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13 August, 18:46

Calculate [H3O ] for the solutions below at 25 °C.

[OH-] = 1.83x10^-7 M

[H3O+] = M

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  1. 13 August, 18:53
    0
    For this question, we can use relations for pH and pOH. We calculate as follows:

    pOH = - log [OH-]

    pOH = - log [ 1.83x10^-7 M]

    pOH = 6.74

    pH + pOH = 14

    pH = 14 - 6.74

    pH = 7.26

    pH = - log [H3O+]

    7.26 = - log[H3O+]

    [H3O+] = 5.46 x 10^-8 M

    Hope this answers the question. Have a nice day.
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