The pH at which Ni(OH)2 begins to precipitate from
a solution containing 10^-3M Ni2+
[ksp of Ni(OH)2 = 10^-11]
(1) pH = 3
(2) pH = 10
(3) pH = 13
(4) pH = 11
Answers
Explanation:
The idea here is that you need to use magnesium hydroxide's solubility product constant to determine what concentration of hydroxide anions, OH−, would cause the solid to precipitate out of solution.
As you know, the dissociation equilibrium for magnesium hydroxide looks like this
Mg(OH)2(s]⇌Mg2+(aq]+2OH−(aq]
The solubility product constant, Ksp, will be equal to
Ksp=[Mg2+]⋅[OH−]2
Rearrange to find the concentration of the hydroxide anions
[OH−]= ⎷Ksp[Mg2+]
Plug in your values to get
[OH−]=√1.0⋅10−110.1=√1.0⋅10−10=10−5M
As you know, you can use the concentration of hydroxide anions to find the solution's pOH
pOH=−log([OH]−)
pOH=−log(10−5)=5
Finally, use the relationship that exists between pOH and pH at room temperature
pOH + pH=14
to find the pH of the solution
pH=14−5=