Question

Question #da59d

Answer 1

The answer is `1.8 * 10^(-7)` `"M"`.

The general reaction looks like this

`Mg_((aq))^(2+) + 2NaOH_((aq)) -> Mg(OH)_(2(s)) + 2Na^(+)(aq)`

The reaction that is of interest is actually this

`Mg(OH)2(s) rightleftharpoons Mg_((aq))^(2+) + 2OH_((aq))^(-)`

In order to determine the maximum concentration of `Mg^(2+)` ions permissible in the `NaOH` solution before a precipitate will be formed, you'd need the value of the solubility product constant, `K_(sp)`.

The expression for `K_(sp)` for this reaction is

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

A precipitate will form when the solution is saturated with ions; if the concentrations of the ions are low enough, a precipitate will not form. The threshold for the formation of a precipitate implies that the above equation be satisfied.

For a saturated solution, on the verge of forming a precipitate, the maximum value for the product of the concentrations of the two ions is equal to `K_(sp)`.

From this point on, any increase in the concentrations of the ions will result in the formation of the precipitate.

Since your reaction takes place in 1 L, 0.01 M `NaOH` solution, and knowing that `NaOH` dissociates completely into `Na^(+)` and `OH^(-)` ions, the concentration of hydroxide ions will be

`[OH^(-)] = [NaOH] = 0.01` `"M"`

This means that the maximum concentration of `Mg^(2+)` ions the solution can hold before a precipitate is formed is

`[Mg^(2+)] = K_(sp)/([OH^(-)]^2) = K_(sp)/(0.01)^2 = K_(sp)/(10^(-4)`

As a numerical value, I've found `K_(sp)` to be equal to `1.8*10^(-11)` (you can use any value given to you, of course), which means that

`[Mg^(2+)] = (1.8 * 10^(-11))/10^(-4) = 1.8 * 10^(-7)` `"M"`