Titration of a Weak Base with a Strong Acid The same principles applied above are also applicable where we have: 1. Before addition of any acid, we have a solution of the weak base and calculation of the pH of the weak base should be performed as in previous sections. 2. After starting addition of the strong acid to the weak base, the salt of the weak base is formed. Therefore, a buffer solution results and you should consult previous lectures to find out how the pH of buffers is calculated.

3. At the equivalence point, the amount of strong acid is exactly equivalent to the weak base and thus there will be 100% conversion of the weak base to its salt. The problem now is to calculate the pH of the salt solution. 4. After the equivalence point, we would have a solution of the salt with excess strong acid. The presence of the excess acid suppresses the dissociation of the salt in water and the pH of the solution controlled by the excess acid only. Now, let us apply the abovementioned concepts on an actual titration of a weak base with a strong acid.

Example Find the pH of a 50 mL solution of 0. 10 M NH3 (kb = 1 Example Find the pH of a 50 mL solution of 0.10 M NH3 (kb = 1.75x10-5) after addition of 0, 10, 25, 50, 60 and 100 mL of 0.10 M HCl. Solution 1. After addition of 0 mL HCl The solution is only 0.10 M in ammonia, therefore we have: NH3 + H2O D NH4+ + OH-

Kb = [NH4+][OH-]/[NH3] 1. 75. 10-5 = x. x / (0 Kb = [NH4+][OH-]/[NH3] 1.75*10-5 = x * x / (0.10 – x) kb is very small that we can assume that 0.10>>x. We then have: 1.75*10-5 = x2 / 0.10 x = 1.3x10-3 M Relative error = (1.3x10-3 /0.10) x 100 = 1.3 % The assumption is valid, therefore: [OH-] = 1.3x10-3 M pOH = 2.88 pH = 11.12

2. After addition of 10 mL HCl A buffer will be formed from the base and its salt Initial mmol NH3 = 0.10 x 50 = 5.0 mmol HCl added = 0.10 x 10 = 1.0 mmol NH3 left = 5.0 – 1.0 = 4.0 [NH3] = 4.0/60 M mmol NH4+ formed = 1.0 [NH4+] = 1.0/60 M NH3 + H2O D NH4+ + OH-

Kb = [NH4+][OH-]/[NH3] 1. 75. 10-5 = (1. 0/60 + x). x / (4 Kb = [NH4+][OH-]/[NH3] 1.75*10-5 = (1.0/60 + x) * x / (4.0/60 – x) kb is very small that we can assume that 1.0/60 >>x. We then have: 1.75*10-5 = 1.0/60 x / 4.0/60 x = 7.0x10-5 Relative error = (7.0x10-5 /1.0/60) x 100 = 0.42 % The assumption is valid, therefore: [OH-] = 7.0x10-5 M pOH = 4.15 pH = 9.85

3. After addition of 25 mL HCl A buffer will be formed from the base and its salt Initial mmol NH3 = 0.10 x 50 = 5.0 mmol HCl added = 0.10 x 25 = 2.5 mmol NH3 left = 5.0 – 2.5 = 2.5, [NH3] = 2.5/75 M mmol NH4+ formed = 2.5, [NH4+] = 2.5/75 M NH3 + H2O D NH4+ + OH-

Kb = [NH4+][OH-]/[NH3] 1. 75. 10-5 = (2. 5/75 + x). x / (2 Kb = [NH4+][OH-]/[NH3] 1.75*10-5 = (2.5/75 + x) * x / (2.5/75 – x) kb is very small that we can assume that 2.5/75 >>x. We then have: 1.75*10-5 = 2.5/75 x / 2.5/75 x = 1.75x10-5 Relative error = (1.75 x10-5 /2.5/75) x 100 = 0.052 % The assumption is valid, therefore: [OH-] = 1.75x10-5 M pOH = 4.76 pH = 9.24

3. After addition of 50 mL HCl A buffer will be formed from the base and its salt Initial mmol NH3 = 0.10 x 50 = 5.0 mmol HCl added = 0.10 x 50 = 2.5 mmol NH3 left = 5.0 – 25.0 = 0 This is the equivalence point mmol NH4+ formed = 5.0, [NH4+] = 5.0/100 = 0.05 M NH4+ D H+ + NH3

Ka = 10-14/1.75x10-5 = 5.7x10-10 Ka = [H+][NH3]/[NH4+] Ka = x * x / (0.05 – x) Ka is very small. Assume 0.05 >> x 5.7*10-10 = x2/0.05 x = 5.33x10-6 Relative error = (5.33x10-6/0.05) x 100 = 0.011 % The assumption is valid and the [H+] = 5.33x10-6 M pH = 5.27

5. After addition of 60 mL HCl Initial mmol of NH3 = 0. 10 x 50 = 5 5. After addition of 60 mL HCl Initial mmol of NH3 = 0.10 x 50 = 5.0 Mmol HCl added = 0.10 x 60 = 6.0 Mmol HCl excess = 6.0 – 5.0 = 1.0 [H+] = 1.0/110 M pH = 2.04 6. After addition of 100 mL HCl Mmol HCl added = 0.10 x 100 = 10.0 Mmol HCl excess = 10.0 – 5.0 = 5.0 [H+] = 5.0/150 M pH = 1.48

Titration of a Polyprotic Acid with a Strong Base Each proton in a polyprotic acid is supposed to titrate separately. However, only those protons which satisfy the empirical relation ka1 > 104 ka2 can result in an observable break at the point of equivalence. For example, carbonic acid shows two breaks in the titration curve. Each one corresponds to a specific proton of the acid. The method of calculation of the pH is similar to that described above but initially for the first proton then the second. Each equivalence point requires a separate indicator to visualize the end point.

There are few points to put in mind when dealing with problems of titration of polyprotic acids with strong bases: 1. Before addition of any base, you only have the polyprotic acid solution and thus calculation of the pH is straightforward as previously described. 2. When we start addition of base, the first proton is titrated and bicarbonate will form. A buffer solution of carbonic acid and carbonate is formed and you should refer to the section on such calculations. 3. When all the first proton is titrated, all carbonic acid is now converted to bicarbonate (an amphoteric protonated salt) and calculation of the pH is achieved using the appropriate root mean square equation.

4. Further addition of base starts titrating the second proton thus some bicarbonate is converted to carbonate and a buffer is formed. Calculate the pH of the resulting buffer in the same way as in step 2. 5. When enough base is added so that the titration of the second proton is complete, all bicarbonate is converted to carbonate and this is the second equivalence point. The pH is calculated for carbonate (unprotonated salt). 6. Addition of excess base will make the solution basic where this will suppress the dissociation of carbonate. The hydrogen ion concentration is calculated from the concentration of excess hydroxide.

After addition of 0 mL NaOH Example Find the pH of a 50 mL solution of a 0.10 M H2CO3 after addition of 0, 25, 50, 75, 100, and 150 mL of 0.10 M NaOH. Ka1=4.3x10-7 and ka2 = 4.8x10-11. Solution After addition of 0 mL NaOH We only have the carbonic acid solution and the pH calculation for such types of solution was discussed earlier and can be worked as below: 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, we can neglect the H+ from the second step and therefore we have: H2CO3 D H+ + HCO3- ka1 = 4.3 x 10-7 Ka1 = x * x/(0.10 – x) Assume 0.10>>x since ka1 is small 4.3*10-7= x2/0.10, x = 2.1x10-4 Relative error = (2.1x10-4/0.10) x 100 = 0.21% The assumption is valid and [H+] = 2.1x10-4 M, pH = 3.68

After addition of 25 mL NaOH A buffer is formed from H2CO3 left and the formed HCO3- Initial mmol H2CO3 = 0.10 x 50 = 5.0 Mmol NaOH added = 0.10 x 25 = 2.5 Mmol H2CO3 left = 5.0 – 2.5 = 2.5 [H2CO3] = 2.5/75 M mmol HCO3- formed = 2.5 [HCO3-] = 2.5/75 M H2CO3 D H+ + HCO3- ka1 = 4.3 x 10-7

The assumption is valid [H+] = 4.3x10-7 M pH = 6.37 ka1 = x(2.5/75 + x)/(2.5/75 – x) ka1 is very small and in presence of the common ion the dissociation will be further suppressed. Therefore, assume 2.5/75>>x. x = 4.3x10-7 M Relative error = {4.3x10-7/(2.5/75)} x 100 = 0.0013% The assumption is valid [H+] = 4.3x10-7 M pH = 6.37

After addition of 50 mL NaOH Initial mmol H2CO3 = 0. 10 x 50 = 5 After addition of 50 mL NaOH Initial mmol H2CO3 = 0.10 x 50 = 5.0 mmol NaOH added = 0.10 x 50 = 5.0 mmol H2CO3 left = 5.0 – 5.0 = 0 This is the first equivalence point mmol HCO3- formed = 5.0 [HCO3-] = 5.0/100 = 0.05 M Now the solution contains only the protonated salt. Calculation of the pH can be done using the relation [H+] = {(ka1kw + ka1ka2[HCO3-])/(ka1 + [HCO3‑]}1/2 [H+] = {(4.3x10-7 * 10-14 + 4.3x10-7 * 4.8x10-11 * 0.0.05)/(4.3x10-7 + 0.0.05)}1/2 [H+] = 4.5x10-9 M pH = 8.34

After addition of 75 mL NaOH Here you should remember that 50 mL of the NaOH will be used in the titration of the first proton. Therefore, it is as if we add 25 mL to the HCO3- solution. We then have: Initial mmol HCO3- = 5.0 Mmol NaOH added = 0.10 x 25 = 2.5 Mmol HCO3- left = 5.0 – 2.5 = 2.5 [HCO3-] = 2.5/125 M mmol CO32- formed = 2.5 [CO32-] = 2.5/125 M Once again we have a buffer solution from HCO3- and CO32-. The pH is calculated as follows: HCO3- D H+ + CO32- ka2 = 4.8 x 10-11

ka1 = x(2.5/125 + x)/(2.5/125 – x) ka1 is very small and in presence of the common ion the dissociation will be further suppressed. Therefore, assume 2.5/125>>x. x = 4.8x10-11 M Relative error = {4.8x10-11/(2.5/125)} x 100 = V. small The assumption is valid, [H+] = 4.8x10-11 M pH = 10.32

After addition of 100 mL NaOH At this point, all carbonic acid was converted into carbonate. The first 50 mL of NaOH were consumed in converting H2CO3 to HCO3-. Therefore, as if we add 50 mL to HCO3- solution and we have: Initial mmol HCO3- = 5.0 mmol NaOH added = 0.10 x 50 = 5.0 mmol HCO3- left = 5.0 – 5.0 = ?? This is the second equivalence point mmol CO32- formed = 5.0 [CO32-] = 5.0/150 M

CO32- + H2O = HCO3- + OH- Kb = kw/ka2 We used ka2 since it is the equilibrium constant describing relation between CO32- and HCO3-. However, in any equilibrium involving salts look at the highest charge on any anion to find which ka to use. Kb = 10-14/4.8x10-13 = 2.1x10-4

Kb = x. x/(5. 0/150 – x) Assume 5. 0/150 >> x 2. 1x10-4 = x2/(5 Kb = x * x/(5.0/150 – x) Assume 5.0/150 >> x 2.1x10-4 = x2/(5.0/150) x = 2.6x10-3 Relative error = (2.6x10-3 /(5.0/150)) x 100 = 7.9% Therefore, assumption is invalid and we have to use the quadratic equation. However, I’ll accept the answer this time. Therefore, [OH-] = 2.6x10-3 M pOH = 2.58 pH = 14 – 2.58 = 11.42

After addition of 150 mL NaOH At this point, all carbonic acid was converted into carbonate requiring 100 mL NaOH. mmol NaOH excess = 0.1 x 50 = 5.0 [OH-] = 5.0/200 pOH = 1.60 pH = 14.00 – 1.60 = 12.40

Titration of a Polybasic base with a Strong Acid Example Find the pH of a 50 mL solution of a 0.10 M Na3PO4 (ka1 = 1.1x10-2, ka2 = 7.5x10-8, ka3 = 4.8x10-13) after addition of 0, 25, 50, 75, 100, 125, 150, and 175 mL of 0.10 M HCl. Solution 1. After addition of 0 mL HCl At this point, we only have the solution of PO43- (an unprotonated salt) and we can find the pH as follows

PO43- + H2O D HPO42- + OH- kb= kw/ka3 We used ka3 since it is the equilibrium constant describing relation between PO43- and HPO42-. However, in any equilibrium involving salts look at the highest charge on any anion to find which ka to use. Kb = 10-14/4.8x10-13 = 0.020

Kb = x. x/0. 10 – x Assume 0. 10 >> x 0. 02 = x2/0. 10 x = 0 Kb = x * x/0.10 – x Assume 0.10 >> x 0.02 = x2/0.10 x = 0.045 Relative error = (0.045/0.10) x 100 = 45% Therefore, assumption is invalid and we have to use the quadratic equation. If we solve the quadratic equation we get: X = 0.036 Therefore, [OH-] = 0.036 M pOH = 1.44 pH = 14 – 1.44 = 12.56

2. After addition of 25 mL HCl A buffer starts forming from phosphate remaining and the hydrogen phosphate produced from the reaction. PO43- + H+ D HPO42- Initial mmol PO43- = 0.10 x 50 = 5.0 Mmol H+ added = 0.10 x 25 = 2.5 Mmol PO43- left = 5.0 – 2.5 = 2.5 [PO43-] = 2.5/75 M mmol HPO42- formed = 2.5 [HPO42-] = 2.5/75 M Now we look at any dissociation equilibrium equation containing both species. This can be obtained from the relation from ka3, for example

HPO42- D PO43- + H+ Ka3 = x(2. 5/75 + x)/(2 HPO42- D PO43- + H+ Ka3 = x(2.5/75 + x)/(2.5/75 – x) Since ka3 is very small, assume 2.5/75 >> x 4.8x10-13 = x(2.5/75)/(2.5/75) x = 4.8x10-13 It is clear that the relative error will be exceedingly small and the assumption is, for sure, valid. [H+] = 4.8x10-13 M pH = 12.32

3. After addition of 50 mL HCl At this point, all PO43- will be converted to HPO42- Initial mmol PO43- = 0.10 x 50 = 5.0 mmol H+ added = 0.10 x 50 = 5.0 mmol PO43- left = 5.0 – 5.0 = ?? This is the first equivalence point mmol HPO42- formed = 5.0 [HPO42-] = 5.0/100 = 0.05 M This is a protonated salt with two charges where we should use ka2 and ka3, i.e. the relation [H+] = {(ka2kw + ka2ka3[HPO42-])/(ka2 + [HPO42‑]}1/2 [H+] = 2.3x10-10 pH = 9.65

4. After addition of 75 mL HCl A second buffer is formed where we have HPO42- + H+ D H2PO4- You should understand that 50 mL were consumed in the conversion of PO43- to HPO42-, thus 25 mL only were added to HPO42- Initial mmol HPO4- = 0.10 x 50 = 5.0 mmol H+ added = 0.10 x 25 = 2.5 mmol HPO42- left = 5.0 – 2.5 = 2.5 [HPO42-] = 2.5/125 mmol H2PO42- formed = 2.5 [H2PO4-] = 2.5/125

Since ka3 is very small, assume 2.5/125 >> x H2PO4- D H+ + HPO42- 2.5/125 Before Equilibrium H+ HPO42- H2PO42- Equation x 2.5/125 + x 2.5/125 – x At Equilibrium Ka2 = x(2.5/125 + x)/(2.5/125 – x) Since ka3 is very small, assume 2.5/125 >> x 7.5x10-8 = x(2.5/125)/(2.5/125) x = 7.5x10-8 It is clear that the relative error will be exceedingly small and the assumption is, for sure, valid [H+] = 7.5x10-8 M pH = 7.12

5. After addition of 100 mL HCl 50 mL of HCl were consumed in converting PO43- into HPO42- Initial mmol HPO42- = 0.10 x 50 = 5.0 mmol H+ added = 0.10 x 50 = 5.0 mmol HPO42- left = 5.0 – 5.0 = 0 This is the second equivalence point [H2PO4-] = 5.0/150 = 0.033 M At this point, all HPO42- will be completely converted into H2PO4- which is a protonated salt where the pH can be calculated from the relation [H+] = {(ka1kw + ka1ka2[H2PO4-])/(ka1 + [H2PO4‑]}1/2 [H+] = 2.5x10-5 M pH = 4.6

6. After addition of 125 mL HCl H2PO4- + H+ D H3PO4 50 mL were consumed in converting PO43- to HPO42- and 50 mL were consumed in converting HPO42- into H2PO4-, therefore as if we add 25 mL to H2PO4- Initial mmol H2PO4- = 0.10 x 50 = 5.0 Mmol H+ added = 0.10 x 25 = 2.5 Mmol HPO42- left = 5.0 – 2.5 = 2.5 [HPO42-] = 2.5/175 M mmol H3PO4 formed = 2.5 [H3PO4] = 2.5/175 M This is a buffer formed from the acid and its conjugate base. The best way to calculate the pH is to use the ka1 expression where:

Since ka1 is very small (!!!), assume 2.5/175 >> x H3PO4 D H+ + H2PO4- 2.5/175 Before Equilibrium H+ H2PO4- H3PO4 Equation x 2.5/175 + x 2.5/175 – x At Equilibrium Ka1 = x(2.5/175 + x)/(2.5/175 – x) Since ka1 is very small (!!!), assume 2.5/175 >> x 1.1x10-2 = x(2.5/175)/(2.5/175) x = 1.1x10-2 M Relative error = {1.1x10-2/(2.5/175)} x 100 = 77% It is clear that the relative error is very large and the assumption is, for sure, invalid and we should use the quadratic equation. [H+] = 5.2x10-3 M pH = 2.29

7. After addition of 150 mL HCl At this point, all PO43- is converted into the acid Initial mmol H2PO4- = 0.10 x 50 = 5.0 Mmol H+ added = 0.10 x 50 = 5.0 Mmol HPO42- left = 5.0 – 5.0 = 0 This is the third equivalence point mmol H3PO4 formed = 5.0 [H3PO4] = 5/200 = 0.025

Therefore, we only have the acid in solution and calculation of the pH is done as follows: H3PO4 D H+ + H2PO4- ka1 = 1.1 x 10-2 H2PO4- D H+ + HPO42- ka2 = 7.5 x 10-8 HPO42- D H+ + PO43- ka3 = 4.8 x 10-13 Since ka1 >> ka2 (ka1/ka2 > 104) the amount of H+ from the second and consecutive equilibria is negligible if compared to that coming from the first equilibrium. Therefore, we can say that we only have:

H3PO4 D H+ + H2PO4- ka1 = 1. 1 x 10-2 Ka1 = x. x/(0. 025 – x) Assume 0 H3PO4 D H+ + H2PO4- ka1 = 1.1 x 10-2 Ka1 = x * x/(0.025 – x) Assume 0.025>>x since ka1 is small (!!!) 1.1*10-2 = x2/0.025 x = 0.017 Relative error = (0.017/0.025) x 100 = 68% The assumption is invalid according to the criteria we set at 5% and thus we have to use the quadratic equation.

After addition of 175 mL HCl 50 mL were consumed in converting PO43- to HPO42-, 50 mL were consumed in converting HPO42- into H2PO4-, and 50 mL HCl were consumed in converting H2PO4- to H3PO4, therefore, 25 mL of excess HCl are added mmol H+ excess = 0.10 x 25 = 2.5 [H+]excess = 2.5/225 = 0.011 M [H+] = [H+]excess + [H+]H3PO4

Since at this point both H+ from excess HCl and phosphoric acid contribute to the overall [H+] We can calculate the [H+] from ka1: Ka1 = x(2.5/225 + x)/(5.0/225 – x) Since ka1 is very small (!!!), assume 2.5/225 >> x 1.1x10-2 = x(2.5/225)/(5.0/225) x = 2.2x10-2 M Relative error = {2.2x10-2/(2.5/225)} x 100 = 198% It is clear that the relative error is very large and the assumption is, for sure, invalid and we should use the quadratic equation.