How do you calculate Ksp from molar solubility?
1 Answer
Here's how you can do that.
Explanation:
Let's take a generic dissociation equilibrium to work with here
"X"_ n "Y"_m rightleftharpoons color(blue)(n) * "X"^(m+) + color(purple)(m) * "Y"^(n-)XnYm⇌n⋅Xm++m⋅Yn−
Now, the molar solubility of this generic salt
Let's assume that you are given a molar solubility equal to
Now, notice that every mole of
This means that the saturated solution will contain
["X"^(m+)] = color(blue)(n) * s[Xm+]=n⋅s
["Y"^(n-)] = color(purple)(m) * s[Yn−]=m⋅s
The solubility product constant,
K_(sp) = ["X"^(m+)]^color(blue)(n) * ["Y"^(n-)]^color(purple)(m)Ksp=[Xm+]n⋅[Yn−]m
Plug in the expressions you have for the concentrations of the two ions in terms of
K_(sp) = (color(blue)(n) * s)^color(blue)(n) * (color(purple)(m) * s)^color(purple)(m)Ksp=(n⋅s)n⋅(m⋅s)m
K_(sp) = color(blue)(n^n) * s^color(blue)(n) * color(purple)(m^m) * s^color(purple)(m)Ksp=nn⋅sn⋅mm⋅sm
This is equivalent to
color(green)(|bar(ul(color(white)(a/a)color(black)(K_(sp) = color(blue)(n^n) * color(purple)(m^m) * s^((color(blue)(n)+color(purple)(m)))color(white)(a/a)|))))
Let's take a numerical example to test out this expression.
Magnesium hydroxide,
"Mg"("OH")_2 , has a molar solubility of1.44 * 10^(-4)"M" in pure water at room temperature. What is theK_(sp) of the salt?
The first thing to do is identify the values of
"Mg"("OH")_ (2(s)) rightleftharpoons "Mg"_ ((aq))^(2+) + 2"OH"_ ((aq))^(-)
As you can see, you have
{(n=1), (m=2) :}
This means that the
K_(sp) = 1^1 * 2^2 * (1.44 * 10^(-4)"M")^((1 + 2))
K_(sp) = 1.2 * 10^(-11)"M"^3
The solubility product constant is usually given without added units, so you'd have
K_(sp) = 1.2 * 10^(-11)
