Titration curve for pH vs. volume of NaoH 12.00 10.00 6.00 2.00 0.00 10.00 25.00

Question

Titration curve for pH vs. volume of NaoH 12.00 10.00 6.00 2.00 0.00 10.00 25.00 30.00 35.00 40.00 45.00 15.00 20.00 0.00 5.00 Volume of 0.1906M NaoH (in mL) Appendix A1: Part 1: Titration of a diprotic cid Volume of Measured NaoH burette volume of NaoH pH reading (mL added (mu) 2.000,5 2.261 2.43 13.3 13.8 2.68 15.7 6.2 2.84 16,8 7,3 3.07 17,8 18.3 3,40 18,8 19.3 4.20 19.4 9,9 5.15 19.9 20,4 5.49 20.3 20.8 5.71 20,7 5.91 21.4 21.9 6.10 22.4 22,9 6.30 23.7 6.50 25.5 6,71 27,6 28.1 6.92 29.9 30.4 7.13 32 32.5 7.32 33,8 34.3 7.53 35.3 35.8 7,89 37,2 8.00 37.5 8,40 38,4 9.29 39.1 9,6 10.23 39.1 39.6 10.55 39.7 40.2 10.6840 1 406 10.82 41.3 41,8 1.03 41,9

Explanation / Answer

2. From the first graph

let 39 ml of acid solution is titrated and it took 39 ml of NaOH to reach IInd Eq point by graph

excess [OH-] after IInd Eq point = 0.1906 M x 0.1 ml/(39 + 39) ml

                                                   = 0.00024 M

pOH = -log[OH-] = 3.612

pH = 14 - pH = 10.388

3.From the graph

let 14.25 ml of base solution is titrated and it took 14.25 ml of HCl to reach IInd Eq point by graph

excess [H+] after IInd Eq point = 0.2568 M x 0.1 ml/(14.25 + 14.25) ml

                                                   = 0.00090 M

pH = -log[H+] = 3.045

[Note. We have assumed Volume of Acid in Ist and volume of base solution in the IInd titration. By putting the correct values and following the same procedure one may get the exact pH values]

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