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    Dwf11396 Posts: 1, Reputation: 1
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    #1

    Apr 22, 2013, 12:26 PM
    Chemistry
    What is the pH of a solution prepared by dissolving 3.1g of NaOH to make 305 mL of solution?
    Gernald's Avatar
    Gernald Posts: 901, Reputation: 93
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    #2

    Apr 22, 2013, 12:49 PM
    First convert grams to moles by dividing by the molar mass.

    3.1g NaOH / (22.9+16+1 g/mol) = 0.0777 mol

    Then divide the moles by the volume in liters.

    305ml = 0.305 L

    0.0777mol /0.305L = 0.2547M

    You know that: pH = -log [H3O+] = -log (1 x 10^-12) = 12 or pOH = -log [OH-]

    So using the second formula: -log (0.2547M) = 0.594

    pH + pOH = 14
    pH + 0.594 = 14
    pH = 13.4 ??


    If you just heard something that is my brain exploding; it's been a LONG time since I've done any of this, so you might want to double check my work.
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