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A 2.00 x 10^-3 mole sample of actylsalicylic acid was dissolved in 15.00 ml of water and titrated with .100 M NaOH. The equivalence point was reached after 20.0ml of NaOH solution had been added. Using following data calculate Ka for acetylsalicylic acid and the PH of solution after total volume of 25.00 ml of NaOH solution had been added

Volume of .1 M NaOH added : 0.00 pH = 2.22

5.00 pH = 2.97

10.00 pH = 3.44

15.00 pH = 3.92

20.00 pH = 8.13

25.00

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