Question

with these titrations can you help with the questions below? And for like number 2: is it something like this? 0.2568 M x 0.1 ml/(14.25 + 14.25) ml and if it is i dont understand why its (14.25 + 14.25) ml twice? can you possibly explain?

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

Answer 1

It could be the volume salt of the diprotic acid (inital) = 14.25 mL (which is not given in the question)

Intial volume of diprotic acid salt = 14.25 mL

and after the second equivalence point, you reach to a volume = 14.25 mL + 14.25 mL

To calculate the molarity of H+ in the solution after addition of 0.2568 M , 0.1 ml HCl; ; use M1V1 = M2V2

M2 = 0.2568 M x 0.1 ml/ 14.25 mL + 14.25 mL

M2 = 0.0009 M

pH = -log[H+] = -log(0.0009) = 3.05