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What is the pH of a solution prepared by dissolving 1.00g of Ba(OH)2 per liter?

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  1. Moles Ba(OH)2 = 1.00 g / 171.342 g/mol =0.00584

    Ba(OH)2 >> Ba2+ + 2 OH-

    Moles OH- = 2 x 0.00584 =0.0117

    [OH-] = 0.0117 mol / 1.0 L

    pOH = - log 0.0117 = 1.93

    pH = 14 - pOH = 14 -1.93  =12.1

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