Asked by Miaow

At 22 °C an excess amount of a generic metal hydroxide M(OH)2, is mixed with pure water. The resulting equilibrium solution has a pH of 10.56. What is the Ksp of the compound at 22 °C?

I have found my [OH-] concentration (3.63x10^-4) but I'm not sure I'm setting up my ICE table or my Ksp expression right:

M(OH)2 ---> M^2+ + 2OH^-
---------------------------------
excess 0 2(3.63x10^-4)
+x +x
----------------------------------
less x (7.26x10^-4)x

Ksp = [M][OH^-]^2

=(x)(7.26x10^-4x +x)

Not real sure where to go from here...
Thank you for any guidance!
Answered by DrBob222
Your 3.63E-4 = (OH^-) is correct but your ICE table needs some work.
..........M(OH)2 ==> M^2+ + 2OH^-
equil.......................3.63E-4
[note: If the pH is 10.58 and OH^- = 3.63E-4, then that's what OH IS. It isn't 2 times that. So
Ksp =- (M^2+)(OH^-)^2
(M^2+) = 1/2 x 3.63E-4 = 1.82E-4
(OH^-) = 3.63E-4
Substitute into Ksp expression and solve for Ksp.
Answered by Miaow
Oh geez, thank you SO much! And so quick!
Answered by jen
How did you find the concentration of [OH-]
Answered by vox
14-pH = pOH

From the pOH, you can calculate [OH-] raising 10 to the opposite pOH power.

[OH-]=10^(-pOH)
Answered by hey
why is OH/2 of M^2+
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