Solutions for Acid-Base Titrations Exercises

Solutions for Acid-Base Titrations Exercises 1. If the acid or base is too weak or too dilute, there is very little change in pH at the equivalence point. This makes it difficult to have a satisfactory end point. 2. . C6H5CH2NH2(aq) + H3O+(aq) 3. K º C6H5CH2NH3+(aq) + H2O(l) The equation for the titration of a weak base, B, with a strong acid is B(aq) + H3O+(l) 6 HB+(aq) + H2O(l) This produces a solution of the conjugate acid, HB+, at the equivalence point so the solution is acidic (pH<7). 4. The titration reaction is HCO2H(aq) + OH-(l) 6 HCO2-(aq) + H2O(l). Since the analyte and titrant concentrations are equal, it will take 50.0 mL of base to reach the equivalence point. Note that for formic acid Ka = 1.80 x 10-4 (pKa = 3.74). 20.0 mL base init mol HCO2H = (0.050 mol HCO2H/L)(0.0500 L) = 0.0025 mol mol OH- added = (0.0200 L)(0.050 mol OH-/L) = 0.0010 mol HCO2H(aq) + OH-(l) 6 HCO2-(aq) + H2O(l) init mol 0.0025 0.0010 0 mol after rxn 0.0015 0 0.0010 A buffer is formed with: [HCO2H] = (0.0015 mol)/(0.0700 L) = 0.0214 M [HCO2-] = (0.0010 mol)/(0.0700 L) = 0.0143 M Use Henderson-Hasselbalch, Solutions for Acid-Base Titrations Exercises 50.0 mL base At the equivalence point all of the formic acid has been converted to its conjugate base so you have a solution of a weak base. [HCO2-] = (0.0025 mol)/(0.100 L) = 0.0250 M HCO2-(aq) + H2O(l) init 0.0250 M equil 0.0250 - x ~ 0.0250 º HCO2H(aq) + OH-(aq) 0 ~10-7 M x ~10-7 + x ~ x where x is the increase in [HCO2H] and is assumed to be << 0.025 M and >> 10-7 M. pOH = -log (1.18 x 10-6) = 5.93 Y pH = 8.07 60.0 mL base At this point 10.0 mL extra base has beeen added and you have a solution of a weak base and strong base. The OH- contribution from the weak base is not significant so the pH depends on the excess OH- present. excess mol OH- added = (0.0100 L)(0.050 mol OH-/L) = 0.00050 mol [OH-] = (0.00050 mol)/(0.110 L) = 0.00455 M pOH = -log (0.00455) = 2.34 5. Y pH = 11.66 This occurs when the titrant volume is one-half that needed to reach the equivalence point. Here, half the base (B) has been converted to its conjugate acid (HB+) and [B] = [HB+] and pH = pKa. At this point a small addition of acid will have the least effect on the [B]/[HB+] ratio and hence on the pH. Solutions for Acid-Base Titrations Exercises 6. When the titrant volume is one-half that needed to reach the equivalence point. half the weak acid (HA) has been converted to its conjugate base (A-), buffer is formed and [HA] = [A-]. The Henderson-Hasselbalch equation looks like this At this point the principal electrolyte in solution is NaA. Half the starting moles of A- are now in a volume 50% larger so [NaA] = 0.0333 M. Thus, μ = 0.0333 M. γHA ~ 1 and Substituting and solving for pKa gives 7. When NaCN is titrated with HClO4, the following reaction occurs. CN-(aq) + H3O+(aq) 6 HCN(aq) + H2O(l) init mol CN- = (0.100 mol CN-/L)(0.0500 L) = 0.00500 mol a) 4.20 mL of 0.438 M HClO4 (mol H+ added = (0.00420 L)(0.438 mol H+/L) = 0.00184 mol) CN-(aq) + H3O+(aq) init mol 0.00500 0.00184 mol after rxn 0.00316 0 A buffer is formed with: 6 HCN(aq) + H2O(l) 0 0.00184 [HCN] = (0.00184 mol)/(0.05420 L) = 0.0339 M [CN-] = (0.00316 mol)/(0.05420 L) = 0.0583 M Solutions for Acid-Base Titrations Exercises HCN(aq) + H2O(l) init 0.0339 M equil ~0.0339 M º H3O+(aq) + CN-(aq) ~10-7 M 0.0583 M x ~0.0583 M b) 11.82 mL of 0.438 M HClO4 (mol H+ added = 0.00518 - c) at equivalence point, 11.42 mL of HClO4 have been added and there is an aqueous solution of HCN and NaClO4 [HCN] = (0.00500 mol)/(0.06142 L) = 0.0814 M HCN(aq) + H2O(l) init more than needed to reach equivalence point; assume pH is determined by the excess strong acid) 0.0814 M equil 0.0814 - x º H3O+(aq) + CN-(aq) ~10-7 M 0 ~x x Y x = 7.10 x 10-6 M = [H3O+] Y pH = 5.15 Solutions for Acid-Base Titrations Exercises 8. The ratio [HIn]/[In-] changes from 10:1 when pH = pKHIn - 1 to 1:10 when pH = pKHIn + 1. This change is generally sufficient to cause a color change. 9. Cresol purple at: pH = 1.0 - red pH = 2.0 - orange (mix of red & yellow) pH = 3.0 - yellow 10. No. The titration produces a solution of the conjugate base at the equivalence point. This solution will have a pH > 7 which is well outside the color transition range for bromocresol green. 11. a) The titration reaction is F- + H3O+ 6 HF + H2O. Since the concentration of HClO4 is twice that of the NaF, you need one-half the volume (V) of the analyte solution. Thus, the formal concentration of HF at the equivalence point is given by: At the equivalence point, HF(aq) + H2O(l) init 0.0200 M equil 0.0200 - x º H3O+(aq) + F-(aq) ~10-7 M 0 ~x x Y x = 3.688 x 10-3 M (x << 0.02 assumption is not valid! - use successive approximations) Solutions for Acid-Base Titrations Exercises x2 = 3.331 x 10-3 x3 = 3.367 x 10-3 x4 = 3.363 x 10-3 M = [H3O+] Y pH = 2.47 b) The equivalence point pH is quite acidic so there will not be a large change in pH around the equivalence point. Note that the pH of the buffer plateau before the equivalence point is ~ 3.2 and the pH of the titrant is ~1.2. A sharp change in indicator color will not be seen in this titration, but you could use a color match for the end point. 12. Sodium carbonate (Na2CO3) is commonly used to standardize HCl solutions. Potassium hydrogen phthalate (KHC8H4O4) is commonly used to standardize NaOH solutions (see Prob. 13 below). 13. Potassium hydrogen phthalate (KHC8H4O4) 204.2 g/mol HC8H4O4- + OH- 6 H2O + C8H4O42- mol OH- = mol KHC8H4O4 = (0.05 mol/L)(0.030 L) = 0.0015 mol mass KHC8H4O4 = (0.0015 mol)(204.2 g/mol) = 0.31 g