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College Chemistry : Solubility Product Constant

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Example Question #13 : Equilibrium

$CaF_{2}(s)\rightleftharpoons\ Ca^{2+}(aq)\ +\ 2F^{-}(aq)$

Express the solubility product constant expression for the given reaction.

Possible Answers:

$K_{sp}=[Ca^{2+}][F^{-}]$

$K_{sp}=\frac{[Ca^{2+}][F^{-}]^{2}}{[CaF_{2}]}$

$K_{sp}=[Ca^{2+}][F^{-}]^{2}$

$K_{sp}=[Ca^{2+}][F^{-}]$

Correct answer:

$K_{sp}=[Ca^{2+}][F^{-}]^{2}$

Explanation:

The equilibrium given tells us how the solid dissolves in solution:

$CaF_{2}(s)\rightleftharpoons\ Ca^{2+}(aq)\ +\ 2F^{-}(aq)$

However, the solubility product constant (Ksp) given tells us the degree by which a solid dissolves in solution. The larger the Ksp, the more soluble a substance is in water. Writing this expression follows the same rules as other equilibrium constant expressions. Therefore solids and water (when it is the solvent) are omitted from this expression. You must raise the concentration of the substances involved to the power of its coefficient.

For the chemical reaction given, the Ksp is:

$K_{sp}=[Ca^{2+}][F^{-}]^{2}$

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Example Question #1 : Solubility Product Constant

$PbCl_{2}(s)\rightleftharpoons\ Pb^{2+}(aq)\ +\ 2Cl^{-}(aq)$

Express the solubility product constant expression for the given reaction.

Possible Answers:

$K_{sp}=\frac{[Pb^{2+}][Cl^{-}]^{2}}{[PbCl_{2}]}$

$K_{sp}=[Pb^{2+}][Cl^{-}]^{2}$

$K_{sp}=[Pb^{2+}][Cl^{-}]$

Correct answer:

$K_{sp}=[Pb^{2+}][Cl^{-}]^{2}$

Explanation:

The equilibrium given tells us how the solid dissolves in solution:

$PbCl_{2}(s)\rightleftharpoons\ Pb^{2+}(aq)\ +\ 2Cl^{-}(aq)$

For the chemical reaction given, the Ksp is:

$K_{sp}=[Pb^{2+}][Cl^{-}]^{2}$

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Example Question #1 : Solubility Product Constant

How many grams of $\textrm{BaF}_{\textrm{2}}$ are dissolved in 75.0 mL of a saturated solution of $\textrm{BaF}_{\textrm{2}}$ ?

$\textrm{BaF}_{\textrm{2}}(s)\rightleftharpoons\textrm{Ba}^{\textrm{2+}}(aq)\textrm{ + 2 F}^{\textrm{ -}}(aq)$

$\textrm{K}_{\textrm{sp}}=1.00\textrm{ x 10}^{-6}$

Possible Answers:

$1.10\textrm{ g BaF}_{2}$

$0.0132\textrm{ g BaF}_{2}$

$0.166\textrm{ g BaF}_{2}$

$0.0829\textrm{ g BaF}_{2}$

$0.132\textrm{ g BaF}_{2}$

Correct answer:

$0.0829\textrm{ g BaF}_{2}$

Explanation:

To calculate how many grams of $\textrm{BaF}_{\textrm{2}}$ are dissolved in the solution, we first solve for the molarity of the solution.

Using the dissociation equation

$\textrm{BaF}_{\textrm{2}}(s)\rightleftharpoons\textrm{Ba}^{\textrm{2+}}(aq)\textrm{ + 2 F}^{\textrm{ -}}(aq)$

We can write out the equation for the solubility product constant as

$\textrm{K}_{\textrm{sp}}=\textrm{[Ba}^{2+}][\textrm{F}^{\textrm{ -}}]^{2}$

$1.00\textrm{ x 10}^{-6}=\textrm{[Ba}^{2+}][\textrm{F}^{\textrm{ -}}]^{2}$

Because there are 2 fluoride ions for every bariumion, we can rewrite the equation as

$[\textrm{Ba}^{2+}]=x$

$1.00\textrm{ x 10}^{-6}=(x)(2x)^{2}$

Now solve for x

$1.00\textrm{ x 10}^{-6}=(x)(4x^{2})$

$1.00\textrm{ x 10}^{-6}=4x^{3}$

$x^{3}=\frac{1.00\textrm{ x 10}^{-6}}{4}=2.50\textrm{ x 10}^{-7}$

$x=\sqrt[3]{2.50\textrm{ x 10}^{-7}}=6.30\textrm{ x 10}^{-3}$

$[\textrm{Ba}^{2+}]=6.30\textrm{ x 10}^{-3}M$

Solve for the concentration of dissolved $\textrm{BaF}_{\textrm{2}}$

$[\textrm{BaF}_{2}]=[\textrm{Ba}^{2+}]=6.30\textrm{ x 10}^{-3}M$

Now that we have the concentration of dissolved $\textrm{BaF}_{\textrm{2}}$ , we can calculate how many grams are dissolved in the solution

$(6.30\textrm{ x 10}^{-3}M)(0.0750\textrm{ L})=4.73\textrm{ x 10}^{-4}\textrm{ mol}$

$(4.73\textrm{ x 10}^{-4}\textrm{ mol})(175.34\frac{\textrm{g}}{\textrm{mol}})=0.0829\textrm{ g BaF}_{2}$

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College Chemistry : Solubility Product Constant

Study concepts, example questions & explanations for College Chemistry

All College Chemistry Resources

Example Questions

Example Question #13 : Equilibrium

Example Question #1 : Solubility Product Constant

Example Question #1 : Solubility Product Constant

All College Chemistry Resources

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