Asked by jen
This question was posted: 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 am looking at it and cannot find how they got the concentration of [OH-]. Everything that needs to be done after that I understand. I just do not get how they got the concentration
I have found my [OH-] concentration (3.63x10^-4)
but I am looking at it and cannot find how they got the concentration of [OH-]. Everything that needs to be done after that I understand. I just do not get how they got the concentration
Answers
Answered by
DrBob222
Your post isn't entirely clear as to what you do and don't understand. Here is how you do it.
........M(OH)2 ==> M^2+ + 2OH^-
The problem tells you pH = 10.56 which makes pOH = 3.44 and from that
pOH = -log(OH^-) = 3.44
Therefore, (OH^-) = 3.63E-4
Ksp = (M^2+)(OH^-)^2 = ?
(M^2+) = 1/2 x 3.63E-4
(OH^-) = 3.63E-4
Ksp = about 2E-11
........M(OH)2 ==> M^2+ + 2OH^-
The problem tells you pH = 10.56 which makes pOH = 3.44 and from that
pOH = -log(OH^-) = 3.44
Therefore, (OH^-) = 3.63E-4
Ksp = (M^2+)(OH^-)^2 = ?
(M^2+) = 1/2 x 3.63E-4
(OH^-) = 3.63E-4
Ksp = about 2E-11
Answered by
jen
I was just having trouble with getting the [OH-] concentration. I had the answer just didn't know how to get to it. Thank you, it helped.
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