Answer to Question #143013 in Inorganic Chemistry for Micheal

Question

Explain how do I calculate the fraction of species in the weak acid H3A?

Answer 1

H₃A <=> H⁺ + H₂A^-

$K_1 = \frac{[H^+][H_2A^-]}{[H_3A]}$

$[H_3A] = \frac{[H^+][H_2A^-]}{K_1K_2K_3}$

H₂A^- <=> H⁺ + HA^-2

$K_2 = \frac{[H^+][HA^{2-}]}{[H_2A^-]}$

$[H_2A^-] = \frac{[H^+]^2[A^{3-}]}{K_2K_3}$

HA^-2 <=> H⁺ + A^-3

$K_3 = \frac{[H^+][A^{3-}]}{[HA^{2-}]}$

$[HA^{2-}] = \frac{[H^+][A^{3-}]}{K_3}$

The equations to calculate the fractions of each species:

$α_1 = \frac{[H_3A]}{[H_3A] + [H_2A^-] + [HA^{2-}] + [A^{3-}]}$

$α_1 = \frac{\frac{[H^+]^3[A^{3-}]}{K_1K_2K_3}}{\frac{[H^+]^3[A^{3-}]}{K_1K_2K_3} + \frac{[H^+]^2[A^{3-}]}{K_2K_3} + \frac{[H^+][A^{3-}]}{K_3} + [A^{3-}]}$

$α_1 = \frac{[H^+]^3}{[H^+]^3 + K_1[H^+]^2 + K_1K_2[H^+] + K_1K_2K_3}$

The denominator is the same for all of the fractional species equations.

$α_2 = \frac{\frac{[H^+]^2[A^{3-}]}{K_2K_3}}{denominator}$

$α_2 = \frac{[H^+]^2}{\frac{[H^+]^3}{K_1} + [H^+]^2 + [H^+]K_2 + K_2K_3}$

$α_3 = \frac{\frac{[H^+][A^{3-}]}{K_3}}{denominator}$

$α_3 = \frac{[H^+]}{\frac{[H^+]^3}{K_1K_2}+\frac{[H^+]^2}{K_2} + [H^+] + K_3}$

$α_4 = 1 – α_1 – α_2 - α_3$