Acid-Base Titrations

In this chapter, we will study titrations of: 1. Strong acid with strong base 2. Strong acid with weak base 3. Strong base with weak acid 4. Strong acid with polybasic bases 5. Strong base with polyprotic acids 6. Strong acid with a mixture of two bases 7. Strong base with a mixture of two acids.

Acid-Base Indicators An acid-base indicator is either a weak acid or base which changes color upon changing from one chemical form to the other, depending on the pH. Indicators are added in a very small amounts in order to decrease the titration error. We can represent the equilibrium of an indicator as follows: HIn (color 1) D H+ + In- (color 2)

Kin = [H+][In-]/[HIn] pkin = pH – log [In-]/[HIn], or pH = pkIn + log [In-]/[HIn] Color 1 can be visually observed, in presence of color 2, if [HIn] is at least 10 times [In-] and color 2 can be visually observed, in presence of color 1, if [In-] is at least 10 times [HIn]. Therefore, the final equation can be rewritten as: pH = pkIn + 1

pH = pkIn + 1 This equation is of extreme significance since it suggests that: 1. There is a pH range of two units only where an indicator can be used. 2. The pkin should be very close to the pH at the equivalence point of the titration. Therefore, one should look at the pH at the equivalence point of the titration in order to select the right indicator.

Once again, the pkin of the indicator should be close to the pH of the equivalence point of the titration of interest. Look at the following titration curve, both indicators have their transition ranges on the break of the curve and thus either indicator can be used for this titration. pH mL Titrant

However, the following titration curve requires a different indicator (the lower one is suitable ):

The break on the titration curve is very important for accurate determination of the end point. As the break becomes steeper and sharper, very small excess of a titrant is needed for good visualization of the end point. We may think of the accuracy of an end point by imagining that the indicator starts changing color when the pH of the solution touches one side of its range but the color is not clear enough unless the pH reaches the other side of the range. Look at the following titration curve:

pH mL Titrant

The distance between the two arrows represents the volume which is needed to undoubtedly observe the end point. The color of the indicator starts changing at a volume corresponding to the first arrow and can be visually seen at a volume corresponding to the second arrow. This volume is added in excess of the required volume and thus an error is obtained corresponding to that volume excess. A better titration curve is seen below where a minimal extra volume is required to visualize the end point:

pH mL Titrant I have intentionally shown the two arrows coincide in order to indicate that sharp inflections or breaks can be very advantageous in titrations where very low errors should be expected.

What makes the break sharper? This is an important question and the answer is rather simple: Two reasons; 1. Concentrations of reactants (analyte and titrant) where as concentration increase, sharpness of the break increase. 2. The dissociation constant where as the dissociation constant increases, the sharpness of the break increases. This suggests that strong acids and bases are expected to have sharper breaks while weak acids are expected to have diffused breaks.

Strong acids can be titrated with strong bases leading to formation of salts and water. Remember that the salt formed from a strong acid and strong base is a combination between the weak conjugate base of the strong acid and the weak conjugate acid of the strong base. Both conjugates are weak and thus do not react with water which means a [H+] = 10-7 M and pH=7 at the equivalence point. Let us now look at an example for such a titration:

Lecture 27 Acid-Base Titrations, Cont… Strong Acids versus Strong Bases Strong Bases/Acids versus Weak Acids/Bases

Example Calculate the pH of a 50 mL 0f 0.10 M HCl after addition of 0, 20, 40, 50, 80, and 100 mL of 0.1 M NaOH. Solution After addition of 0 mL NaOH we only have HCl. [H+] = 0.10 M , same as HCl concentration since HCl is a strong acid; pH = 1.0

After addition of 20 mL NaOH mmol H+ left = 0.1*50 – 0.1*20 = 3.0 [H+] = 3.0/70 = 0.043 pH = 1.34 After addition of 40 mL NaOH mmol H+ left = 0.1*50 – 0.1*40 = 1.0 [H+] = 1.0/80 = 0.0125 pH = 1.90 Same steps are used for calculation of pH at any point before equivalence.

After addition of 50 mL NaOH mmol H+ left = 0.1*50 – 0.1*50 = 0 Therefore, this is the equivalence point and the H+ will only be produced from dissociation of water H2O D H+ + OH- [H+] = [OH-] = 10-7 M pH = 7

After addition of 80 mL NaOH mmol OH- excess = 0.1*80 – 0.1*50 = 3.0 [OH-] = 3.0/130 = 0.023 pOH = 1.63 pH = 14 – 1.63 = 12.37 The same procedure is used for calculation of any point after equivalence is reached.

Titration of a Weak Acid with a Strong Base Weak acids could be titrated against strong bases where the following points should be remembered: 1. Before addition of any base, we have a solution of the weak acid and calculation of the pH of the weak acid should be performed as in previous sections. 2. After starting addition of the strong base to the weak acid, the salt of the weak acid is formed. Therefore, a buffer solution results and you should consult previous lectures to find out how the pH is calculated for buffers.

3. At the equivalence point, the amount of strong base is exactly equivalent to the weak acid and thus there will be 100% conversion of the acid 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 base. The presence of the excess base suppresses the dissociation of the salt in water and the pH of the solution comes from the excess base only. Now, let us apply the abovementioned concepts on an actual titration of a weak acid with a strong base.

Example Find the pH of a 50 mL solution of 0.10 M HOAc (ka = 1.75x10-5) after addition of 0, 10, 25, 50, 60 and 100 mL of 0.10 M NaOH. Solution After addition of 0 mL NaOH: we have the acid only where: HOAc D H+ + OAc-

Ka = [H+][OAc-]/[HOAc] Ka = x * x / (0.10 – x) Ka is very small. Assume 0.10 >> x 1.75*10-5 = x2/0.10 x = 1.3x10-3 Relative error = (1.3x10-3/0.10) x 100 = 1.3% The assumption is valid The [H+] = 1.3x10-3 M pH = 2.88

2. After addition of 10 mL NaOH initial mmol HOAc = 0.10 x 50 = 5.0 mmol NaOH added = 0.10 x 10 = 1.0 mmol HOAc left = 5.0 – 1.0 = 4.0 [HOAc] = 4.0/60 M mmol OAc- formed = 1.0 [OAc-] = 1.0/60 M HOAc D H+ + OAc-

Ka = x (1.0/60 – x)/ (4.0/60 – x) Assume 21.0/60 >> x 1.75x10-5 = x (1.0/60)/4.0/60 x = 1.75x10-5 x 1.0/4.0 x = 7.0 x10-5 Relative error = {7.0 x10-5/(1.0/60)} x 100 = 0.42% The assumption is valid pH = 4.15

3. After addition of 25 mL NaOH initial mmol HOAc = 0. 10 x 50 = 5 3. After addition of 25 mL NaOH initial mmol HOAc = 0.10 x 50 = 5.0 mmol NaOH added = 0.10 x 25 = 2.5 mmol HOAc left = 5.0 – 2.5 = 2.5 [HOAc] = 2.5/75 M mmol OAc- formed = 2.5 [OAc-] = 2.5/75 M HOAc = H+ + OAc-

Ka = x (2. 5/75 – x)/ (2. 5/75 – x) Assume 2. 5/75 >> x 1 Ka = x (2.5/75 – x)/ (2.5/75 – x) Assume 2.5/75 >> x 1.75x10-5 = x (2.5/75)/(2.5/75) x = 1.75x10-5 Relative error = {1.75x10-5/(2.5/75)} x 100 = 0.052% The assumption is valid pH = 4.76 Note here that at 25 mL NaOH (half the way to equivalence point) pH = pka . It is therefore possible to find the ka of a weak monoprotic acid by calculating half the volume of strong base needed to reach the equivalence point. At this volume ka = [H+].

4. After addition of 50 mL NaOH initial mmol HOAc = 0. 10 x 50 = 5 4. After addition of 50 mL NaOH initial mmol HOAc = 0.10 x 50 = 5.0 mmol NaOH added = 0.10 x 50 = 5.0 mmol HOAc left = 5.0 –5.0 = 0 This is the equivalence point mmol OAc- formed = 5.0 [OAc-] = 5.0/100 = 0.05 M Now the solution we have is a 0.05 M acetate OAc- + H2O D HOAc + OH-

Kb = kw/ka Kb = 10-14/1.75x10-5 = 5.7x10-10 Kb = [HOAc][OH-]/[OAc-] Kb = x * x/(0.05 – x) Kb is very small and we can fairly assume that 0.05>>x 5.7x10-10 = x2/0.05 x = 5.33 x 10-6 Relative error = (7.6x10-6/0.05) x100 = 0.011% The assumption is valid. [OH-] = 5.33x10-6 M pOH = 5.27 pH = 14 – 5.27 = 8.73

5. After addition of 60 mL NaOH Initial mmol of HOAc = 0.10 x 50 = 5.0 mmol NaOH added = 0.10 x 60 = 6.0 mmol NaOH excess = 6.0 – 5.0 = 1.0 [OH-] = 1.0/110 pOH = 2.04 pH = 14 – 2.04 = 11.96 6. After addition of 100 mL NaOH Initial mmol of HOAc = 0.10 x 50 = 5.0 mmol NaOH added = 0.10 x 100 = 10.0 mmol NaOH excess = 10.0 – 5.0 = 5.0 [OH-] = 5.0/150 pOH = 1.48 pH = 14 – 1.48 = 12.52

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 = 5.0 mmol NH3 left = 5.0 – 5.0 = 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

Acid-Base Titrations, Cont… Titration of Polyprotic Acids Lecture 28 Acid-Base Titrations, Cont… Titration of Polyprotic Acids

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 (carbonic acid as an example): Before addition of any base, you only have the polyprotic acid solution and thus calculation of the pH is straight-forward as previously described. Dividing ka1/ka2 allows for omitting 2nd equilibrium.

2. When we start addition of base, the first proton is titrated and bicarbonate will form. A buffer solution of carbonic acid and bicarbonate 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. 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

ka2 = x(2.5/125 + x)/(2.5/125 – x) ka2 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 M pOH = 1.60 pH = 14.00 – 1.60 = 12.40

Taking in consideration the OH- ions from carbonate equilibrium (kb = 2.1*10-4) X = 2.1*10-4 RE = (2.1*10-4/(5/200))*100 = 0.84% Therefore, the hydroxide from carbonate is negligible.

Acid-Base Titrations, Cont… Titration of a Polybasic Base Lecture 29 Acid-Base Titrations, Cont… Titration of a Polybasic Base

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 HPO42- = 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 M mmol H2PO4- formed = 2.5 [H2PO4-] = 2.5/125 M

Since ka3 is very small, assume 2.5/125 >> x H2PO4- D H+ + HPO42- 2.5/125 Before Equilibrium H+ HPO42- H2PO4- 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 only 25 mL of HCl were added to H2PO4- Initial mmol H2PO4- = 0.10 x 50 = 5.0 mmol H+ added = 0.10 x 25 = 2.5 mmol H2PO4- left = 5.0 – 2.5 = 2.5 [H2PO4-] = 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 H2PO4- left = 5.0 – 5.0 = 0 This is the third equivalence point mmol H3PO4 formed = 5.0 [H3PO4] = 5/200 = 0.025 M

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 > 102) 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

We can calculate the [H+] from ka1: Ka1 = x(2. 5/225 + x)/(5 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. Therefore, at this point both H+ from excess HCl and phosphoric acid contributes to the overall [H+]

Acid-Base Titrations, Cont…. Titration of Mixtures of acids and Bases Lecture 30 Acid-Base Titrations, Cont…. Titration of Mixtures of acids and Bases

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-

The figure reflects a larger HCl concentration than HOAc

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 M 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 M [HOAc] = 0.10 x 50 / 75 = 5.0/75 M 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 M 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

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 mmol HOAC left = 5.0 – 2.5 = 2.5 [HOAc] = 2.5/125 M mmol OAc- formed = 2.5 [OAc-] = 2.5/125 M 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 = 0 This is the second equivalence point mmol OAc- formed = 5.0 [OAc-] = 5.0/150 M 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 M 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.

The figure describes a titration of NaOH and carbonate with HCl where a smaller NaOH concentration is used

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 M 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 M [CO32-] = 0.10x50/75 = 5.0/75 M 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 M [OH-] = 2.5/75 M 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 M 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 M mmol HCO3- formed = 2.5 [HCO3-] = 2.5/125 M 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 M

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 M pH = 1.7 You should be able to calculate [H+]H2CO3

How to determine the composition of the final solution? Find the pH of a 20 mL solution of 0.10 M Na2CO3 after addition of 50 mL of 0.10 M HCl.

How to determine the composition of the final solution? Find the pH of a 20 mL solution of 0.10 M H3PO4 after addition of 50 mL of 0.10 M NaOH.

Acid-Base Titartions, Cont… Complexometric Reactions Lecture 31 Acid-Base Titartions, Cont… Kjeldahl Analysis Complexometric Reactions

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: 113

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%