Asked by Peighton
The Ksp of manganese(II) hydroxide, Mn(OH)2, is 2.00 × 10-13. Calculate the molar solubility of this compound.
Answers
Answered by
DrBob222
...............Mn(OH)2 ==> Mn^2+ + 2OH^-
I..............solid.................0.............0
C............solid..................x............2x
E............solid...................x.............2x
Write the expression for Ksp for Mn(OH)2, plug in the E line above, and solve for x =-(Mn^2+)= [Mn(OH)2] and 2x = (OH^-)
Post your work if you get stuck.
I..............solid.................0.............0
C............solid..................x............2x
E............solid...................x.............2x
Write the expression for Ksp for Mn(OH)2, plug in the E line above, and solve for x =-(Mn^2+)= [Mn(OH)2] and 2x = (OH^-)
Post your work if you get stuck.
Answered by
Anonymous
Ksp=[Products]/[Reactants]
Mn(OH)2 -----------> Mn^+ + 2OH^-
..............Mn(OH)2 ==> Mn^2+ + 2OH^-
I..............solid.................0.............0
C............solid..................x............2x
E............solid...................x.............2x
Ksp=[x][2x]^2
Ksp=4x^3
Ksp/4=x^3
Solve for x.
Mn(OH)2 -----------> Mn^+ + 2OH^-
..............Mn(OH)2 ==> Mn^2+ + 2OH^-
I..............solid.................0.............0
C............solid..................x............2x
E............solid...................x.............2x
Ksp=[x][2x]^2
Ksp=4x^3
Ksp/4=x^3
Solve for x.
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