GRE Subject Test: Chemistry - Solubility and Precipitation

The $K_{sp}$ of is $1.8*10^{-14}$ . What is the molar solubility of in water?

$1.3*10^{-7} \textup{ M}$

$4.3*10^{-5}\textup{ M}$

$3.1*10^{-2} \textup{ M}$

$0.011\textup{ M}$

$3.22\textup{ M}$

Explanation

The equation for the dissolution of in water is:

$CdCO_{3(s)}\rightleftharpoons Cd^{2+}_{(aq)}+CO_{3}^{2-}_{(aq)}$

The $K_{sp}$ for the above equation is:

$K_{sp}=[Cd^{2+}][CO_{3}^{2-}]=1.8 *10^{-14}$

Due to the molar ratios of the species of , the concentration of $Cd^{2+}$ and $CO_3^{2-}$ should be equal when dissolved:

$[Cd^{2+}]=[CO_{3}^{2-}]=X$

Plugging into the $K_{sp}$ equation gives:

$K_{sp}=X* X=X^{2}=1.8*10^{-14}$

$X=\sqrt{1.8* 10^{-14}}=1.3*10^{-7}$

Therefore, the solubility of in water is $1.3*10^{-7}\textup{ M}$

For the titration of $0.001 M\ Ag^{2+}$ with solution of $Cl^-_{(aq)}$ , the end point occurs upon addition of of the $Ag^{2+}$ solution. Determine the initial concentration of chloride that was present in the original solution.

$0.0013\ M$

$1.321\ M$

$0.0026\ M$

$0.063\ M$

Explanation

The equivalence point is when enough titrant has been added so that the number of moles of titrant equals the number of moles of analyte. We must first determine the number of moles of silver ions at the equivalence point.

At the equivalence point the following occurs in solution:

$moles\ of\ Ag^{2+}=moles\ of\ Cl^{-}$

The number of moles of silver ions at equivalence point is calculated as follows:

$0.200\ L\times \frac{0.001\ moles}{L}=2.0\times 10^{-4}\ moles\ Ag^{2+}$

$Molarity=\frac{moles\ of\ solute}{Liters\ of\ solution}$

$Liters\ of\ solution= Original\ volume\ of\ Cl^{-}\ solution$

Plugging these values into the equation gives:

$\frac{2\times10^{-4}\ moles}{0.150\ L}=0.0013\ M\ Cl^{-}$

Write the solubility product, $K_{sp}$ , for $BaSO_4_{(s)}$ .

$K_{sp} = [Ba^{2+}][SO_{4^}^{}^{2-}]$

$K_{sp} = [Ba][SO_{4}]^{2-}$

$K_{sp} = [Ba][SO_{4}]$

$K_{sp} = [Ba]^{2+}[SO_{4}]^{2-}$

$K_{sp} = [Ba]^{4+}[SO_{4}]^{2-}$

Explanation

The solubility product constant ( $K_{sp}$ ) is an expression that describes the extent to which a compound is soluble in an aqueous solution. It describes the equilibrium between a solid and its constituent ions in a solution. The equilibrium constant for $BaSO_4_{(s)}$ can be written as:

$K_{eq} = \frac{[Ba^{2+}][SO_{4^}^{}^{2-}]}{[BaSO_{4}(s)]}$

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

$K_{eq}*[BaSO_{4}]=[Ba^{2+}][SO_{4^}^{}^{2-}]$

The product, $K_{eq}*[BaSO_{4}]$ , can be considered a constant expression called the solubility product constant ( $K_{sp}$ ) and the equation can be written in the form:

$K_{sp} = [Ba^{2+}][SO_{4^}^{}^{2-}]$

Write the $K_{sp}$ expression for the following equilibria:

$Mg(OH)_{2(s)} \leftrightarrow Mg^{2+(aq)} + 2 OH^{-}_{(aq)}$

$Ksp = [Mg^{2+}][OH^{-}]^{2}$

$K_{sp} = [Mg]^{+}[OH]^{-}$

$K_{sp} = [Mg^{+}][OH^{-}]$

$K_{sp} = [Mg^{+}]2[OH^{-}]$

None of these

Explanation

The solubility product constant ( $K_{sp}$ ) is an expression that describes the extent to which a compound is soluble in an aqueous solution. It describes the equilibrium between a solid and its constituent ions in a solution.The equilibrium between undissolved $Mg(OH)_{2(s)}$ and its ions dissolved in solution is:

$Mg(OH)_{2(s)} \leftrightarrow Mg^{2+(aq)} + 2 OH^{-}_{(aq)}$

The equilibrium constant for $Mg(OH)_{2(s)}$ can be written as:

$K_{eq} = \frac{[Mg^{2+}][OH^{-}]^{2}}{[Mg(OH)_{2}(s)]}$

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

$K_{eq}[Mg(OH)_{2}](s)=[Mg^{2+}][OH^{-}]^{2}$

The product, $K_{eq}[Mg(OH)_{2}]$ , can be considered a constant expression called the solubility product constant ( $K_{sp}$ ) and the equation can be written in the form:

$K_{sp} = [Mg^{2+}][OH^{-}]^{2}$

The Ksp of is $5.0*10^{-9}$ . What is the molar solubility of in water?

$7.1*10^{-5}$

$2.3*10^{-3}$

$3*10^{-3}$

$5.3*10^{-8}$

$8.5*10^{-6}$

Explanation

The equation for the dissolution of in water is below:

$BaSO_{4}(s)\rightleftharpoons Ba^{2+}(aq)\ +\ SO_{4}^{2-}(aq)$

The Ksp for the above equation is:

$Ksp=[Ba^{2+}][SO_{4}^{2-}]=5.0*10^{-9}$

Due to the solubility of of in water, the concentration of $Ba^{2+}$ and $SO_4^{2-}$ - should be equal:

$[Ba^{2+}]\ =\ [SO_{4}^{2-}]=X$

Plug into the Ksp equation:

$Ksp=X*X=X^{2}=5.0*10^{-9}$

Therefore:

$X=\sqrt{5.0\ *10^{-9}}=7.1*10^{-5}$

The solubility of in water is $7.1*10^{-5}$

$Pb(OH)_{2}$ has a $K_{sp}$ equal to $1.2\times10^{-15}$ . What would be the numerical expression used to determine the molar solubility (S) of $Pb(OH)_{2}$ ?

$\sqrt[3]{6.0\times10^{-16}}$

$\sqrt[3]{4.0\times10^{-16}}$

$\sqrt[2]{4.0\times10^{-16}}$

$\sqrt[2]{1.2\times10^{-15}}$

Explanation

The equation for the dissolution of $Pb(OH)_{2}$ in water is below:

$Pb(OH)_2_{(s)}\rightleftharpoons\ Al^{2+}_{(aq)}\ +\ 2OH^{-}_{(aq)}$

The Ksp expression for the above chemical equation is:

$K_{sp}=S\times\2S^{2}=2S^{3}$

Rearranging this equation to solve for solubility (S) gives:

$\frac{K_{sp}}{2}=S^{3}$

$\sqrt[3]{\frac{K_{sp}}{2}}=S$

Plugging the value for Ksp into the equation gives:

$S=\sqrt[3]{\frac{1.2\times10^{-15}}{2}}=\sqrt[3]{6.0\times10^{-16}}$

Calculate the molar solubility of with $K_{sp} = 3.6\times 10^{-8}$ in a solution containing $0.01\textup{ M}$ .

$3.6\times10^{-6}\textup{ M}$

$5.1\times10^{-8}\textup{ M}$

$2.1\times10^{-3}\textup{ M}$

$1.0\times10^{-2}\textup{ M}$

$5.8\times10^{-4}\textup{ M}$

Explanation

The equation for the dissolution of in water is below: $PbF_{2(s)}\rightleftharpoons Pb^{2+}(aq)+2F^{-}(aq)$

The $K_{sp}$ for the above equation is:

$Ksp=[Pb^{2+}][F^{-}]^{2}=3.6\times10^{-8}$

Due to the solubility of in water, the concentration of $Pb^{2+}$ can be set to be:

$[Pb^{2+}]=X$

Therefore concentration will be double:

$F^{-}=2X$

Using an ICE table to process the data:

Screen shot 2015 11 09 at 9.29.39 pm

Plugging these variable into the $K_{sp}$ equation gives:

$Ksp=[X]\times [0.010+2X^{2}]=3.6\times10^{-8}$

$2X^{2}<< 0.010$

Therefore, the $K_{sp}$ equation can be approximated to:

$Ksp=[X]\times [0.010]=3.6\times10^{-8}$

$[X]=\frac{3.6\times10^{-8}}{[0.010]}=3.6\times10^{-6}M$

Therefore, the solubility of in water is $3.6\times10^{-6}M$ .

Calculate the molar solubility of $Mn(OH)_{2}$ with a $K_{sp} = 1.6\times 10^{-13}$ if enough is added to raise the pH of the solution to pH 10.

$1.6\times10^{-9}\textup{ M}$

$7.9\times10^{-3}\textup{ M}$

$1.2\times10^{-7}\textup{ M}$

$0.1992\textup{ M}$

$3.2\textup{ M}$

Explanation

The equation for the dissolution of $Mn(OH)_{2}$ in water is below: $Mn(OH)_{2(s)}\rightleftharpoons Mn^{2+}(aq)+2OH^{-}(aq)$

The $K_{sp}$ for the above equation is:

$Ksp=[Mn^{2+}][OH^{-}]^{2}=1.3\times10^{-15}$

Due to the solubility of $Mn(OH)_{2}$ in water, the concentration of $Mn^{2+}$ can be set to:

$[Mn^{2+}]=X$

Therefore the concentration of $OH^{-}$ should be double that of $Mn^{2+}$ :

$[OH^{-}]=2X$

The solution was adjusted to pH 10 using , therefore:

$[OH^{-}]=10^{-pOH}=10^{-4}=1.0\times10^{-4}$

Using an ICE table to process the data:

Screen shot 2015 11 13 at 4.35.54 pm

Plugging into the $K_{sp}$ equation gives:

$Ksp=[X]\times [0.0001+2X^{2}]=1.6\times10^{-13}$

$2X^{2}<< 0.0001$

Therefore, the $K_{sp}$ equation can be approximated to:

$Ksp=[X]\times [0.0001]=1.6\times10^{-13}$

$[X]=\frac{1.6\times10^{-13}}{[0.0001]}=1.6\times10^{-9}M$

Therefore, the solubility of $Mn(OH)_{2}$ in water is $1.6\times10^{-9}M$ .

$Al(OH)_{3}$ has a $K_{sp}$ equal to $1.3\times10^{-33}$ . What would be the numerical expression used to determine the molar solubility (S) of $Al(OH)_{3}$ ?

$\sqrt[4]{4.3\times10^{-34}}$

$\sqrt[4]{6.5\times10^{-34}}$

$\sqrt[3]{2.2\times10^{-33}}$

$\sqrt[3]{4.3\times10^{-33}}$

Explanation

The equation for the dissolution of $Al(OH)_{3}$ in water is below:

$Al(OH)_3_{(s)}\rightleftharpoons\ Al^{3+}_{(aq)}\ +\ 3OH^{-}_{(aq)}$

The Ksp expression for the above chemical equation is:

$K_{sp}=S\times\3S^{3}=3S^{4}$

Rearranging this equation to solve for solubility (S) gives:

$\frac{K_{sp}}{3}=S^{4}$

$\sqrt[4]{\frac{K_{sp}}{3}}=S$

Plugging the value for Ksp into the equation gives:

$S=\sqrt[4]{\frac{1.3\times10^{-33}}{3}}=\sqrt[4]{4.3\times10^{-34}}$

Which salt is insoluble in water?

$Al(OH)_{3}$

$CuCl_{2}$

$CuSO_{4}$

$NaNO_{3}$

Explanation

Solubility rules must be followed for substances in aqueous media. Below are some of the solubility rules and they must be followed in the order given. For example, rule 1 should have precedence over rule 2.

1. All alkali metal $(Li, Na, K, Rb, \text{and}\ Cs)$ and compounds are soluble.