2. Protonated Salts These are usually amphoteric salts which react as acids and bases. For example, NaH2PO4 in water would show the following equilibria: H2PO4- D H+ + HPO42- H2PO4- + H2O D OH- + H3PO4 H2O D H+ + OH- [H+]solution = [H+]H2PO4- + [H+]H2O – [OH-]H2PO4- [H+]solution = [HPO42-] + [OH-] – [H3PO4]

Now make all terms as functions in either H+ or H2PO4-, then we have: [H+] = {ka2 [H2PO4-]/[H+]} + kw/[H+] –{[H2PO4-][H+]/ka1} Rearrangement gives [H+] = {(ka1kw + ka1ka2[H2PO4-])/(ka1 + [H2PO4‑]}1/2 At high salt concentration and low ka1 this relation may be approximated to: [H+] = {ka1ka2}1/2 Where; the pH will be independent on salt concentration but only on the equilibrium constants.

Example Find the pH of a 0.10 M NaHCO3 solution. ka1 = 4.3 x 10-7, ka2 = 4.8 x 10-1 Solution HCO3- D H+ + CO32- HCO3- + H2O D OH- + H2CO3 H2O D H+ + OH- [H+] = {(ka1kw + ka1ka2[HCO3-])/(ka1 + [HCO3-]}1/2 [H+] = {(4.3x10-7 * 10-14 + 4.3x10-7 * 4.8x10-11 * 0.10)/(4.3x10-7 + 0.10)}1/2 [H+] = 4.5x10-9 M pH = 8.34

The same result can be obtained if we use [H+] = {ka1ka2}1/2 [H+] = {4.3x10-7 * 4.8x10-11}1/2 = 4.5x10-9 M This is since the salt concentration is high enough. Now look at the following example and compare:

Example Find the pH of a 1.0x10-4 M NaHCO3 solution. ka1 = 4.3 x 10-7, ka2 = 4.8 x 10-11 Solution [H+] = {(4.3x10-7 * 10-14 + 4.3x10-7 * 4.8x10-11 * 1.0x10-4)/(4.3x10-7 + 1.0x10-4)}1/2 [H+] = 7.97x10-9 M pH = 8.10 Substitution in the relation [H+] = {ka1ka2}1/2 will give [H+] = {4.3x10-7 * 4.8x10-11}1/2 = 4.5x10-9 M, which is incorrect You can see the difference between the two results.

Protonated Salts with multiple charges HPO42- is a protonated salt which behaves as an amphoteric substance where the following equilibria takes place: HPO42- D H+ + PO43- HPO42- + H2O D H2PO4- + OH- H2O D H+ + OH- [H+] = [H+]HPO4- + [H+]water – [OH-]HPO4- [H+] = [PO43-] + [OH-] – [H2PO4-] [H+] = ka3 [HPO42-]/[H+] + kw/[H+] – [H+][HPO42-]/ka2 Rearrangement of this relation gives [H+] = {(ka2kw + ka2ka3 [HPO42-])/(ka2 + [HPO42-])}1/2 Approximation, if valid, gives: [H+] = (ka2ka3)1/2

Example Find the pH of a 0.20 M Na2HPO4 solution. Ka1 = 1.1x10-2, ka2 = 7.5x10-8, ka3 = 4.8x10-13. Solution HPO42- is doubly charged so we use ka2 and ka3 as the relation above [H+] = {(ka2kw + ka2ka3 [HPO42-])/(ka2 + [HPO42-])}1/2 [H+] = {(7.5x10-8 * 10-14 + 7.5x10-8 * 4.8x10-13 * 0.20)/(7.5x10-8 + 0.20)}1/2 = 2.0x10-10 M pH = 9.70

Using the approximated expression we get: [H+] = (7.5x10-8 * 4.8x10-13)1/2 = 1.9x10-10 M pH = 9.72 This small difference is because ka2kw is not very small as compared to the second term and thus should be retained.

pH Calculations for Mixtures of Acids The key to solving such type of problems is to consider the equilibrium of the weak acid and consider the strong acid as 100% dissociated as a common ion.

Example Find the pH of a solution containing 0.10 M HCl and 0.10 M HOAc. ka = 1.8x10-5 Solution HOAc D H+ + OAc-

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

Example Find the pH of a solution containing 0.10 M HCl and 0.10 M H3PO4. Ka1 = 1.1x10-2, ka2 = 7.5x10-8, ka3 = 4.8x10-13. Solution It is clear from the acid dissociation constants that ka1>>ka2 and thus only the first equilibrium contributes to the H+ concentration. Now treat the problem as the previous example: H3PO4 D H2PO4- + H+

Ka = (0.10 + x) x/(0.10 – x) Assume 0.1 >> x since ka is small (!!!) 1.1x10-2 = 0.10 x/0.10 x = 1.1x10-2 Relative error = (1.1x10-2/0.10) x100 = 11 % The assumption is therefore invalid and we have to solve the quadratic equation. Result will be X = 9.2x10-3 Therefore [H+] = 0.10 + 9.2x10-3 = 0.11 pH = 0.96

Example Find the pH of a solution containing 0.10 M HCl and 0.10 M HNO3. Solution [H+] = [H+]HCl + [H+]HNO3 Both are strong acids which are 100% dissociated. Therefore, we have [H+] = 0.10 + 0.10 = 0.2 pH = 0.70

In some situations we may have a mixture of two weak acids In some situations we may have a mixture of two weak acids. The procedure for pH calculation of such systems can be summarized in three steps: For each acid, decide whether it is possible to neglect dissociations beyond the first equilibrium if one or both are polyprotic acids. If step 1 succeeds to eliminate equilibria other than the first for both acids, compare ka1 values for both acids in order to check whether you can eliminate either one. You can eliminate the dissociation of an acid if ka for the first is 100 times than ka1 for the second (a factor of 100 is enough since the acid with the larger ka suppresses the dissociation of the other). Perform the problem as if you have one acid only if step 2 succeeds.

Example Find the pH of a solution containing 0.10 M H3PO4 (ka1 = 1.1x10-2, ka2 = 7.5x10-8, ka3 = 4.8x10-13) and 0.10 M HOAc (ka = 1.8x10-5). Solution It is clear for the phosphoric acid that we can disregard the second and third equilibria since ka1>>> ka2. Therefore we treat the problem as if we have the following two equilibria: H3PO4 D H+ + H2PO4- HOAc D H+ + OAc-

Now compare the ka values for both equilibria: Ka1/ka = 1.1x10-2/1.8x10-5 = 6.1x102 Therefore the first equilibrium is about 600 times better than the second. For the moment, let us neglect H+ from the second equilibrium as compared to the first. Solution for the H+ is thus simple H3PO4 D H2PO4- + H+

Ka = x * x/(0.10 – x) Assume 0.1 >> x since ka is small (!!!) 1.1x10-2 = x2 /0.10 x = 0.033 Relative error = (0.033/0.10) x100 = 33 % The assumption is therefore invalid and we have to solve the quadratic equation. Result will be X = 0.028

Now let us calculate the H+ coming from acetic acid which is equal to [OAc-] HOAc D H+ + OAc- Ka = [H+][OAc-]/[HOAc] 1.8x10-5 = 0.028 * [OAc-]/0.10 [OAc-] = 6.4x10-5 = [H+]HOAc Relative error = (6.4x10-5/0.028) x100 = 0.23% Therefore, we are justified to disregard the dissociation of acetic acid in presence of phosphoric acid. Never calculate the H+ concentration from each acid and add them up. This is incorrect.

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:

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 and 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