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GRE Subject Test: Chemistry : Solubility and Precipitation

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Example Question #6 : Solubility

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?

Possible Answers:

$\small 0.05M$

$\small 0.108M$

$\small 0.035M$

$\small 0.071M$

Correct answer:

$\small 0.108M$

Explanation:

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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Example Question #41 : Analytical Chemistry

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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:

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

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Example Question #41 : Analytical Chemistry

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

Possible Answers:

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

$K_{sp} = [Ag^{2+}][CN^{2-}]$

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

$K_{sp} = [Ag][CN]$

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

Correct answer:

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

Explanation:

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

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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Example Question #1 : Solubility And Precipitation

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

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

Possible Answers:

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

None of these

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

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

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

Correct answer:

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

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 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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Example Question #2 : Solubility And Precipitation

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?

Possible Answers:

11 M

5 M

1.8 M

16 M

Correct answer:

1.8 M

Explanation:

The pH of the solution is 8.5 , therefore the pOH would be 5.5 considering the relationship:

14 = pH + pOH

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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Example Question #42 : Analytical Chemistry

The K_sp of BaSO_4 is $5.0*10^{-9}$ . What is the molar solubility of BaSO_4 in water?

Possible Answers:

$3*10^{-3}$

$5.3*10^{-8}$

$8.5*10^{-6}$

$2.3*10^{-3}$

$7.1*10^{-5}$

Correct answer:

$7.1*10^{-5}$

Explanation:

The equation for the dissolution of BaSO_4 in water is below:

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

The K_spfor the above equation is:

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

Due to the solubility of BaSO_4 of in water, the concentration of $Ba^{2+}$ and $SO_4^{2-}$ ^- should be equal:

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

Plug into the K_sp equation:

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

Therefore:

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

The solubility of BaSO_4 in water is $7.1*10^{-5}$

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Example Question #1 : Solubility And Precipitation

The K_sp of CuBr is $5.3*10^{-9}$ . What is the molar solubility of CuBr inwater?

Possible Answers:

$3.4*10^{-5}$

$5.6*10^{-9}$

$7.3*10^{-5}$

$2.2*10^{-3}$

$7.7*10^{-3}$

Correct answer:

$7.3*10^{-5}$

Explanation:

The equation for the dissolution of CuBr in water is below:

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

The K_sp for the above equation is:

$Ksp=[Cu^{+}][Br^{-}]=5.3*10^{-9}$

Due to the solubility of CuBr of in water, the concentration of Cu^+ and Br^- should be equal:

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

Plugging X into the K_sp equation gives:

$Ksp=X*X=X^{2}=5.3*10^{-9}$

Therefore:

$X=\sqrt{5.3\ *10^{-9}}=7.3*10^{-5}$

The solubility of CuBr in water is $7.3*10^{-5}$

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Example Question #1 : Solubility And Precipitation

Which salt is insoluble in water?

Possible Answers:

$Al(OH)_{3}$

$CuCl_{2}$

AgCl

$CuSO_{4}$

$NaNO_{3}$

Correct answer:

AgCl

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 NH_4^+ compounds are soluble.

2. Nitrate (NO_3^-) , perchlorate (ClO_4^-) , chlorate, (ClO_3^3) , and acetate (CH_3COO^-) 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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Example Question #1 : Solubility And Precipitation

The $K_{sp}$ of CdCO_3 is $1.8*10^{-14}$ . What is the molar solubility of CdCO_3 in water?

Possible Answers:

$3.22\textup{ M}$

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

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

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

$0.011\textup{ M}$

Correct answer:

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

Explanation:

The equation for the dissolution of CdCO_3 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 CdCO_3 , 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 CdCO_3 in water is $1.3*10^{-7}\textup{ M}$

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Example Question #51 : Analytical Chemistry

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

Possible Answers:

$1.3\times10^{-13}\textup{ M}$

$0.5421\textup{ M}$

$4.11\textup{ M}$

$2.3\times10^{-3}\textup{ M}$

$3.1\times10^{-3}\textup{ M}$

Correct answer:

$1.3\times10^{-13}\textup{ M}$

Explanation:

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 NaOH , therefore:

pH +pOH=14

pOH=14-12=2

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

Using an ICE table to process the data:

Screen shot 2015 11 13 at 4.39.13 pm

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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GRE Subject Test: Chemistry : Solubility and Precipitation

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #6 : Solubility

Example Question #41 : Analytical Chemistry

Example Question #41 : Analytical Chemistry

Example Question #1 : Solubility And Precipitation

Example Question #2 : Solubility And Precipitation

Example Question #42 : Analytical Chemistry

Example Question #1 : Solubility And Precipitation

Example Question #1 : Solubility And Precipitation

Example Question #1 : Solubility And Precipitation

Example Question #51 : Analytical Chemistry

All GRE Subject Test: Chemistry Resources

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