Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 decade ago

Calculate the [H3O ] for 0.018 M Ca(OH)2.

Answer in units of M.

2 Answers

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  • 1 decade ago
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    Ca(OH)2 in water dissociates into its ions, so

    Ca(OH)2 --> Ca^2+(aq) + 2OH-

    Because we have a 2:1 mole ratio between Ca(OH)2 and OH-, then [OH-] = 0.036M

    We know that

    [H3O+][OH-] = Kw = 1E-14

    Thus, [H3O+] = 1E-14 / [OH-] = 1E-14 / 0.036 = 2.78E-13 M

    [Answer: see above]

  • 1 decade ago

    To keep it simple, I will assume this is at 25C, in water, and first year chemistry (a little more than 101):

    Calcium Hydroxide is not very soluble in water. From teh solubility product:

    Ca(OH)2 --> Ca(2+) + 2OH(-)

    Ksp = [Ca][OH]^2 = 5.2 X 10-6

    Set up your ICE table:

    Ca(2+) OH-

    Initial 0 0

    Change + x + 2x

    Equilibrium x 2x

    5.2 X 10-6 = (x)(2x)^2 = 4x^2

    Solving for x, x = 0.011 M

    So, in this simple calculation, you have too much Calcium hydroxide, only 0.011 M is actually in solution, the rest is precipitated on the bottom.

    Since x = 0.011 M, then [OH] = 2x, so [OH] = 0.022

    In any aqueous solution,

    [H][OH] = Kw = 1 X 10-14

    1 X 10-14 = (x)(0.022)

    therefore [H+] = 4.54 X 10^-13 and we all know [H] is actually [H3O]

    (pH = 12.34)

    IN REALITY:

    You would need to take a systemmatic approach to solving this.

    1. Write out relevant equations

    2. Write out the mass balance

    3. Write out the charge balance

    4. Write out the relevant equilibria

    5. Solve for [H]

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