How do you calculate Ksp from molar solubility?

Answer 1

Here's how you can do that.

Here, let's work with a generic dissociation equilibrium.

#"X"_ n "Y"_m rightleftharpoons color(blue)(n) * "X"^(m+) + color(purple)(m) * "Y"^(n-)#
Now, the molar solubility of this generic salt #"X"_n"Y"_m# tells you the number of moles of salt that can be dissolved in one liter of solution to form a saturated solution.
Let's assume that you are given a molar solubility equal to #s# #"mol L"^(-1)# for this salt in water at room temperature. This tells you that you can dissolve #s# moles of #"X"_n"Y"_m# per liter of solution at this temperature.
Now, notice that every mole of #"X"_n"Y"_m# that dissolves produces #color(blue)(n)# moles of #"X"^(m+)# cations and #color(purple)(m)# moles of #"Y"^(n-)# anions.

Thus, the saturated solution will have the following contents:

#["X"^(m+)] = color(blue)(n) * s#
#["Y"^(n-)] = color(purple)(m) * s#
The solubility product constant, #K_(sp)#, for this dissociation equilibrium looks like this
#K_(sp) = ["X"^(m+)]^color(blue)(n) * ["Y"^(n-)]^color(purple)(m)#
Plug in the expressions you have for the concentrations of the two ions in terms of #s# to find
#K_(sp) = (color(blue)(n) * s)^color(blue)(n) * (color(purple)(m) * s)^color(purple)(m)#
#K_(sp) = color(blue)(n^n) * s^color(blue)(n) * color(purple)(m^m) * s^color(purple)(m)#

This is the same as

#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)|))))#

To test this expression numerically, let's look at an example.

Magnesium hydroxide, #"Mg"("OH")_2#, has a molar solubility of #1.44 * 10^(-4)"M"# in pure water at room temperature. What is the #K_(sp)# of the salt?
The first thing to do is identify the values of #n# and #m# by writing the dissociation equilibrium for magnesium hydroxide
#"Mg"("OH")_ (2(s)) rightleftharpoons "Mg"_ ((aq))^(2+) + 2"OH"_ ((aq))^(-)#

You can observe that you have

#{(n=1), (m=2) :}#
This means that the #K_(ps)# of magnesium hydroxide is
#K_(sp) = 1^1 * 2^2 * (1.44 * 10^(-4)"M")^((1 + 2))#
#K_(sp) = 1.2 * 10^(-11)"M"^3#

Typically, the solubility product constant is provided without additional units, so you would have to

#K_(sp) = 1.2 * 10^(-11)#
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Answer 2

To calculate the solubility product constant (Ksp) from molar solubility, you use the following steps:

  1. Write the balanced chemical equation for the dissolution of the compound in water.
  2. Use stoichiometry to determine the concentration of each ion in solution.
  3. Write the equilibrium expression for the dissolution reaction.
  4. Substitute the equilibrium concentrations (in terms of molar solubility) into the equilibrium expression.
  5. Solve for Ksp.

The general equation for the dissolution of a sparingly soluble salt ( A_{\text{m}}B_{\text{n}} ) can be written as:

[ A_{\text{m}}B_{\text{n}}(s) \rightleftharpoons m A^{m+}(aq) + n B^{n-}(aq) ]

And the equilibrium expression for this dissolution reaction is:

[ K_{\text{sp}} = [A^{m+}]^m[B^{n-}]^n ]

Where ( K_{\text{sp}} ) is the solubility product constant, and ( [A^{m+}] ) and ( [B^{n-}] ) are the equilibrium concentrations of the ions in solution.

By substituting the concentrations in terms of molar solubility (( s )) into the equilibrium expression, you can solve for ( K_{\text{sp}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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