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High School Chemistry : Calculating Solubility

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Example Question #192 : High School Chemistry

Barium fluoride is an insoluble salt.

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

Possible Answers:

$4.9*10^{-3}M$

$1.8*10^{-2}M$

$3.5*10^{-3}M$

$3.6*10^{-2}M$

Correct answer:

$1.8*10^{-2}M$

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 , and fluoride ions increase by .

$BaF_2\rightarrow Ba^{2+}+2F^-$

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

$K_{sp}=[Ba^{2+}][F^-]^2=2.4*10^{-5}$

$[Ba^{2+}]=x\ \text{and}\ [F^-]=2x$

$2.4*10^{-5}=[x][2x]^2$

$2.4*10^{-5}=4x^3$

$x = 1.8*10^{-2}M$

The solubility for barium fluoride is $1.8*10^{-2}M$ .

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Example Question #1 : Calculating Solubility

For PbCl_2 , $K_{sp}=1.2*10^{-5}$

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

Possible Answers:

8.44 g

1.00 g

2.21 g

6.11 g

3.88g

Correct answer:

1.00 g

Explanation:

Definition of $K_{sp}$ :

$K_{sp}=[X]^a[Y]^b$ for X_aY_b .

For PbCl_2 :

$K_{sp}=1.2*10^{-5}=[Pb^{+2}][Cl^{-}]^2$

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

$1.2*10^{-5}=[X][2X]^2$

$1.2*10^{-5}=4X^3$

Solve for the unknown variable:

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

$1.44*10^{-2}M=X$

Multiply times the given volume:

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

Multiply times the molar mass of PbCl_2 :

$.0036moles*\frac{278.1grams}{moles}=1.00 g$

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High School Chemistry : Calculating Solubility

Study concepts, example questions & explanations for High School Chemistry

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Example Question #192 : High School Chemistry

Example Question #1 : Calculating Solubility

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