Question

The pH of a 0.74 M solution of alloxanic acid (HC₄H₃N₂O₅) is measured to be 3.39.

Calculate the acid dissociation constant Ka of alloxanic acid.

Be sure your answer has the correct number of significant digits.

Answer 1

In equilibrium, for any acid:

HA <-> H+ + A-

where HA is the molecular acid, H+ the proton and A- the conjguate base

then

Ka = [H+][A-]/[HA]

Assume

[H+] = [A-] = -x

[HA] = M-x = 0.74 - x

calculate x

x = from pH

pH = -log(H)

[H+] = 10^-pH = 10^-3.39 = 0.00040738 M

then

[H+] = [A-] = x = 0.00040738

[HA] = M-x = 0.74 - x = 0.74-0.00040738 = 0.73959

substitut ein Ka

Ka = (0.00040738 )(0.00040738) /(0.73959 ) = 2.2439*10^-7

sig fig --> 3

Ka = 2.24*10^-7