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21 July, 05:40

The pH of a 1.0M solution of butanoic acid HC4H7O2 is measured to be 2.41. Calculate the acid dissociation constant Ka of butanoic acid.

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  1. 21 July, 07:05
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    Ka = 1.52 E-5

    Explanation:

    CH3 - (CH2) 2-COOH ↔ CH3 (CH2) 2COO - + H3O+

    ⇒ Ka = [H3O+][CH3) CH2) 2COO-] / [CH3 (CH2) 2COOH]

    mass balance:

    ⇒ C CH3 (CH2) 2COOH = [CH3 (CH2) 2COO-] + [CH3 (CH2) 2COOH] = 1.0 M

    charge balance:

    ⇒ [H3O+] = [CH3 (CH2) 2COO-]

    ⇒ Ka = [H3O+]² / (1 - [H3O+])

    ∴ pH = 2.41 = - Log [H3O+]

    ⇒ [H3O+] = 3.89 E-3 M

    ⇒ Ka = (3.89 E-3) ² / (1 - 3.89 E-3)

    ⇒ Ka = 1.519 E-5
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