Department of Chemistry University of Texas at Austin

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1 Polyprotic and Special Cases Calculations Supplemental Worksheet KEY For the following polyprotic acid questions: Citric acid (H3C6H5O6) Ka1 = 8.4 x 10 4 Ka2 = 1.8 x 10 5 Ka3 = 4.0 x 10 6 Oxalic acid (H2C2O4) Ka1 = 6.5 x 10 2 Ka2 = 6.1 x Calculate the ph of a 2L solution containing 0.4 M oxalic acid. Then calculate the ph of the solution if you titrate in 400 ml of 1 M NaOH. First we need to calculate the ph of the original solution. We can perform our RICE table in molarity units instead of mole units because we are not changing the volume yet. R. H2C2O4 (aq) + H2O (l) HC2O4 (aq) + H3 (aq) I. 0.4 M M 0 M C. x x +x E. 0.4M x x x Ka1 = 6.5 x 10 2 = [HC 2 ][H3 ] = [x][x] [H 2 C 2 ] [0.4M x] This first Ka is actually not too small so let s not make approximations about the x in the denominator yet ± ( )(0.4 x) = x 2 x x = x = 2 x = M (discard negative solution) ph= log (H 3 ) = log (x) = log ( M) 0.88 Now we can perform the titration: mol of H2C2O4 initially present = (0.4M)(2L)= 0.8 mol mol NaOH added = (1M)(0.4L) = 0.4 mol We need to perform the neutralization in moles because we are changing the volume as we titrate. R. H2C2O4 (aq) + NaOH (aq) NaHC2O4 (aq) + H2O (l) I. 0.8 mol 0.4mol 0 M C. 0.4mol 0.4mol +0.4mol E. 0.4mol 0mol 0.4mol Now we need to recognize that we have 0.4mol of oxalic acid in solution and 0.4mol of its conjugate base. This is a buffer solution and we can use Henderson- Hasselbalch to calculate the ph of the solution. We could convert these mole values to molarity by divided by 2.4 L (new total volume) but in the HH equation the two mole values are divided thereby negating the volume factor (try solving it both ways if you re unsure).

2 A ph=pk a1 +log HA ph= log ( mol ) +log = mol 2. Given the solution from problem number 1, many ml of NaOH need to be added to reach a buffer solution containing equal parts HC2O4 and C2O4 2? You can think through this problem in many different ways. One way is to systematically consider the steps in the titration (drawing a titration curve may help). The initial solution contains 0.8 mol of H2C2O4 (0.4M*2L). If all 0.8 mol of H2C2O4 are neutralized using 0.8 mol of NaOH, then the solution would contain 0.8 mol of HC2O4. To create a buffer solution with equal parts HC2O4 and C2O4 2 we would need to neutralize half of the HC2O4 in solution. So we would need to add another 0.4mol of NaOH to create a buffer solution of 0.4 mol of HC2O4 and 0.4 mol of C2O4 2. In sum: We need to titrate in 0.8 mol of NaOH to neutralize all of the initial H2C2O4. Then, we need to add 0.4 mol of NaOH to neutralize half of the HC2O4. We need 1.2 mol of the NaOH solution. V = mol M = 1.2mol 1 mol = 1.2L NaOH = 1200mL NaOH L 3. What is the ph of the solution describe in problem number 1 when 1.4L of NaOH has been added to the initial oxalic acid solution? mol of H2C2O4 initially present = (0.4M)(2L)= 0.8 mol mol NaOH added = (1M)(1.4L) = 1.4 mol We need to perform the neutralization in moles because we are changing the volume as we titrate. R. H2C2O4 (aq) + NaOH (aq) NaHC2O4 (aq) + H2O (l) I. 0.8 mol 1.4mol 0 M C. 0.8mol 0.8mol +0.8mol E. 0mol 0.6mol 0.8mol In solution we still have remaining strong base! We need to perform a second neutralization problem with NaOH and NaHC2O4. R. NaHC2O4 (aq) + NaOH (aq) Na2C2O4 (aq) + H2O (l) I. 0.8 mol 0.6mol 0 M C. 0.6mol 0.6mol +0.6mol E. 0.2mol 0mol 0.6mol

3 Now we re left with a solution that contains HC2O4 and C2O4 2. Let s use the HH equation to solve for the ph of our buffer solution: [A ] ph = pk a2 +log [HA] 0.6mol ph= log ( ) + log = log (3) mol 4. Sketch a titration curve of the complete titration oxalic acid with NaOH. Label the important points on the curve and list the dominant species present at each of these points. Excess Strong Base Region Dominant Species: OH Second Equivalence Point Dominant Species: C 2 2 and OH ph pk a2 Second Buffer Zone Dominant Species: HC 2, C 2 2 First Equivalence Point Dominant Species: HC 2 pk a1 First Buffer Zone Dominant Species: H 2 C 2, HC 2 Initial Weak Acid Solution Dominant Species: H 2 C 2 ml of NaOH added We know that both buffer solutions are acidic because the pka values of both Ka vaues are less than 7 (H3 is also a dominant species). At the second equivalence point however, the only thing in solution is the weak base C2O4 2 which means the solution will be basic (OH is also a dominant species). 5. Calculate the three pka values for citric acid (H3C6H5O6).

4 pk a1 = log (K a1 )= log ( )= 3.1 pk a2 = log (K a2 ) = log ( )= 4.7 pk a3 = log (K a3 ) = log ( )= Describe the protonation state of citric acid in your stomach (ph 2). Stomach acid is quite acidic. The ph of stomach acid is less than the first pka of citric acid which means that the citric acid would be fully protonated and have a neutral charge. 7. Describe the dominant species in a solution of citric acid that has been titrated with a strong base until a ph meter reads with ph 5.4. What type of solution is this? When the ph is 5.4 we have an acidic buffer solution with equal parts HC6H5O6 2 and C6H5O6 3. The buffer solution is acidic (ph = 5.4) so we know there is more H3 in solution than OH. (PSA: The real titration of citric acid becomes complicated because its buffer regions overlap! You can still calculate everything you need for these overlapping buffer zones but we won t go into that now.) 8. Citric acid is often used a chelating agent meaning it is used to bind to certain ions in a solution. For example, citric acid can be used to chelate calcium ions and reduce the hardness of water. Explain why citric acid might be a better chelating agent than acetic acid (pka = 4.7). Citric acid has four protonation states, three of which are charged (negatively). Acetic acid has two protonation states, only one of which is charged (negatively). Therefore, citric acid has a wider range of ph values in which it can interact with other charged species in solution such as calcium ions. Furthermore, fully deprotonated citric acid has a 3 charge, which allows it to more flexibly interact with ions with higher positive charges such as Fe You are titrating a 1L solution of 1M citric acid with 3M KOH. You want to know when you ve just reached a solution containing C6H5O6 3 without excess strong base using an indicator. Select from the following indicators to use in your titration and explain your choice: Thymol Blue pka =1.65 Bromophenol Blue pka =4.10 Methyl Red pka = 5.00 Bromothymol Blue pka = 7.30 Phenophthalein pka = 9.50 The point of interest here is the third equivalence point of this titration. At this point we would have just titrated all of the citric acid completely into its fully deprotonated form

5 C6H5O6 3. Beyond the third equivalence point we start pouring in excess strong base. So we need an indicator that will change colors near the ph of the third equivalence point. We can start by eliminating all by the last two indicators on the list by looking at our pka values for citric acid. At ph values lower than 3.1 we have mostly fully protonated citric acid (H3C6H5O6). At ph 3.1 we have an acidic buffer of H3C6H5O6 and H2C6H5O6. When the ph meter reads 4.7, we have a different acidic buffer of H2C6H5O6 and HC6H5O6 2. When the ph meter reads 5.4 we have the third acidic buffer of HC6H5O6 2 and C6H5O6 3. The ph meter will read higher than 5.4 by the time we ve titrated the citric acid completely into its fully deprotonated form C6H5O6 3. Thymol Blue, Bromophenol Blue and Methyl Red exhibit color changes in ph environments near their pka values. They would ve changed colors before the desired point. So we need to evaluate Bromothymol Blue and Phenophthalein. Which of these indicators will change colors near the third equivalence point of this titration? To answer this question we need to calculate the ph of the third equivalence point where the only citric acid species present in solution is the weak base C6H5O6 3. There is initially 1 mol citric acid in solution (1M*1L). The needed mol of KOH to completely deprotonate is 3 mol. We need to add 1 L of the 3L KOH solution to the original solution. This would bring the final volume up to 2 L and the concentration of C6H5O6 3 at the third equivalence point to 0.5M (1 mol 2L). R. C6H5O6 3 (aq) + H2O (l) HC6H5O6 2 (aq) + OH (aq) I. 0.5 M M 0 M C. x x +x E. 0.5M x x x K b = K w = = !! K a K! =! [!]! [!] (we can ignore the x in the denominator because Kb is very small) [!.!!!!] [!.!!] x = [OH ] = 3.53 x 10 5 M poh = log(oh ) = log(3.53 x 10 5 M) = 4.5 ph = 14 poh = 9.5 You should use Phenophthalein as an indicator because its color change occurs very near the third equivalence point of the titration (around 9.50). 10. Explain how and why your calculations for the ph of 0.7 M HNO3 and 7 x 10 8 M HNO3 are different. In your explanation calculate the ph of these solutions. For a very dilute solution of a strong acid the hydronium ion concentration due to the auto- ionization of water is non- negligible. Therefore, we need to account for the two sources of

6 the hydronium ion. The strong acid dissociates completely so that accounts for M of the hydronium ion concentration. The auto- ionization of water then adds another amount and we can represent the hydronium ion concentration with this equality: [H3 ] = x Kw is always the product of the hydronium ion concentration and hydroxide ion concentration. Water always auto- ionizes in equal parts so if x represents the amount of hydronium ion concentration from the auto- ionization then the hydroxide ion concentration from the auto- ionization is also x. Kw = [H3 ][OH ] = (x + 7 x 10 8 )(x) = x 2 + ( )x = 0 Using the quadratic equation we can solve for x and therefore for the ph of the solution. x = 7 10!! ± !!" !!" 2 x = !! M [H3 ] = x = !! ph = log([h3 ]) = log( ) = 6.85 For non- dilute solutions of strong acid the calculations are much simpler. We assume that the strong acid dissociates completely and that the hydronium ion concentration comes directly from the strong acid dissociation: [H3 ] = 0.7 M ph = log([h3 ]) = log(0.7m) = 0.15

Titration of a Weak Acid with a Strong Base

Titration of a Weak Acid with a Strong Base Weak Acid w/ Strong Base Overall: INITIAL ph: Weak acids do not fully dissociate we need to do an ICE table to determine initial ph. We expect it to be weakly