Calculating Solubility - High School Chemistry

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Question

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?

Answer

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.

$K_{sp}=[Ba^{2+}][F^-]^2=2.410^{-5}$

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

$2.410^{-5}=[x][2x]^2$

$2.410^{-5}=4x^3$

$x = 1.810^{-2}M$

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

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Question

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

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

Answer

Definition of $K_{sp}$ :

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

For :

$K_{sp}=1.210^{-5}=[Pb^{+2}][Cl^{-}]^2$

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

$1.210^{-5}=[X][2X]^2$

$1.210^{-5}=4X^3$

Solve for the unknown variable:

$({310^{-6}})^{\frac{1}{3}}=X$

$1.4410^{-2}M=X$

Multiply times the given volume:

$\frac{1.4410^{-2}moles}{Liter}.250L=.0036moles$

Multiply times the molar mass of :

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

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