Question

The acid dissociation constant Ka of alloxanic acid HC4H3N2O5 is ×2.2410−7. Calculate the pH of a 0.57M solution of alloxanic acid. Round your answer to 1 decimal place.

Answer 1

Answer

pH = 3.4

Explanation

Ionization equilibrium of alloxanic acid is

HC4H3N2O5(aq) + H2O(l) <-------> C4H3N2O5^-(aq) + H2O(l)

K_a = [C43N2O5^-] [ H3O⁺] / [HC4H3N2O5] = 2.24 ×10^-7

Initial concentration

[HC4H3N25] = 0.57

[C4H3N2O5^-] = 0

[H3O⁺] = 0

change in concentration

[HC4H3N2O5] = -x

[C4H3N2O5^-] = + x

[H3O⁺] = +x

so,

x²/( 0.57 - x) = 2.24 ×10^-7

x = 0.0003572

[H3O⁺] = 0.0003572M

pH = -log( 0.0003572M)

pH = 3.4