What is the #"pH"# of a solution of magnesium hydroxide given that #K_(sp) = 1.8 * 10^(-11)# ?

1 Answer
Feb 28, 2016

Answer:

#"pH" = 10.52#

Explanation:

In order to find the pH of a solution, you must determine the concentration of hydronium ions, #"H"_3"O"^(+)#, either directly or indirectly.

When you're dealing with a Bronsted - Lowry acid, you will be solving for the concentration of hydronium ions directly.

When you're dealing with a Bronsted - Lowry base, like you are here, you will be solving for the concentration of hydronium ions indirectly, i.e. by solving for the concentration of hydroxide ions, #"OH"^(-)#.

Magnesium hydroxide, #"Mg"("OH")_2#, us insoluble in aqueous solution. This means that when you place magnesium hydroxide in water, an equilibrium will be established between the undissolved solid and the dissolved ions.

#"Mg"("OH")_text(2(s]) rightleftharpoons "Mg"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)#

The solubility product constant, #K_(sp)#, for this equilibrium looks like this

#K_(sp) = ["Mg"^(2+)] * ["OH"^(-)]^color(red)(2)#

What you need to do here is use the #K_(sp)# to find the molar solubility, #s#, of magnesium hydroxide in aqueous solution at #25^@"C"#.

To do that, set up an ICE table

#" ""Mg"("OH")_text(2(s]) " "rightleftharpoons" " "Mg"_text((aq])^(2+) " "+" " color(red)(2)"OH"_text((aq])^(-)#

#color(purple)("I")" " " " " "-" " " " " " " " " " " " " "0" " " " " " " " " " 0#
#color(purple)("C")" " " " " "-" " " " " " " " " " " "(+s)" " " " " "(+color(red)(2)s)#
#color(purple)("E")" " " " " "-" " " " " " " " " " " " " "s" " " " " " " " " "color(red)(2)s#

Remember, the concentration of the solid is assumed to be constant.

Here #K_(sp)# will be equal to

#K_(sp) = s * (color(red)(2)s)^color(red)(2) = 4s^3#

Rearrange to solve for #s#

#s = root(3)(K_(sp)/4) = root(3)((1.8 * 10^(-11))/4) = 1.65 * 10^(-4)#

This means that the concentration of hydroxide ions in a saturated solution of magnesium hydroxide at #25^@"C"# will be

#["OH"^(-)] = color(red)(2) * 1.65 * 10^(-4)"M" = 3.30 * 10^(-4)"M"#

Now, at #25^@"C"#, the concentration of hydronium ions and the concentration of hydroxide ions have the following relationship

#color(blue)(K_W = ["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-4))" "#, where

#K_W# - the ion product for water's auto-ionization

Plug in your value to get

#["H"_3"O"^(+)] = 10^(-14)/(3.30 * 10^(-4)) = 3.03 * 10^(-11)"M"#

This means that the pH of the solution will be

#color(blue)("pH" = - log(["H"_3"O"^(+)]))#

#"pH" = - log(3.03 * 10^(-11)) = color(green)(10.52)#

Alternatively, you can use the equation

#color(blue)("pH " + " pOH" = 14)#

Here

#color(blue)("pOH" = - log(["OH"^(-)]))#