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22 September, 12:39

What is the concentration of a solution of (a) KOH for which the pH is 11.89; (b) Ca (OH) 2 for which the pH is 11.68

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  1. 22 September, 15:14
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    [KOH] = 7.76*10⁻³ M

    [Ca (OH) ₂] = 2.39*10⁻³ M

    Explanation:

    KOH → K⁺ + OH⁻

    pH = - log [H⁺]

    14 = pH + pOH

    pOH = - log [OH⁻]

    10^-pOH = [OH⁻]

    14 - 11.89 = 2.11 → pOH

    2.11 = - log [OH⁻]

    10⁻²°¹¹ = [OH⁻] → 7.76*10⁻³ M

    As ratio is 1:1, [KOH] = 7.76*10⁻³ M

    14 - 11.68 = 2.32 → pOH

    10⁻²°³² = [OH⁻] → 4.78*10⁻³ M

    Ca (OH) ₂ → Ca²⁺ + 2OH⁻

    Ratio is 2:1, so I will have the half of base.

    4.78*10⁻³ / 2 = 2.39*10⁻³ M
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