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14 December, 08:33

Assuming complete dissociation, what is the pH of a 4.82 mg/L Ba (OH) 2 solution?

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  1. 14 December, 08:35
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    Ba (OH) ₂ is a strong base and completely ionises into its corresponding ions

    Ba (OH) ₂ - - - > Ba²⁺ + 2OH⁻

    1 mol of Ba (OH) ₂gives 2 mol of OH⁻ ions.

    molarity of Ba (OH) ₂ is 4.82 x 10⁻³ g/L / 171.3 g/mol - 2.81 x 10⁻⁵ mol/L

    since 1 mol of Ba (OH) ₂gives 2 mol of OH⁻ ions.

    therefore [OH⁻] = 2[Ba (OH) ₂]

    [OH⁻] = 2 x 2.81 x 10⁻⁵ mol/L = 5.62 x 10⁻⁵ M

    pOH can be calculated using the OH⁻ concentration

    pOH = - log [OH⁻]

    pOH = - log (5.62 x 10⁻⁵ M)

    pOH = 4.25

    pH can be calculated as follows

    pH + pOH = 14

    pH = 14 - 4.25

    pH = 9.75

    pH of solution is 9.75
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