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 10.18. What is the Ksp of the salt at 22 °C? Ksp =
POH = 14 - pH
= 14 - 10.18
= 3.82
use:
pOH = -log [OH-]
3.82 = -log [OH-]
[OH-] = 1.514*10^-4 M
M(OH)2 gives 2 OH-
So, solubility , s = 1.514*10^-4/2
s = 7.568*10^-5 M
At equilibrium:
M(OH)2 <---->
M2+
+ 2
OH-
s 2s
Ksp = [M2+][OH-]^2
Ksp = (s)*(2s)^2
Ksp = 4(s)^3
Ksp = 4(7.568*10^-5)^3
Ksp = 1.734*10^-12
Answer: 1.73*10^-12
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