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21 September, 12:23

What is the Kb value for a 0.17M solution of unknown with a pH of 9.45?

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  1. 21 September, 12:46
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    Kb = 4.673 E-9

    Explanation:

    B (aq) + H2O (l) ↔ BH + (aq) + OH - (aq) Kb = [BH+][OH-] / [B] pH + pOH = 14

    ∴ C B = 0.17 M

    ∴ pH = 9.45

    ⇒ pOH = 14 - 9.45 = 4.55

    ⇒ pOH = - Log [OH-] = 4.55

    ⇒ [OH-] = 2.8184 E-5 M

    mass balance:

    ⇒ C B = 0.17 M = [B] + [BH+]

    charge balance:

    ⇒ [OH-] = [BH+] ... [H3O+] is neglected, it come frome the water

    ⇒ [B] = 0.17 - [BH+]

    ⇒ [B] = 0.17 - [OH-] = 0.17 - 2.8184 E-5

    ⇒ [B] = 0.16997 M

    ⇒ Kb = [OH-]² / [B]

    ⇒ Kb = (2.8184 E-5) ² / (0.16997)

    ⇒ Kb = 4.673 E-9
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