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30 April, 15:16

Given a diprotic acid, H2A, with two ionization constants of Ka1 = 2.4x10^-4 and Ka2 = 4.9x10^-11, calculate the pH for 0.182 M solution of NaHA.

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  1. 30 April, 18:03
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    pH = 6.964 which rounds to 7.0 if 2 sig figs

    Explanation:

    Ka₁ * Ka₂ = x

    2.4x10⁻⁴ * 4.9x10⁻¹¹ = 1.176x10⁻¹⁴

    √1.176x10⁻¹⁴ = 1.084435337x10⁻⁷

    pH = - log[H⁺]

    pH = - log[1.084435337x10⁻⁷]

    pH = 6.964796339
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