Answer to Question #143013 in Inorganic Chemistry for Micheal

Question #143013

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


1
Expert's answer
2020-11-09T14:05:13-0500

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:

"\u03b1_1 = \\frac{[H_3A]}{[H_3A] + [H_2A^-] + [HA^{2-}] + [A^{3-}]}"


"\u03b1_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-}]}"


"\u03b1_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.

"\u03b1_2 = \\frac{\\frac{[H^+]^2[A^{3-}]}{K_2K_3}}{denominator}"


"\u03b1_2 = \\frac{[H^+]^2}{\\frac{[H^+]^3}{K_1} + [H^+]^2 + [H^+]K_2 + K_2K_3}"


"\u03b1_3 = \\frac{\\frac{[H^+][A^{3-}]}{K_3}}{denominator}"


"\u03b1_3 = \\frac{[H^+]}{\\frac{[H^+]^3}{K_1K_2}+\\frac{[H^+]^2}{K_2} + [H^+] + K_3}"


"\u03b1_4 = 1 \u2013 \u03b1_1 \u2013 \u03b1_2 - \u03b1_3"


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