Asked by Anon
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 10.49. What is the Ksp of the salt at 22 °C?
Answers
Answered by
DrBob222
pH + pOH = pKw = 14.
Solve for pOH. Then
pOH = -log(OH^-)
(OH^-) = about 3E-4 but that's only approximate.
.........M(OH)2 ==> M^2+ + 2OH^-
equl......solid....x.......2x
The problem tell you 2x = (OH^-) = 3E-4
Ksp = (Mg^2+)(OH^-)^2
Ksp = (x)(3E-4)^2
Remember to recalculate that 3E-4 number.
Solve for pOH. Then
pOH = -log(OH^-)
(OH^-) = about 3E-4 but that's only approximate.
.........M(OH)2 ==> M^2+ + 2OH^-
equl......solid....x.......2x
The problem tell you 2x = (OH^-) = 3E-4
Ksp = (Mg^2+)(OH^-)^2
Ksp = (x)(3E-4)^2
Remember to recalculate that 3E-4 number.
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