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11 January, 06:59

Like all equilibrium constants, Kw varies somewhat with temperature. Given that Kw is 3.31 * 10-13 at some temperature, compute the pH of a neutral aqueous solution at that temperature.

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  1. 11 January, 07:27
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    Since Kw = [H⁺][OH⁻], and the concentration of both substances are the same, the equation is now Kw=[H⁺]²

    So,

    3.31x10⁻¹³ = [H⁺]²

    Take the square root = 5.75x10⁻⁷

    Then take the negative log to find the pH:

    -log (5.75x10⁻⁷) = 6.25
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