A 2.00 x 10^-3 mole sample of actylsalicylic acid was dissolved in 15.00 ml of w

Question

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              ?

Explanation / Answer

Chemical reaction:
C9H8O4 (aq) + NaOH (aq) ---------->NaC9H8O4 (aq) +H2O (l)

Given that it took 20.00 mL of NaOH to neutralize given acid.Thus after adding 10 mL of NaOH half waf point is reached at whichnumber of mole sof acid is equal to the number of moles of base. Soa buffer with this situation is having pH equal to its dissociationconstant of the acid. Given at 10.00 mL pH = 3.44
pH = pKa
pKa = 3.44
Ka = 3.6 x 10-4

Part2:
Number of moles of acetylsalicylic acid = 2.00 x10-3 moles
Concentration = 2.00 x 10-3 moles / 0.015L
= 0.133 M

C9H8O4 (aq) + NaOH (aq) ---------->NaC9H8O4 (aq) +H2O (l)
Before rxn ( moles) 2.00 x10-3 25 mL * 0.1 M
0.0025

Afterran 0 0.0005 2.00 x 10-3

Here hydroxide ion and NaC9H8O4,both are basic in nature. But hydroxide being strong base itsupress the dissociation ofNaC9H8O4.

[OH-] = 0.0005 moles / (0.015 + 0.025 )L
= 0.0125 M
pOH = 1.9
pH = 14 - 1.9
= 12.1

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