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GRE Subject Test: Chemistry : GRE Subject Test: Chemistry

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

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1 Diagnostic Test 152 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Analyzing Solids

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

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

Possible Answers:

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

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

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

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

None of these

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 : Analyzing Solids

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

16 M

1.8 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 #2 : Analyzing Solids

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

Possible Answers:

$3*10^{-3}$

$7.1*10^{-5}$

$8.5*10^{-6}$

$5.3*10^{-8}$

$2.3*10^{-3}$

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 #311 : Gre Subject Test: Chemistry

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

Possible Answers:

$5.6*10^{-9}$

$2.2*10^{-3}$

$7.3*10^{-5}$

$7.7*10^{-3}$

$3.4*10^{-5}$

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 #2 : Solubility And Precipitation

Which salt is insoluble in water?

Possible Answers:

$CuSO_{4}$

$Al(OH)_{3}$

AgCl

$NaNO_{3}$

$CuCl_{2}$

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 #2 : 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:

$0.011\textup{ M}$

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

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

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

$3.22\textup{ M}$

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 #3 : Solubility And Precipitation

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:

$0.5421\textup{ M}$

$4.11\textup{ M}$

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

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

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

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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Example Question #51 : Analytical Chemistry

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

Possible Answers:

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

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

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

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

Correct answer:

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

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:

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

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Example Question #11 : Solubility And Precipitation

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

Possible Answers:

$0.1992\textup{ M}$

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

$3.2\textup{ M}$

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

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

Correct answer:

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

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 NaOH , therefore:

pH +pOH=14

pOH=14-10=4

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

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

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Example Question #52 : Analytical Chemistry

Calculate the molar solubility of PbF_2 with $K_{sp} = 3.6\times 10^{-8}$ in a solution containing $0.01\textup{ M}$ NaF .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

The equation for the dissolution of PbF_2 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 PbF_2 in water, the concentration of $Pb^{2+}$ can be set to be:

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

Therefore F^- 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 PbF_2 in water is $3.6\times10^{-6}M$ .

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GRE Subject Test: Chemistry : GRE Subject Test: Chemistry

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

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #1 : Analyzing Solids

Example Question #2 : Analyzing Solids

Example Question #2 : Analyzing Solids

Example Question #311 : Gre Subject Test: Chemistry

Example Question #2 : Solubility And Precipitation

Example Question #2 : Solubility And Precipitation

Example Question #3 : Solubility And Precipitation

Example Question #51 : Analytical Chemistry

Example Question #11 : Solubility And Precipitation

Example Question #52 : Analytical Chemistry

All GRE Subject Test: Chemistry Resources

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