Anonymous

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

Relevance

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

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

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]