Question #da59d

1 Answer
Jan 7, 2015

The answer is 1.8 * 10^(-7)1.8107 "M"M.

The general reaction looks like this

Mg_((aq))^(2+) + 2NaOH_((aq)) -> Mg(OH)_(2(s)) + 2Na^(+)(aq)Mg2+(aq)+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))^(-)Mg(OH)2(s)Mg2+(aq)+2OH(aq)

In order to determine the maximum concentration of Mg^(2+)Mg2+ ions permissible in the NaOHNaOH solution before a precipitate will be formed, you'd need the value of the solubility product constant, K_(sp)Ksp.

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

K_(sp) = [Mg^(2+)] * [OH^(-)]^2Ksp=[Mg2+][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)Ksp.

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 NaOHNaOH solution, and knowing that NaOHNaOH dissociates completely into Na^(+)Na+ and OH^(-)OH ions, the concentration of hydroxide ions will be

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

This means that the maximum concentration of Mg^(2+)Mg2+ 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)[Mg2+]=Ksp[OH]2=Ksp(0.01)2=Ksp104

As a numerical value, I've found K_(sp)Ksp to be equal to 1.8*10^(-11)1.81011 (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)[Mg2+]=1.81011104=1.8107 "M"M