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1 July, 03:17

Phosphoric acid, H 3 P O 4 (aq), is a triprotic acid, meaning that one molecule of the acid has three acidic protons. Estimate the pH and the concentrations of all species in a 0.100 M phosphoric acid solution.

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  1. 1 July, 07:13
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    H3PO4 = 0.30 M

    H2PO4 - = 4.59 x 10^-2M

    HPO2^4 - = 6.17 x 10^-8 M

    PO3^4 - = 6.44 x 10^-19 M

    H + = 4.59 x 10^-2 M

    Phosphoric acid can be classified as a weak acid which dissciates to give three acid where the products of one dissociation equilibrium become components of the subsequent dissociation equilibrium. Below are their equilibrium equations:

    H3PO4 (aq) ⇌ H2PO4 - (aq) + H + (aq) H2PO4 - (aq) ⇌ HPO2^4 - (aq) + H + (aq)

    HPO2^4 - (aq) ⇌ PO3^4 - (aq) + H + (aq)

    The pH for the concentrations are:

    Ka = 10^-pKa,

    Ka1 = 10-2.16

    = 6.17 x 10^-3

    Ka2 = 10-7.21

    = 6.17 * 10^-8

    Ka3 = 10-12.32

    = 4.79 * 10

    The following concentrations are:

    Solving for [H2PO4^-] and [H+],

    Ka1 = {[H2PO4^-]*[H+]}/[H3PO4]

    6.92 x 10^-3 = ((x) * (x)) / 0.35 - x

    2.42 x 10^-3 - 6.92 x 10^-3 = x2

    Solving the above using general quadratic formula,

    x = 4.59 x 10^-2M

    x = [H2PO4^-] = [H+]

    Solving for [HPO4^2-],

    Ka2 = {[HPO4^2-]*[H+]}/[H2PO4^-]

    6.17 x 10^8 - = ((x) * (4.59 x 10-2)) / 4.59 x 10^-2

    x = 6.17 x 10^-7

    x = [HPO4^2-]

    Solving for [PO4^3-],

    Ka3 = [PO4^3-]*[H+]}/[HPO4^2-]

    4.79 x 10^-13 = ((x) * (4.5 x 10^-*)) / 6.17 x 10^-8

    x = 6.44 x 10^-19M

    x = [PO4^3-]
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