High School Chemistry : Calculating Solubility

Study concepts, example questions & explanations for High School Chemistry

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Example Questions

Example Question #1 : Calculating Solubility

Barium fluoride is an insoluble salt.

If the solubility product constant of barium fluoride is \(\displaystyle 2.4*10^{-5}M\), what is the solubility of barium fluoride?

Possible Answers:

\(\displaystyle 1.8*10^{-2}M\)

\(\displaystyle 4.9*10^{-3}M\)

\(\displaystyle 3.6*10^{-2}M\)

\(\displaystyle 3.5*10^{-3}M\)

Correct answer:

\(\displaystyle 1.8*10^{-2}M\)

Explanation:

If given the solubility product constant for a salt, we can determine the solubility for the salt as well. In order to do this, we need to use an ICE table and the equilibrium constant expression.

Initial: When the salt is added to water, there are no ions initially in solution. Because the salt is a solid, its concentration is irrelevant.

Change: When one molecule of barium fluoride dissolves, one barium ion and two fluoride ions are introduced to the solution. As a result, barium ions increase by \(\displaystyle x\), and fluoride ions increase by \(\displaystyle 2x\).

\(\displaystyle BaF_2\rightarrow Ba^{2+}+2F^-\)

End: The equilibrium expression for this salt is \(\displaystyle [Ba^{2+}][F^{-}]^{2}\), which will allow us to calculate the solubility based on the change in ion concentration.

\(\displaystyle K_{sp}=[Ba^{2+}][F^-]^2=2.4*10^{-5}\)

\(\displaystyle [Ba^{2+}]=x\ \text{and}\ [F^-]=2x\)

\(\displaystyle 2.4*10^{-5}=[x][2x]^2\)

\(\displaystyle 2.4*10^{-5}=4x^3\)

\(\displaystyle x = 1.8*10^{-2}M\)

The solubility for barium fluoride is \(\displaystyle 1.8*10^{-2}M\).

Example Question #1 : Calculating Solubility

For \(\displaystyle PbCl_2\)\(\displaystyle K_{sp}=1.2*10^{-5}\)

Determine the maximum amount of grams of \(\displaystyle PbCl_2\) that will dissolve in \(\displaystyle .250L\) of water at \(\displaystyle 25^\circ C\).

Possible Answers:

\(\displaystyle 8.44 g\)

\(\displaystyle 1.00 g\)

\(\displaystyle 3.88g\)

\(\displaystyle 6.11 g\)

\(\displaystyle 2.21 g\)

Correct answer:

\(\displaystyle 1.00 g\)

Explanation:

Definition of \(\displaystyle K_{sp}\):

\(\displaystyle K_{sp}=[X]^a[Y]^b\) for \(\displaystyle X_aY_b\).

For \(\displaystyle PbCl_2\):

\(\displaystyle K_{sp}=1.2*10^{-5}=[Pb^{+2}][Cl^{-}]^2\)

Due to the chemical formula, there will be twice as many chloride ions as lead ions.

\(\displaystyle 1.2*10^{-5}=[X][2X]^2\)

\(\displaystyle 1.2*10^{-5}=4X^3\)

Solve for the unknown variable:

\(\displaystyle ({3*10^{-6}})^{\frac{1}{3}}=X\)

\(\displaystyle 1.44*10^{-2}M=X\)

Multiply times the given volume:

\(\displaystyle \frac{1.44*10^{-2}moles}{Liter}*.250L=.0036moles\)

Multiply times the molar mass of \(\displaystyle PbCl_2\):

\(\displaystyle .0036moles*\frac{278.1grams}{moles}=1.00 g\)

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