Question

What is the pH of a 0.35 mol/L solution of facetylsalicylic acid 2

Answer 1

Soll - What is the pH of a 0,35 mol/L solution of acetylsalicylic and 2 =) Calculation - Given: concentration of solution = 0.35 mol = 0,35m. Acetylsalicylic auid is week oeid not dissociate completely in water, ka for cilety/scalicylic acid = 3.0 X 10 =) Dissociation of acetylsalicylic acid in water of their ICE table is as follows- H Cg HyO4 + H2O cus cas) 300x104 Hoot + Cg Hy84 (49) car) I 0.35 m C - tze tae E 0.35-ze ze re :ka= ex ze [Hot] [eghou I CH Cg H704 0.35-ze .:: 3:0x1024 ze? - 0.35 3.0x104 = 1666.66 0.35-2 2 r4 - 3.0x 10 ze - 0.35 0.35 3.0x1.5 > Loo 2e²=(3.0x10h)x 20.35) Then ze is very small and neglect it. 2 2 = l'OS X 104

:: 2 = V C1.05x104 -2 N 20 = 2 0.0102 21.02 X 10 m. [Hoot .: [Hz oty = 1.027 10 m. => calculate ph from [Hot] pH = -log [H3ot - pH - clog (1.02% 10 Jm. :: pH = 1:99 200 -2 pH = 2:0 =) pH of 0.35 mol/L.solution of acetylsalicylic acid = 2.0