Question

STIONS 26- 31 REFER TO THE FOLLOWING CHART IT SHOWS RESULTS OF A TITRATION OF 20 ML OF LCI WITH 0.75 M NaOH. YOUR WORK HCT WITH 0.75 M NaOH. SHOW Logger Pro Column Data: Volume in mlph First Derivative Second Derivative 0.02 002 0.05 0.02 0.01 0.06 0.06 0.07 0.10 0.03 0.04 0.03 0.05 011 0 00 0.13 0.18 0 33 24. Circle the part of the curve at the equivalence point. 0.00 2.25 200 228 3.00 2.32 4.00 2.37 500 2.43 6.00 2.49 6.50 2.54 7.00 2.59 7.50 2.65 800 2.72 8.50 2.81 9.00 2.93 9.50 3.10 9.70 3.20 10.00 3.35 10 20 3.60 10.40 3.98 10.50 4.17 10.55 4.40 10.60 4.79 10.65 5.31 10.70 5.91 10.75 6.49 10.80 7.17 10.85 8.36 10.90 886 10.95 9.12 11.00 9.30 11.10 9.56 11.30 9.91 11.50 10.19 12.00 10.53 13.00 10.95 14.00 11.16 15.00 11.29 16.00 11.40 18.00 11.52 20.00 11.61 0.13 0.16 0.21 029 0.39 0.52 0 80 1.55 1.00 2.80 6.20 9.10 11.20 11.80 12.50 18.70 16.90 7.60 25. Find and write the NaOH volume before and after the drop producing the largest pH increase was added. 4.40 0.83 2.06 2.75 4.17 28 67 63.00 50.00 27.00 14.00 69.00 43.00 -111.00 -125.00 -46.67 -15.78 4.53 -2.87 -1.53 -0.38 -0.17 -0.10 -0.05 -0.02 -0.01 0.00 26. Determine the volume of NaOH added at the equivalence point. To do this, add the two NaOH values determined above and divide by two. 293 203 27. Calculate the number of moles of NaOH used. 1.58 0.89 0.51 0.32 0.17 0.12 0.08 0.05 0.05 28. Using the equation for the neutralization

Answer 1

Equivalence point =12 ml is end of sudden rise in pH. Veq=14 +V)/2 =(10+12)/2 =11 mL V=10ml is start of sudden rise in pH Number of moles of NaOH used = molarity of NaOH used X volume of NaOH used in litres = 0.75 (mol/L) X 0.011L =0.00825 mol.

Comment in case of any doubt.