Question

E H M N o R D dph/dvol Activity: 2.88 g of an unknown diprotic acid H2A were weighted and dissolved in delonised water to a final volume of 250 ml. 25 ml of this H2A solution was titrated with a standardized solution of NaOH 0.100 M. The table shows the titration data of pH as a function of NaOH added. Show your calculation below. B C 1 vol Naomi рн Avg Vol 2 0.00 1.52 3 5.00 1.74 4 7.50 1.85 5 10.00 1.97 6 12.50 2.1 7 15.00 2.24 18.00 2.44 9 20.00 2.62 10 22.50 2.95 11 24.00 3.37 12 24.50 3.08 13 25.00 4.11 14 25.50 4.54 15 26.00 16 30.00 5.63 17 40.00 6.41 18 45.00 6.83 19 49.00 7.61 20 49.50 21 50.00 9.38 50.50 10.82 51.00 11.12 24 55.00 11.8 60.00 12.07 26 1) Plot pH Yaxis) vs Vol NaOH (Xaxis), label the plot as H2A titration and add axis titles. 2) Fill out the column Avg Vol by calculating the average volume (vol 2 /vol 1)/2 Fill out the column dpt/dvol-2-ph/Vol2-Voll) Plot dph/dvol Yaxis) vs Avg Vol (Xaxis), label the plot as ist derivative and add axis titles 3) From the ist derivative plot, select the Ave volume of Neow at the apex of the peaks representing the two equivalence points. The Avy volume of NICH at ist equivalence point is The Aug volume of NaOH at 2nd equivalence point is – mi Which of these two volumes is expected to be more accurate? Decide based on the sharpest change on ph from the H2A thration plot. Use this Ave volume for the calculation in 04. answer 4) The number of moles of NaOH up to the equivalence point selected is: (show your calculation 5) Depending on your selection in Q3, use Eq3or5 from the lab handout to calculate the moles of H2A present in the titrated sample (25 mi).

PLEASE HELP WITH THE WHITE BOX QUESTIONS & THE HIGHLIGHTED YELLOW AREA. BOLD THE ANSWERS PLEASE