Chem 5 Fall 2014 Mini quiz 5 – Solutions Mon If you take acetic acid, CH3COOH, and turn one of the hydrogens (one of the three that come early in the formula) in chlorine, you have chloroacetic acid, CH2ClCOOH. This change boosts Ka up from its acetic acid value (1.76 × 10–5) to a significantly higher value: → H+(aq) + CH2ClCOO–(aq) CH2ClCOOH(aq) ← Ka = 1.4 × 10–3, pKa = 2.85 (a) Calculate the pH of a solution formed from 0.25 L of 0.80 M CH2ClOOH to which 0.15 mol NaOH(s) has been dissolved. Let’s see how much acid we have (which we’ll abbreviate as simply HA), assuming none of it dissociates: o = 0.80 M 0.25 L = 0.20 mol nHA Next, we realize that the hydroxide ions, OH–, coming from NaOH, will turn HA into A– until all the added OH– is gone (because it is the limiting reagent). Thus, at equilibrium, we have nHA = 0.20 mol – 0.15 mol = 0.05 mol nA– = 0.15 mol We find the pH easily from the Henderson-Hasselbalch equation: pH = pKa + log10 [A–] = 2.85 + log10 0.15 = 3.33 [HA] 0.05 Note that we could use the ration of moles of A– and HA. We did not need to convert these into concentrations, because both species are in the same volume! (b) Suppose an additional 0.10 mol NaOH(s) is added to the solution in part (a). Find the new pH for this solution. This extra OH– will react with the remaining 0.05 mol of HA, turning all of it into A– and leaving behind 0.10 mol – 0.05 mol = 0.05 mol unreacted OH–. This OH– will determine the pH: [OH–] = 0.05 mol = 0.20 M 0.25 L Tue Thus, pOH = –log10(0.20) = 0.70, or pH = 14 – pOH = 13.3 Propionic acid is CH3CH2COOH, which is related to acetic acid, CH3COOH, but with a –CH2– bit stuck in the molecule. It, too, is a weak acid, dissociating to H+ and the propionate anion, CH3CH2COO–, as shown below: → H+(aq) + CH3CH2COO–(aq) CH3CH2COOH(aq) ← 1 Ka = ? Chem 5 Fall 2014 Mini quiz 5 – Solutions (a) When a 0.15 mol sample of propionic acid is dissolved in 300 mL of water and 40 mL of a 0.500 M NaOH solution is added to it, the pH is measured to be 4.06. What is Ka for propionic acid? The NaOH solution provides (0.040 L)(0.500 M) = 0.020 mol OH–. We start with 0.15 mol of acid (let’s call it HA to save time), and the OH– turns HA into A– and leaves us with 0.15 mol – 0.020 mol = 0.13 mol HA and 0.020 mol A–. Since we know the pH, we can use the Henderson-Hasselbalch equation to find pKa and thus Ka: pKa = pH – log10 [A–] = 4.06 – log10 0.02 = 4.87 or Ka = 10–pKa = 1.34 × 10–5 [HA] 0.13 (b) What would the pH be at the equivalence point of this titration? At the equivalence point, all the HA has been turned into A–. The pH then is determined by hydrolysis of A–, but note as well that we need to find the final total volume of the solution at the equivalence point. We started with 0.15 mol HA, and to reach this amount of OH–, we should add a volume of our OH– solution given by 0.15 mol/0.500 M = 0.300 L. Thus, the total volume at the equivalence point is 0.300 L + 0.300 L = 0.600 L. This volume holds 0.15 mol A–, which hydrolyzes according to – 10–14 → HA(aq) + OH–(aq) Kb = [HA][OH ] = K W = A–(aq) + H2O(l) ← = 7.46 × 10–10 Ka 1.34 × 10–5 [A–] We also recognize that [HA] = [OH–] at the equivalence point, and [A–] = 0.15 mol/0.600 L = 0.25 M. This lets us calculate [OH–], and thus pOH and pH: Kb = [HA][OH–] [A–] 2 = [OH–] = 7.46 × 10–10 0.25 or [OH–] = 1.37 × 10–5 so that pOH = –log10(1.37 × 10–5) = 4.86 and pH = 14 – pOH = 9.14. Wed Oxalic acid, H2C2O4, is a solid diprotic acid that has a fairly large first acid dissociation constant Ka1 but a very small second acid dissociation constant Ka2: → H+(aq) + HC2O–(aq) H2C2O4(aq) ← 4 – 2– + → HC2O (aq) ← H (aq) + C2O (aq) 4 4 Ka1 = 5.9 × 10–2, pKa1 = 1.23 Ka2 = 6.4 × 10–5, pKa2 = 4.19 (a) If a solution of oxalic acid is titrated with OH–, at some point it is observed that the solution has a pH = 5.0. In the blanks below, enter the three species H2C2O4, HC2O–4, and C2O2– 4 in order of decreasing concentration. You do not have to calculate anything in this part of the problem! ________________ > ________________ >> ________________ 2 Chem 5 Fall 2014 Mini quiz 5 – Solutions Compare the solution’s pH to the two pKa values: pH = 5.0 is just a bit higher than pKa2. That tells us that we’re just past the second half-equivalence point, where [C2O2– 4 ]= – 2– – [HC2O4]. Thus, [C2O4 ] > [HC2O4] .>> [H2C2O4]. (b) If this solution started with 0.050 mol oxalic acid dissolved in 100 mL of water and then was titrated with a solution of sodium hydroxide with [OH–] = 0.500 M, calculate the pH when 60 mL of this solution has been added to the original oxalic acid solution. With 60 mL = 0.060 L of 0.500 M OH– added, we have added a total of (0.500 M)(0.060 L) = 0.030 mol OH–. We started with 0.050 mol H2C2O4, and now 0.030 mol of it is gone: turned into HC2O–4 by the OH–. We thus have 0.020 mol H2C2O4 and 0.030 mol HC2O–4, which are governed by the first equilibrium. Henderson-Hasselbalch to the rescue! [HC2O–4] = 1.23 + log10 0.030 = 1.41 pH = pKa1 + log10 [H2C2O4] 0.020 3