Chemistry [OH-] and [H3O+] Questions

What is the [OH-] of a 4.0x10^-4M solution of Ca(OH)2?


Given the following [H3O+] values, determine the pH of each solution
A. 1.0x10^-7 M
B. 1.0x10^-3 M
C. 1.0x10^-12 M
D. 1.0x10^-5 M

Thank you for your help.. I'm so lost. If you can explain some, that'd be great :( .

Ca(OH)2 solution Molarity is 4.0x10^-4 Therefore the OH- = 8x10^-4
pOH = - log of the {OH-} = -log of 8x10^-4 = 4.00 - 0.90 = 3.10 pH = 14.00 -3.10 = 11.90
Take a look at your log chart. The log of 1.0 = 0.00
The pH's are 3.00, 2.00 and 5.00

For your first question: Ca(OH)2 ----> Ca+2 + 2OH-

Ca(OH)2 is a strong base, disassociates 100%. Two mols of OH- are produced so you have to multiply the concentration by 2. 4.0x10^-4M x 2 = 8.0x10^-4 = [OH-]

For the A-D all you have to do is take the negative log of the [H+] concentration.

So for example -log(1.0x10^-7) = pH

8 * 10^-4 M in [OH-]

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pH = -log10[H3O+]

log10[10^x]=x
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A 7
B3
C12
D 5

yep that easy