Calculate the concentration in M (unsure)?

Question

How do I calculate the concentration of Ca(OH)2 solution with a pH of 11, in M?

11 = -log[H3O+]

[H3O+] = (10^-11)/(10^-14)

= 10^3

(10^3)/2 = 500M (which I'm sure it's wrong).

I think the answer given was 5 x 10^-14 M.

Sorry if my working out is stupid. My teacher is a completely useless and hasn't taught us - I've tried to learn it myself :(

Thanks a lot for any help. I appreciate it.

Answer 1

You are dealing with a base, Ca(OH)2. Under these conditions it is easier to work with pOH and not pH

Remember the equation:

pH+pOH = 14

You have pH = 11, Therefore pOH = 3.00

[OH-] = 10^-pOH

[OH-] = 10^-3.00

[OH] = 0.001moles OH per litre

But Ca(OH)2 ionises to produce 2 OH -ions per mol

Ca(OH)2 <=> Ca  2+   +  2OH-

1mol Ca(OH)2 gives 2mol OH-

You must therefore divide the [OH-] by 2 to get the molarity of Ca(OH)2

0.001/2 = 0.0005

Molarity of Ca(OH)2 = 0.0005M