Question #da59d

1 Answer
Jan 7, 2015

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"#