Question

1. What is the pH of a 0.35 mol/l solution of acetylsalicylic acid (C9H8O4)? 2 points Your answer

Answer 1

Answer) 2.0 is the pH of acetylsalicylic acid.

Explanation:

Acetylsalicylic acid is a weak acid. Write the dissociation reaction of acetylsalicylic acid as follows:

Concentration Cq Hg Oy (aq) H₂O H₂ot Cq H₂ Oy 0.35 Initial change Final farge negligible large +x 0.35-x X x² Ka = [H,ot][C, H, O, J [CoHgOy 0.35-x

The equilibrium constant is written as:

х К. : [н,o'][с, н, 0 ] [снооч ) 0.35 – Х. Ka Cq Hq Oy is weak acid, 0.35-x ~ 0.35 Ka б. 35

pKa of acetylsalicylic acid is 3.5

pKa = - log Ka

Ka = 10^-pKa

Ka = 10^-3.5

-3.5 2 10 x 0.35 -3.5 = 0.35 X 10 g? = 0.000110679 22 tz x = 10.000110679 x=0.0105 M = [H,04]

Calculate the pH as follows:

pH = - log[H_{​​​​​​3}O⁺] = - log(0.0105) = 1.97 ~ 2.0