Question

What is the pH of a 0.28 M solution of ascorbic acid

Answer 1

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!