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30 August, 12:54

What is the [oh-] in a sample of lime juice with a ph of 2.0?

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  1. 30 August, 13:41
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    As we know,

    pH + pOH = 14

    Finding pOH,

    pH + pOH = 14

    2 + pOH = 14

    pOH = 14 - 12

    pOH = 12

    Concentration of [OH⁻] is calculate as,

    pOH = - log [OH⁻]

    12 = - log [OH⁻]

    -12 = log [OH⁻]

    [OH⁻] = 10⁻¹² Taking antilog

    [OH⁻] = 1 * 10⁻¹²
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