What is the pH of solution of Vitamin C ( molecular formula C6H8O6) that contains 6.39 mg of the vitamin per ml of water ?
Vitamin C is a polyprotic acid with acid equilibrium constants equal to : K a1 = 1.20 × 10-2 and K a2 = 1.60 × 10-12
Use only the first ionization of the acid to determine the pH here.
When I do it, I get the quadratic equation:
1.20e-2 = x^2 / .03631 - x
And I get x = .02087
Then since x = [H3O+], I did:
pH = -log (0.02087) = 1.68
When I submit my answer I keep getting it incorrect. What am I doing wrong? Help please?
Vitamin C is a polyprotic acid with acid equilibrium constants equal to : K a1 = 1.20 × 10-2 and K a2 = 1.60 × 10-12
Use only the first ionization of the acid to determine the pH here.
When I do it, I get the quadratic equation:
1.20e-2 = x^2 / .03631 - x
And I get x = .02087
Then since x = [H3O+], I did:
pH = -log (0.02087) = 1.68
When I submit my answer I keep getting it incorrect. What am I doing wrong? Help please?