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11 November, 15:31

Assuming complete dissociation, what is the ph of a 3.24 mg/l ba (oh) 2 solution?

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  1. 11 November, 17:30
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    We need to first find the molarity of Ba (OH₂) solution.

    A mass of 3.24 mg is dissolved in 1 L solution.

    Ba (OH) ₂ moles dissolved - 3.24 x 10⁻³ g/171.3 g/mol = 1.90 x 10⁻⁵ mol

    dissociaton of Ba (OH) ₂ is as follows;

    Ba (OH) ₂ - - > Ba²⁺ + 2OH⁻

    1 mol of Ba (OH) ₂ dissociates to form 2OH⁻ ions.

    Therefore [OH⁻] = (1.90 x 10⁻⁵) x2 = 3.8 x 10⁻⁵ M

    pOH = - log[OH⁻]

    pOH = - log (3.8 x 10⁻⁵)

    pOH = 4.42

    pH + pOH = 14

    therefore pH = 14 - 4.42

    pH = 9.58
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