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8 March, 14:59

Calculate [OH-] for a solution formed by adding 4.80 mL of 0.180 M KOH to 17.0 mL of 6.5*10-2 M Ca (OH) 2.

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  1. 8 March, 18:04
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    [OH⁻] = 0.167 mol/L

    Explanation:

    Let's apply Kps to solve this, as the Ca (OH) ₂is only slighty solube in water. We imagine that Temperature of working, is 25°C.

    Ca (OH) ₂ → Ca²⁺ + 2OH⁻ Kps = 5.02x10⁻⁶ (25°C)

    s 2s

    Kps = 2s³

    5.02x10⁻⁶ = 2s³

    5.02x10⁻⁶ / 2 = s³ → ∛ (5.02x10⁻⁶ / 2) = s

    s = 0.013

    So [OH⁻] = 0.027 mol/L

    Now, we add KOH

    [OH⁻] = 3.74x10⁻³ moles / 0.0218 L = 0.14 mol/L

    Finally, the concentration of [OH⁻] will be the sum of, hydroxide from Ca (OH) ₂ and KOH

    0.027 mol/L + 0.14 mol/L = 0.167 mol/L
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