Calculate the mass of the metallic ion that can remain at equilibrium in a solution having [OH-] of 8.5 x 10^-5 M: a) Mg(OH)2 , Ksp: 6.8 x 10^-13 So, what I did was I plugged in the numbers in the Ksp expression: Ksp = [Mg][OH]^2 6.8 x 10^-13 = [Mg](8.5 x 10^-5)^2 When I found [Mg], I substituted it in Ksp = (X)(2X)^2 to verify, but I got a different Ksp... What did I do wrong? And should I assume the volume is 1 L to find the mass? Please help!! `Mg(OH)_2 rarr Mg^(2+)+2OH^-`

`K_(sp) = [Mg^(2+)][OH^-]^2`

What we need to know here is `K_(sp)`  is a property that will not change unless the temperature changes. So if you add more `OH^- ` or what ever`K_(sp)`  is a constant under given temperature conditions.

`K_(sp) = [Mg^(2+)][OH^-]^2`

`6.8xx10^(-13) = [Mg^(2+)](8.5xx10^(-5))^2`

`[Mg^(2+)] =...

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`Mg(OH)_2 rarr Mg^(2+)+2OH^-`

`K_(sp) = [Mg^(2+)][OH^-]^2`

What we need to know here is `K_(sp)`  is a property that will not change unless the temperature changes. So if you add more `OH^- ` or what ever`K_(sp)`  is a constant under given temperature conditions.

`K_(sp) = [Mg^(2+)][OH^-]^2`

`6.8xx10^(-13) = [Mg^(2+)](8.5xx10^(-5))^2`

`[Mg^(2+)] = 9.41xx10^10^(-5)M`

As you have said `K_(sp) = [x][2x]^2`  can be applied when Pure `Mg(OH)_2`  solid partially dissolved in water and come to equilibrium. Here there is no such condition because we already have OH^- in the solution. So you cant substitute the `[Mg^(2+)]`  to the old `K_(sp)` and get the same answer for `K_(sp).`

`[Mg^(2+)] = 9.41xx10^10^(-5)`

When you consider 1L of solution

Number of `Mg^(2+) ` moles present `= 9.41xx10^10^(-5)`

Molar mass of Mg = 24.3g/mol

Weight of `Mg^(2+)`  ions in the solution

`= 24.3xx9.41xx10^10^(-5)`

`= 2.29xx10^(-3)g`

So we would have 0.0029g of metallic ion in the solution.

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