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Post Lab Questions and Calculations 1. How many equivalence points were found? Two equivalence points were found. 2. What is the volume of the first equivalence point? 3. What is the volume of the second equivalence point? 4. Are the two equivalence points consistent, i.e. the second is twice the first? 5. Determine the midpoint to the 1st equivalence point and use this to determine pkai and Kat- 6. Determine the midpoint between the first and second equivalence points and determine pka. and Kaz. 7. Write the balanced net ionic equation for the two successive neutralization reactions. a) b)
M 21 10 17 10 M 22 10 23 24 27 29 11 12 13 14 15 1222** 20 P 2 3 6 + R 6 4 S 10 A Volume E 3.95 7.02 5.01 231 18.5 6.3 14 8.6 99 1500 IZE EX CE be 117 112 TE 12.4 12.65 33 13.55 3,43 W 3.30 303 0250 PRE 14 2.08 0.358 0.68 370 902 0241 920 3.41 0244 kis pH 114 FFERE 4.08 0.240 Find Latest 0.076 0.009 164 1.00 0.002 2013 0.005 0.014 2.05 0.112 0.031 214 0.154 0.048 0,200 0.050 0.00 224 0.294 2.90 0341 0042 0.300 0014 331 0.361 0.092 344 0.302 148 pri 12.36 4.05 4.03 4.75 13.21 4711 511 HP B TTR 434 447 0132 0284 -0.001 0.29 0.054 0.268 -0.009 www 0040 0.014 0015 411 0379 4TE 0251 0.023 0.298 BATH 0.300 0342 0122 0.381 0207 0.422 0,360 0760 1.658 3.400 HAL 0.003 0641 10.90 1764 0.027 0013 0005 GLOD BIRD 960 0 Second Derivative Fin Derivative -10 10 -10- 10 10 Vo Volum (m) 20 20 (4x2.98 Ay0.00) 10 Volume (mk) 20
pl → Titration of a Polyprotic Acid Your objectives are: • To observe the shape of the titration curve of a polyprotic acid • To determine the values of K and Ka for oxalic acid A polyprotic acid, such as H.C.O, or H₂PO, will have more than one equivalence point. A theoretical titration curve for the titration of 10 mL of 0.1 M phosphoric acid with 0.1 M NaOH is shown be- low. Because the initial acid and the titrating base concentrations are equal and the initial acid volume was 10 mL, the first equivalence point should appear at 10 mL added base. The graph shows this point clearly, along with the clear second equivalence point at 20 mL added base. But the third equivalence point (at 30 mL added base) is not distinct. This point is washed out because Kas is close to Kw. the water dissociation constant. Phosphate ion is a fairly strong base and competes with OH for hydrogen ions, at high pH. Note that the half-equivalence points are found in each buffer region, midway between adjacent equivalence points. 14 12 10 N 0 10 ml of 0.1 M H3PO4 titrated with 0.1 M NaOH 10 20 30 Volume Base/al. → Figure 1. Titration curve of phosphoric acid. Equivalence points HPO H₂O HPO, HO 522 updated pre/postlab + HPO,+HO PO,³ + H₂O* For Phosphoric Acid, there are four forms of the phosphate, H₂PO4 (phosphoric acid itself), H₂PO4¹ (dihydrogen phosphate anion), HPO4 (hydrogen phosphate anion), and PO. (phos- phate anion). And there are three equilibrium reactions among these species. H₂PO, +H₂O H₂PO + H₂O Kaj Ya 40 Kay" 50 H-POH₂O¹75 10-3 [H₂PO4] HPO [H₂O+] 623 10-3 [H₂PO₂ no hul [POZ][H₂0¹1 10-15 HPOT Vla D. 09 25 M
one of the equili Typically only two of the sequential species dominate the solution, reactions provides sufficient description of the solution. Which set dominates is pH depens Relative Fraction 1.0 0.8 0.6 0.4 0.2 H₂PO4 0.0 + 0 pKat N H₂PO4 pH=pK,+ log pK PH [conjugatebase] [acid] 8 HPO4 10 pk 12 PO 14 Figure 2. Relative fraction of phosphate species versus pH Since Phosphoric Acid is a weak acid, as we add NaOH, we convert some of the H3PO4 into H₂PO4 and create a buffer system. So between the start and the first equivalence point is a buffer solution. The pH of a buffer solution is described by the Henderson-Hasselbalch equa- tion. Proced 1. Calih This equation is based on assumptions that only allows it to be used for buffer solutions. In this equation we see that when [Acid] = [Conjugate Base] that the pH = pka. So once we identify the equivalence point, we can then identify the pH at half the equivalence volume (or the mid- point volume between two successive equivalence points). For Sulfuric acid, the first hydrogen is a strong acid, the second is a stronger weak acid. This may make the first equivalence point hard to see. If we see two equivalences points, it is only between these two that we have a buffer and can use the midpoint to get pH = pKaz. We may only see one equivalence point, that would be the second equivalence point, if so the midpoint to determine Kaz is 3/4 the volume used to get to the equivalence point.
Procedure: 1. Calibrate the pH probe. 2. Condition a clean, 50.0 mL buret with 0.10 M NaOH and fill. Record your initial volume. 3. Weigh out 0.001 mole oxalic acid into a 150 ml beaker. Then add 30 mL of DI water and stir to dissolve. *oolol 10.03-09.003 g 4. Place the beaker on a stir plate and position the rinsed pH probe at the side of the beaker away from the stir bar. Turn the stir plate onto a low to moderate setting. 5. Position the NaOH buret over the beaker. Add a volume of NaOH, let the pH stabilize, then record the pH. Start with approximately 1 mL additions and decrease to approximately 0.2 mL additions through each equivalence point. Expand volume additions between equiva- lence points and after the second equivalence point. Go to about 30 mL total NaOH addi- tions 6. After the titration, create a table in Logger Pro of pH and volume added NaOH. 7. Plot pH versus Volume. 8. Plot the first derivative, This makes the equivalence points more visible as peaks. In LoggerPro, add a Calculated Column. Place the cursor in the Expression box, then click the Function button, look under Calculus to find the derivative. Select that, then select the "Y" (pH) Done 9. Plot the second derivative graph. 10. Print out all three graphs. (If working with a partner or group, please print only one set of graphs to share)
Table 1. Oxalic Acid-Sodium Hydroxide Titration Suggested vol Actual Total NaOH to add, volume ml added, NaOH, mL 0-00 ML 1.00ML 1.00mL 1.00 mL 1.00ML 1.00 mL 5.01 mL 1.00ML 6.00mL 1.00mL 7.02mL 8.00 mL 8.20mL 8.40mL 8.60mL 8.80ML 9.00 mL 1.00 mL 0.20mL 0.20 mL 0.20ml 0.20 mL 0.20mL 0.20mL 0.20mL 0.20 mL 0.00 mL 1.00 mL 2.00mL 3.05mL 4.00 mL S22 updated pre/postab Measured pH Suggested vol, Actual total NaOH, mL volume, NaOH, mL 1.77 1.84 1.92 2.02 2.14 2.31 2.54 2.90 3.27 3.35 3.42 3.46 3.53 3.59 3.63 20mL 9.40mL 3.69 9.68L 3.76 0.20mL 10.00ML 3.83 4.08 0.20mL 11.01 mL 0.20mL 11.20 mL 4.13. 0.20 ML 0.20mL 020 mL 0.20 mL 0.20mL 0.20ml 0.20 m2 0.20 mL 0.20 m2 0.20 mL 0.20 ML 020mL 0.20mL 0.20 mL 0.20mL 0.20mL 0.20mL 1.00 mL 1.00 mL 11.40mL 11.60 mL 12.00 mL 12.40mL 12-65mL 14.00 ML 14,20 mL 14.40 mL 14.61 m2 14.80 m2 12.30 ML 13.00mL 13.20 m2 13.40mL 4.96 13.55mL 5.13 13.80ML 5.4.5 6.13 9.16 10.28 10.74 10.93 15.00 ML 11.08 16.00 ML 11.50 17.00 mL 11.70 18.00 mL 11.83 1.00mL 1.00ML 19.00ml 1.00L 20.00 mL 1.00-L 21.00L Measured pH 1.00mL 22.00 ML 1.00mL 23.00 ML 1.00mL 24-00 1.00PL 25.00ML 1.00 26.00ml 4.19 4.23 4.34 4.47 4.57 4.63 4.71 4.81 11.92 12.00 12.06 12.11 12.15 12-19 12.22 12.25 4 2 200. Su tre as ib hi id: al 10 fic OL ni hi on lic PH a NaOH AL 100 27.00 12.28 2.00 28.00-12.31 11.00 29.00mL 12-33 (1.00 30-00-4 12.35 20
20 29 30 21 22 23 24 25 Can 20 27 . U 20 7 . 18 17 18 . 10 11 12 13 14 16 EL Volume THE 2700 07.02 14 M 64 PAWI 3.00 8.01 2.25 AN e 20 04 ved 10 11.01 112 11K 11.6 210 124 12.65 12.4 13 13.06 132 134 p pri 12.36 KE SON 334 6011 130 0.009 LOL 0.117 EBANY 16 REER REE Help BUC 3.71 200 Dine 254 KOME 0.004 IMG 331 REE HID bords 463 471 4.01 4.00 ASE HU ONE 4.00 2.42 0308 2013 341 8.254 4001 10.20 0144 3.50 0.200 4.000 361 10.35 -0.04 0.254 4017 0241 0.244 0.005 BEE 0.014 koro 421 0.204 434 0.305 447 0.342 4.57 10.301 0422 APON THES 0014 400 0.249 0014 413 0279 0015 100 0.021 DOTS CON 0.125 0207 3200 0503 0.001 0.000 3.800 0.700 1.658 ANDRUNG WAS 10- Volume ink) Volume (n) 200.00 10 Volume ( 30