What is the ph of a 0.28 m solution of ascorbic acid

Hey there!

Values Ka1 and Ka2 :

Ka1 => 8.0*10⁻⁵

Ka2 => 1.6*10⁻¹²

H2A + H2O > H3O⁺  + HA⁻

 Ka2 is very less so I am not considering that dissociation.

Now Ka = 8.0*10⁻⁵ = [H3O⁺] [HA⁻] / [H2A]

lets concentration of H3O⁺  = X then above equation will be

8.0*10−5 = [x] [x] / [0.28 -x

8.0*10−5 = x² /  [0.28 -x ]

x² + 8.0*10⁻⁵x  - 2.24 * 10⁻⁵

solve the quardratic equation

X =0.004693 M

pH = -log[H⁺]

pH = - log [ 0.004693 ]

pH = 2.3285

Hope that helps!

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.