Solubility and Precipitation - GRE Subject Test: Chemistry

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Question

Consider the following balanced equation for the solubility of barium hydroxide in an aqueous solution.

$\small Ba(OH)_{2}_{(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2OH^{-}_{(aq)}$

$\small K_{sp} = 5*10^{-3}$

What is the solubility of barium hydroxide?

Answer

In order to solve for the solubility of barium hydroxide, we need to establish an ICE table for the reaction.

I. Initially, there are no barium ions or hydroxide ions in the solution, so we can call their concentrations zero at the beginning of the reaction.

C. For every molecule of dissolved barium hydroxide, there will be one ion of barium and two ions of hydroxide. As a result, the increases in each ion can be designated $\small +x$ and $\small +2x$ , respectively.

E. Finally, we plug these values into the equilibrium expression and set it equal to the solubility product constant.

$\begin{matrix} Ba(OH)_2 & Ba^{2+} &OH^- \ /& 0 & 0\ / & x & 2x \end{matrix}$

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

$\small 5*10^{-3} = [x][2x]^{2}$

$\small x = 0.108M$

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Question

How many moles of fluoride ions are present in of a completely saturated solution of lead (II) fluoride?

$PbF_2(s)\rightarrow Pb^{2+}(aq)+2F^-(aq)$

$K_{sp}=3.7*10^{-8}$

Answer

Recall that the solubility product constant is given by the equation below, for a reaction in the following format.

$aA\rightarrow cC+dD$

$K_{sp}=[C]^c[D]^d$

Using our balanced reaction, we can find the solubility product equation for the dissociation of lead (II) fluoride.

$PbF_2(s)\rightarrow Pb^{2+}(aq)+2F^-(aq)$

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

For each molecule of lead (II) fluoride that dissolves, it produces one lead ion and two fluoride ions. We can conclude that the concentration of fluoride ions in solution will be twice the concentration of lead ions.

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

Use these variables in the solubility product equation, along with the given value from the question.

$3.710^{-8}=(x) (2x)^2$

Now we can solve for the value of $\small x$ .

$3.710^{-8}=4x^3$

$x = \sqrt[3]{\frac{3.710^{-8}}{4}}=2.110^{-3}$

Remember that this value is equal to the concentration of lead ions in solution and half the concentration of fluoride ions in solution.

$[Pb^{2+}]=x=2.110^{-3}M$

$[F^-]=2x=2(2.110^{-3}M)=4.2*10^{-3}M$

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Question

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

Answer

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:

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

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Question

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

Answer

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 $AgCN_{(s)}$ can be written as:

$K_{eq} = \frac{[Ag^{+}][CN^{-}]}{[AgCN(s)]}$

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

$K_{eq}[AgCN(s)]=[Ag^{+}][CN^{-}]$

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

$K_{sp} = [Ag^{+}][CN^{-}]$

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Question

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

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

Answer

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:

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

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Question

Considering that $K_{sp} = 1.8 * 10^{-11}$ for the equilibrium reaction:

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

What would be the $Mg^{2+}$ concentration for a solution buffered at a pH of 8.5?

Answer

The pH of the solution is , therefore the pOH would be considering the relationship:

Therefore $[OH]=3.2x10^{-6}$ based on the equation $[OH] = 10^{-pOH}$

The solubility product expression for $Mg(OH)_{2(s)}$ is : $K_{sp} = [Mg^{2+}][OH^{-}]^{2}$

By inserting the knowns in to the expression, gives:

$1.8 * 10^{-11} = [Mg^{2+}][3.2 * 10^{-6}]^{2}$

Rearrange this equation

$[Mg^{2+}] = \frac{1.8 * 10^{-11}}{(3.2 * 10^{-6})^{2}} = 1.8M$

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Question

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

Answer

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:

$Ksp=[Ba^{2+}][SO_{4}^{2-}]=5.010^{-9}$

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:

$Ksp=XX=X^{2}=5.010^{-9}$

Therefore:

$X=\sqrt{5.0\ 10^{-9}}=7.110^{-5}$

The solubility of in water is $7.110^{-5}$

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Question

The Ksp of is $5.3*10^{-9}$ . What is the molar solubility of in water?

Answer

The equation for the dissolution of in water is below:

$CuBr(s)\rightleftharpoons Cu^{+}(aq)\ +\ Br^{-}(aq)$

The Ksp for the above equation is:

$Ksp=[Cu^{+}][Br^{-}]=5.310^{-9}$

Due to the solubility of of in water, the concentration of and should be equal:

$[Cu^{+}]\ =\ [Br^{-}]=X$

Plugging X into the Ksp equation gives:

$Ksp=XX=X^{2}=5.310^{-9}$

Therefore:

$X=\sqrt{5.3\ 10^{-9}}=7.310^{-5}$

The solubility of CuBr in water is $7.310^{-5}$

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Question

Which salt is insoluble in water?

Answer

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.

2. Nitrate , perchlorate , chlorate, , and acetate salts are soluble.

3. Silver, lead, and mercury salts are insoluble.

Thus, we see that silver chloride is insoluble due to rule 3.

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Question

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

Answer

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:

$K_{sp}=X X=X^{2}=1.810^{-14}$

$X=\sqrt{1.8 10^{-14}}=1.310^{-7}$

Therefore, the solubility of in water is $1.310^{-7}\textup{ M}$

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Question

Calculate the molar solubility of $Co(OH)_{2}$ which has a $Ksp = 1.3\times 10^{-15}$ .

Answer

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

$Co(OH)_{2(s)}\rightleftharpoons Co^{2+}(aq)+2OH^{-}(aq)$

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

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

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

$[Co^{2+}]=x$

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

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

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

$Ksp=x\cdot (2x)^{2}$

Now, we can solve for the x-variable by setting this value equal to the given $K_{sp}$ .

$4^{3}=1.3\times10^{-15}$

$\frac{1.3\times 10^{-15}}{4}=x^{3}$

$x=\sqrt[3]{\frac{1.3\times 10^{-15}}{4}}=6.87534\times10^{-6}$

Therefore, the solubility of $Co(OH)_{2}$ in water is $6.88\times10^{-6}M$ .

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Question

Calculate the molar solubility of $Co(OH)_{2}$ with $Ksp = 1.3\times 10^{-15}$ if enough is added to raise the pH of the solution to pH 12.

Answer

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

$Co(OH)_{2(s)}\rightleftharpoons Co^{2+}(aq)+2OH^{-}(aq)$

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

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

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

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

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

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

The solution was adjusted to pH 12 using , therefore:

$[OH^{-}]=10^{-pOH}=10^{-2}=1.0\times10^{-2}$

Using an ICE table to process the data:

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

$Ksp=[X]\times [0.01+2X^{2}]=1.3\times10^{-15}$

$2X^{2}<< 0.01$

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

$Ksp=[X]\times [0.01]=1.3\times10^{-15}$

$[X]=\frac{1.3\times10^{-15}}{[0.01]}=1.3\times10^{-13}M$

Therefore, the solubility of $Co(OH)_{2}$ in water is $1.3\times10^{-13}M$ .

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Question

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.

Answer

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:

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

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Question

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

Answer

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:

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

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Question

Calculate the molar solubility of with $K_{sp} = 3.6\times 10^{-8}$ in a solution containing $0.10\textup{ M}$ $Pb(NO_{3})_{2}$ .

Answer

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, taking into account the concentration of $Pb^{2+}$ from and $Pb(NO_{3})_{2}$ can be set to:

$[Pb^{2+}]=X+0.10$

The concentration will be double the concentration of the $Pb^{2+}$ coming from :

$F^{-}=2X$

Using an ICE table to process the data:

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

$Ksp= [0.10+X]\times [2X^{2}]=3.6\times10^{-8}$

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

$Ksp=[0.10]\times [2X]^{2}=3.6\times10^{-8}$

$[2X]^{2}=\frac{3.6\times10^{-8}}{[0.10]}$

$[X]^{2}=\frac{3.6\times10^{-8}}{[0.10]\times 2}$

$[X]=\sqrt{\frac{3.6\times10^{-8}}{[0.10]\times 2}}=4.2\times10^{-4}\textup{ M}$

Therefore, the solubility of in water is $4.2\times10^{-4}\textup{ M}$ .

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Question

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

Answer

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:

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

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Question

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

Answer

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:

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

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Question

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.

Answer

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:

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

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Question

Consider the following balanced equation for the solubility of barium hydroxide in an aqueous solution.

$\small Ba(OH)_{2}_{(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2OH^{-}_{(aq)}$

$\small K_{sp} = 5*10^{-3}$

What is the solubility of barium hydroxide?

Answer

In order to solve for the solubility of barium hydroxide, we need to establish an ICE table for the reaction.

I. Initially, there are no barium ions or hydroxide ions in the solution, so we can call their concentrations zero at the beginning of the reaction.

C. For every molecule of dissolved barium hydroxide, there will be one ion of barium and two ions of hydroxide. As a result, the increases in each ion can be designated $\small +x$ and $\small +2x$ , respectively.

E. Finally, we plug these values into the equilibrium expression and set it equal to the solubility product constant.

$\begin{matrix} Ba(OH)_2 & Ba^{2+} &OH^- \ /& 0 & 0\ / & x & 2x \end{matrix}$

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

$\small 5*10^{-3} = [x][2x]^{2}$

$\small x = 0.108M$

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Question

How many moles of fluoride ions are present in of a completely saturated solution of lead (II) fluoride?

$PbF_2(s)\rightarrow Pb^{2+}(aq)+2F^-(aq)$

$K_{sp}=3.7*10^{-8}$

Answer

Recall that the solubility product constant is given by the equation below, for a reaction in the following format.

$aA\rightarrow cC+dD$

$K_{sp}=[C]^c[D]^d$

Using our balanced reaction, we can find the solubility product equation for the dissociation of lead (II) fluoride.

$PbF_2(s)\rightarrow Pb^{2+}(aq)+2F^-(aq)$

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

For each molecule of lead (II) fluoride that dissolves, it produces one lead ion and two fluoride ions. We can conclude that the concentration of fluoride ions in solution will be twice the concentration of lead ions.

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

Use these variables in the solubility product equation, along with the given value from the question.

$3.710^{-8}=(x) (2x)^2$

Now we can solve for the value of $\small x$ .

$3.710^{-8}=4x^3$

$x = \sqrt[3]{\frac{3.710^{-8}}{4}}=2.110^{-3}$

Remember that this value is equal to the concentration of lead ions in solution and half the concentration of fluoride ions in solution.

$[Pb^{2+}]=x=2.110^{-3}M$

$[F^-]=2x=2(2.110^{-3}M)=4.2*10^{-3}M$

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