..........M(OH)2 ==> M^2+ + 2OH^-
E.........solid......x.......2x
pH = 10.14 pOH = 14-10.14 = 3.86
pOH = -log(OH^-)
Solve for OH^- = about 1.4E-4 but that's approximate. That means (M^2+) = 1/2 that.
Ksp = (Mg^2+)(OH^-)^2
Substitute and solve for Ksp.
At 22C(Celsius) an excess amount of a generic metal hydroxide M(OH)2 is mixed with pure water. The resulting equilibrium solution has a pH of 10.14. What is the Ksp of the salt at 22C? Help Please
2 answers
Mr. DrBob222,
would you mind explaining that a little more? I do not completely understand what is happening in the problem. Also, how do you know that Mg^2+ is half of OH^-?
would you mind explaining that a little more? I do not completely understand what is happening in the problem. Also, how do you know that Mg^2+ is half of OH^-?
1 answer