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Today, 08:12

What is the PH of a 0.05M solution of the strong base Ca (OH) 2 at 25C

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  1. Today, 08:13
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    pH = 14 - pH = 14 - 1 = 13

    Explanation:

    0.05M Ca (OH) ₂ = > 0.05M Ca⁺ (aq) + 2 (0.05M) OH⁻ (aq)

    pOH = - log[OH⁻] = - log (0.100) = 1

    pH = 14 - pH = 14 - 1 = 13
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