmol NaOH excess = 0.10 x 50 = 5.0 [OH-] = 5.0/200 = 0.025 pOH = 1.6 pH = 12.4

Titration of a Mixture of Bases Using a Strong Acid The stronger base will be titrated first since most hydroxide comes from it. In a later stage, the weaker base will be titrated. Example Find the pH of a 50 mL solution containing 0.10 M NaOH and 0.10 M Na2CO3 (ka1 = 4.3x10-7, ka2 = 4.8x10-11) after addition of 0, 25, 50, 75, 100, 150, and 200 mL of 0.10 M HCl.

1. After addition of 0 mL HCl The presence of NaOH with carbonate suppresses the association of the carbonate with water and the OH- in solution will be mainly due to NaOH. We can be sure of that by the following calculation: CO32- + H2O D HCO3- + OH-

Kb = 10-14/4. 8x10-11 = 2. 1x10-4 Kb = (0. 10 + x) x/(0 Kb = 10-14/4.8x10-11 = 2.1x10-4 Kb = (0.10 + x) x/(0.10 – x) Assume 0.10>>x since kb is small 2.1*10-4= x x = 2.1x10-4 Relative error = (2.1 x10-4/0.10) x 100 = 0.21% The assumption is valid. [OH-]CO32- = 2.1x10-4 M [OH-] = 0.10 + 2.1x10-4 = 0.1 pOH = 1.0 pH = 14 -1.0 = 13.0

2. After addition of 25 mL HCl Initial mmol NaOH = 0. 10 x 50 = 5 2. After addition of 25 mL HCl Initial mmol NaOH = 0.10 x 50 = 5.0 Mmol HCl added = 0.10 x 25 = 2.5 Mmol NaOH left = 5.0 – 2.5 = 2.5 [OH-] = 2.5/75 [CO32-] = 0.10x50/75 = 5.0/75 CO32- + H2O D HCO3- + OH-

Kb = 10-14/4. 8x10-11 = 2. 1x10-4 Kb = (2. 5/75 + x) x/((5 Kb = 10-14/4.8x10-11 = 2.1x10-4 Kb = (2.5/75 + x) x/((5.0/75) – x) Assume 2.5/75>>x since kb is small 2.1*10-4= (2.5/75) x/(5.0/75) x = 4.2x10-4 Relative error = (4.2 x10-4/(2.5/75)) x 100 = 1.26% The assumption is valid. [OH-]CO32- = 4.2x10-4 M [OH-] = 2.5/75 + 4.2 x10-4 = 2.5/75 [OH-] = 2.5/75 pOH = 1.48 pH = 14 -1.48 = 12.52

3. After addition of 50 mL HCl Initial mmol NaOH = 0. 10 x 50 = 5 3. After addition of 50 mL HCl Initial mmol NaOH = 0.10 x 50 = 5.0 mmol HCl added = 0.10 x 50 = 5.0 mmol NaOH left = 5.0 – 5.0 = 0 This is the first equivalence point where all the hydroxide is consumed and we have a solution of carbonate only. The pH is calculated for that solution as follows: mmol CO32- = 0.10 x 50 = 5.0 [CO32-] = 5.0/100 = 0.05 CO32- + H2O D HCO3- + OH-

Kb = 10-14/4. 8x10-11 = 2. 1x10-4 Kb = x2 /(0. 05 – x) Assume 0 Kb = 10-14/4.8x10-11 = 2.1x10-4 Kb = x2 /(0.05 – x) Assume 0.05 >> x since kb is small , x = 3.2x10-3 Relative error = (3.2 x10-3/0.05) x 100 = 6.4% The assumption is invalid but I'll accept it and the [OH-] = 3.2x10-3 M pOH = 2.49 pH = 14 -2.49 = 11.51

4. After addition of 75 mL HCl The first 50 mL were consumed in the reaction with NaOH and only 25 mL will react with the carbonate Initial mmol CO32- = 0.10 x 50 = 5.0 mmol HCl added = 0.10 x 25 = 2.5 mmol CO32- left = 5.0 – 2.5 = 2.5 [CO32-] = 2.5/125 mmol HCO3- formed = 2.5 [HCO3-] = 2.5/125 This is a buffer and the pH of the solution could be calculated as follows: CO32- + H2O D HCO3- + OH-

Kb = 10-14/4. 8x10-11 = 2. 1x10-4 Kb = (2. 5/125 + x) x/(2 Kb = 10-14/4.8x10-11 = 2.1x10-4 Kb = (2.5/125 + x) x/(2.5/125 – x) Assume 2.5/125 >> x since kb is small , X = 2.1x10-4 Relative error = (2.1 x10-4/(2.5/125)) x 100 = 1.1% The assumption is valid and [OH-] = 2.1x10-4 M pOH = 3.68 pH = 14 – 3.68 = 10.32

5. After addition of 100 mL HCl Remember that 50 mL of HCl were consumed in the reaction with NaOH. Therefore, only 50 mL of added HCl will react with carbonate. mmol CO32- left = 0.10 x 50 – 0.10 x 50 = 0 This is a second equivalence point where all CO32- is converted to HCO3- mmol HCO3- formed = 5.0 [HCO3-] = 5.0/150 M [H+] = {(ka1kw + ka1ka2[HCO3-])/(ka1 + [HCO3‑]}1/2 [H+] = {(4.3x10-7 * 10-14 + 4.3x10-7 * 4.8x10-11 *(5.0/150))/(4.3x10-7 + (5.0/150))}1/2 [H+] = 4.5x10-9 M pH = 8.34

6. After addition of 150 mL HCl Remember that the first 50 mL of HCl are needed to react with NaOH, the second 50 mL are required to convert CO32- into HCO3-. Therefore, only 50 mL are available to react with the HCO3-. Initial mmol HCO3- = 5.0 mmol HCl added = 0.10 x 50 = 5.0 mmol HCO3- left = 5.0 – 5.0 = 0 This is a third equivalence point where all HCO3- is converted to H2CO3. The pH can be calculated as follows mmol H2CO3 formed = 5.0 [H2CO3] = 5/200 = 0.025

H2CO3 D H+ + HCO3- ka1 = 4. 3 x 10-7 HCO3- D H+ + CO32- ka2 = 4 H2CO3 D H+ + HCO3- ka1 = 4.3 x 10-7 HCO3- D H+ + CO32- ka2 = 4.8 x 10-11 Since ka1 is much greater than ka2 Ka1 = x * x/(0.025 – x) Assume 0.025>>x since ka1 is small 4.3*10-7= x2/0.025, x = 1.04x10-4 Relative error = (1.04x10-4/0.025) x 100 = 0.41% The assumption is valid and [H+] = 1.0 x10-4 M pH = 4.00

7. After addition of 200 mL HCl Remember that 50 mL of the added HCl will react with NaOH, another 50 mL of HCl will be consumed in converting the carbonate to bicarbonate. A third 50 mL portion of HCl will be consumed in converting the bicarbonate to carbonic acid. We then have 50 mL of excess HCl only mmol HCl = 0.10 x 50 = 5.0 [H+] = 5/250 = 0.02 pH = 1.7 You should be able to calculate [H+]H2CO3

Kjeldahl Analysis An application of acid-base titrations that finds an important use in analytical chemistry is what is called Kjeldahl nitrogen analysis. This analysis is used for the determination of nitrogen in proteins and other nitrogen containing compounds. Usually, the quantity of proteins can be estimated from the amount of nitrogen they contain. The Kjeldahl analysis involves the following steps:

1. Digestion of the nitrogen containing compound and converting the nitrogen to ammonium hydrogen sulfate. This process is accomplished by decomposing the nitrogen containing compound with sulfuric acid. 2. The solution in step 1 is made alkaline by addition of concentrated NaOH which coverts ammonium to gaseous ammonia , and the solution is distilled to drive the ammonia out. 3. The ammonia produced in step 2 is collected in a specific volume of a standard acid solution (dilute) where neutralization occurs. 4. The solution in step 3 is back-titrated against a standard NaOH solution to determine excess acid. 5. mmoles of ammonia are then calculated and related to mmol nitrogen.

Example A 0.200 g of a urea (FW = 60, (NH2)2CO) sample is analyzed by the Kjeldahl method. The ammonia is collected in a 50 mL of 0.05 M H2SO4. The excess acid required 3.4 mL of 0.05 M NaOH. Find the percentage of the compound in the sample. Solution 2 NH3 + H2SO4 = (NH4)2SO4 mmol H2SO4 reacted = mmol H2SO4 taken – mmol H2SO4 back-titrated

mmol H2SO4 reacted = mmol H2SO4 taken – mmol H2SO4 back-titrated mmol H2SO4 back-titrated = 1/2 mmol NaOH ½ mmol ammonia = mmol H2SO4 reacted 1/2 mmol ammonia = 0.05 x 50 – 1/2 x 0.05 x 3.4 mmol ammonia = 2(0.05 x 50 – 1/2 x 0.05 x 3.4) mmol urea = 1/2 mmol ammonia mmol urea = 0.05 x 50 – 1/2 x 0.05 x 3.4 = 2.415 mg urea = 2.415x60 = 144.9 % urea = (144.9/200)x100 = 72.5%

Modified Kjeldahl Analysis In conventional Kjeldahl method we need two standard solutions, an acid for collecting evolved ammonia and a base for back-titrating the acid. In a modified procedure, only a standard acid is required. In this procedure, ammonia is collected in a solution of dilute boric acid, the concentration of which need not be known accurately. The result of the reaction is the borate which is equivalent to ammonia. NH3 + H3BO3 g NH4+ + H2BO3- Boric acid is a very weak acid which is not titratable while borate is a strong base which can be titrated with a standard HCl solution.

Example A 0.300 g feed sample is analyzed for its protein content by the modified Kjeldahl method. If 25 mL of 0.10 M HCl is required for the titration what is the percent protein content of the sample (mg protein = 6.25 mg N). Solution mmol N = mmol HCl mmol N = 0.10 x 25 = 2.5 mmol N = 2.5 mg N = 2.5 x 14 = 35 mg protein = 35 x 6.25 = 218.8 % protein = (218.8/300) x 100 = 72.9%