What is the #"pH"# of a solution of magnesium hydroxide given that #K_(sp) = 1.8 * 10^(-11)# ?
1 Answer
Explanation:
In order to find the pH of a solution, you must determine the concentration of hydronium ions,
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,
Magnesium hydroxide,
#"Mg"("OH")_text(2(s]) rightleftharpoons "Mg"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)#
The solubility product constant,
#K_(sp) = ["Mg"^(2+)] * ["OH"^(-)]^color(red)(2)#
What you need to do here is use the
To do that, set up an ICE table
#" ""Mg"("OH")_text(2(s]) " "rightleftharpoons" " "Mg"_text((aq])^(2+) " "+" " color(red)(2)"OH"_text((aq])^(-)#
Remember, the concentration of the solid is assumed to be constant.
Here
#K_(sp) = s * (color(red)(2)s)^color(red)(2) = 4s^3#
Rearrange to solve for
#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
#["OH"^(-)] = color(red)(2) * 1.65 * 10^(-4)"M" = 3.30 * 10^(-4)"M"#
Now, at
#color(blue)(K_W = ["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-4))" "# , where
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"^(-)]))#