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GRE Subject Test: Chemistry : Analyzing Reactions

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Example Questions

Example Question #21 : Analytical Chemistry

What is the [H^+] of a 0.12M solution of HCN that has a pH of 5.1?

Possible Answers:

$7.94*10^{-6}M$

4.22M

$1.30*10^{-6} M$

1.36 M

$1.01*10^{-5} M$

Correct answer:

$7.94*10^{-6}M$

Explanation:

We can calculate the hydrogen ion concentration by using the following equation:

$[H^{+}]=10^{-pH}$

Plug in the given value for pH and solve:

$[H^{+}]=10^{-5.1}=7.94*10^{-6} M$

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Example Question #291 : Gre Subject Test: Chemistry

Based on the equilibrium above, what does $NH_{4}^{+}$ act as?

Possible Answers:

an anion

a catalyst

an acid

a base

radical

Correct answer:

an acid

Explanation:

An acid is a substance that can donates a proton. The conjugate acid of a base is formed when the base accepts a proton. In this case, $NH_{4}^{+}$ is the conjugate acid to the base $NH_{3}$ . This is because $NH_{3}$ accepts a hydrogen ion from the water molecule to form $NH_{4}^{+}$ , the conjugate acid.

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Example Question #32 : Analytical Chemistry

Given the following equilibria, what is the hydroxide ion concentration of a solution containing $0.06\textup{ M}$ $NH_{3}$ ?

$NH_{3}+H_{2}O\rightleftharpoons\ NH_{4}^{+}+OH^{-}$

The $K_{b}$ for $NH_{3}$ is $1.75\times10^{-5}$

Possible Answers:

$0.06\textup{ M}$

$0.0010 \textup{ M}$

$0.106\textup{ M}$

$2.13\textup{ M}$

$0.344\textup{ M}$

Correct answer:

$0.0010 \textup{ M}$

Explanation:

Below is the acid-base equilibria of $NH_{3}$ in an aqueous solution:

$NH_{3}+H_{2}O\rightleftharpoons\ NH_{4}^{+}+OH^{-}$

A solution of ammonia ( $NH_{3}$ ) is basic based on the chemical equation given. The $K_{b}$ for this reaction is:

$Kb= \frac{[NH_{}4]^{+}][OH^{-}]}{[NH_{3}]}=1.75\times10^{-5}$

Based on the chemical equation, $[OH^{-}]=[NH_{4}^{+}]$ .

We can make the concentration of these species equal to :

$[OH^{-}]=[NH_{4}^{+}]=X$

The $NH_{3}$ concentration is equal to:

$NH_{3}=0.06-[OH^{-}]$

$NH_{3}$ has a low $K_{b}$ so we can assume the following:

$NH_{3}=0.06-[OH^{-}]\approx0.06$

Plugging the values into the base dissociation constant equation gives:

$Kb= \frac{[NH_{}4]^{+}][OH^{-}]}{[NH_{3}]}=\frac{X^{2}}{0.06}=1.75\times10^{-5}$

Solve for :

$X^{2}= (0.06)\times(1.75\times10^{-5})=1.05\times10^{-6}$

$X=\sqrt{1.05\times 10^{-6}}=0.0010\ M$

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Example Question #32 : Analytical Chemistry

Given the following equilibria, what is the hydrogen ion concentration of a solution containing $0.01\textup{ M}$ $C_{6}H_{5}COOH$ (benzoic acid)?

$C_{6}H_{5}COOH(aq)\rightleftharpoons\ C_{6}H_{5}COO^{-}(aq)+H^{+}(aq)$

The $K_{a}$ for $NH_{3}$ is $6.3\times10^{-5}$ .

Possible Answers:

$6.9\times10^{-9}\textup{ M}$

$0.01\textup{ M}$

$7.9\times10^{-4}\textup{ M}$

$1.5\times10^{-7}\textup{ M}$

$3.6\times10^{-4}\textup{ M}$

Correct answer:

$7.9\times10^{-4}\textup{ M}$

Explanation:

Below is the acid-base equilibria of $C_{6}H_{5}COOH$ in an aqueous solution:

$C_{6}H_{5}COOH(aq)\rightleftharpoons\ C_{6}H_{5}COO^{-}(aq)+H^{+}(aq)$

A solution of benzoic acid ( $C_{6}H_{5}COOH$ ) is acidic based on the chemical equation given. The $K_{a}$ for this reaction is:

$K_{a}= \frac{[C_{6}H_{5}COO^{-}][H^{+}]}{[C_{6}H_{5}COOH]}=6.3\times10^{-5}$

Based on the chemical equation, $[H^{+}]=[C_{6}H_{5}COO^{-}]$ .

We can make the concentration of these species equal to :

$[H^{+}]=[C_{6}H_{5}COO^{-}]=X$

The $C_{6}H_{5}COOH$ concentration is equal to:

$[C_{6}H_{5}COOH]=0.01-[H^{+}]$

$C_{6}H_{5}COOH$ has a low $K_{a}$ so we can assume the following:

$C_{6}H_{5}COOH=0.01-[H^{+}]\approx0.01$

Plugging the values into the base dissociation constant equation gives:

$Kb= \frac{[C_{6}H_{5}COO^{-}][H^{+}]}{[C_{6}H_{5}COOH]}=\frac{X^{2}}{0.01}=6.3\times10^{-5}$

Solve for :

$X^{2}= (0.01)\times(6.3\times10^{-5})=6.3\times10^{-7}$

$X=\sqrt{6.3\times10^{-7}}=7.9\times10^{-4}\textup{ M}$

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Example Question #292 : Gre Subject Test: Chemistry

Given the following equilibria, what is the hydronium ion concentration of a solution containing $0.025\textup{ M}$ HOCl ?

$HOCl(aq)+H_{2}O\rightleftharpoons\ OCl^{-}(aq)+H_{3}O^{+}(aq)$

The $K_{a}$ for HOCl is $3.0\times10^{-8}$ .

Possible Answers:

$7.9\times10^{-4}\textup{ M}$

$2.7\times10^{-5}\textup{ M}$

$5.5\times10^{-2}\textup{ M}$

$8.4\times10^{-9}\textup{ M}$

$0.332\textup{ M}$

Correct answer:

$2.7\times10^{-5}\textup{ M}$

Explanation:

Below is the acid-base equilibria of HOCl in an aqueous solution:

$HOCl(aq)+H_{2}O_{(l)}\rightleftharpoons\ OCl^{-}(aq)+H_{3}O^{+}(aq)$

A solution of HOCl is acidic based on the chemical equation given. The Ka for this reaction is:

$K_{a}= \frac{[OCl^{-}][H_{3}O^{+}]}{[HOCl]}=3.0\times10^{-8}$

Based on the chemical equation, $[H_{3}O^{+}]=[OCl^{-}]$ .

We can make the concentration of these species equal to :

$[H_{3}O^{+}]=[OCl^{-}]=X$

The HOCl concentration is equal to:

$[HOCl]=0.025-[H^{+}]$

HOCl has a low $K_{a}$ so we can assume the following:

$HOCl=0.025-[H_{3}O^{+}]\approx0.025$

Plugging the values into the base dissociation constant equation gives:

$K_{a}= \frac{[OCl^{-}][H_{3}O^{+}]}{[HOCl]}=\frac{X^{2}}{0.01}=3.0\times10^{-8}$

Solve for :

$X^{2}= (0.025)\times(3.0\times10^{-8})=7.5\times10^{-10}$

$X=\sqrt{7.5\times10^{-10}}=2.7\times10^{-5}\textup{ M}$

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Example Question #1 : Oxidation Reduction Analysis

Which molecule would be considered a Lewis acid?

Possible Answers:

$OH^{-}$

Cu(s)

$Zn^{2+}$

$O_{2}$

$NH_{3}$

Correct answer:

$Zn^{2+}$

Explanation:

A lewis acid is an electron pair acceptor. $Zn^{2+}$ would be considered a lewis acid based on the definition. Because $Zn^{2+}$ is electron-deficient based on its oxidation number, it is able to accept an electron pair.

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Example Question #1 : Oxidation Reduction Analysis

Which of the following would be considered a Lewis base?

Possible Answers:

Zn(s)

$Br^{-}$

$CH_{2}Cl_{2}$

$Mn^{2+}$

$H^{+}$

Correct answer:

$Br^{-}$

Explanation:

A lewis base is an electron pair donor. $Br^{-}$ would be considered a lewis base based on the definition. Because $Br^{-}$ is electron rich based on its oxidation number, it is able to donate an electron pair to an electron deficient ion.

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Example Question #2 : Oxidation Reduction Analysis

Which of the following types of reactions best describes the following reaction:

$Ce^{4+}+Fe^{2+}\rightarrow\ Ce^{3+}+Fe^{3+}$

Possible Answers:

Double replacement

Combustion

Redox reaction

Combination

Catalytic

Correct answer:

Redox reaction

Explanation:

A redox reaction is also known as an oxidation/reduction reaction. This type of reaction involves the transfer of electrons from on reactant to another. In this reaction, an electron is transferred from $Fe^{2+}$ to $Ce^{4+}$ forming the products $Ce^{3+}$ and $Fe^{3+}$ . The substance that gains an electron is referred to as being reduced. The substance that loses an electron is referred to as being oxidized.

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Example Question #4 : Oxidation Reduction Analysis

What is the oxidation state of manganese in $KMnO_{4}$ ?

Possible Answers:

Correct answer:

Explanation:

In this compound oxygen has an oxidation number of . There are 4 oxygens, therefore the total oxidation number for the oxygens in $KMnO_{4}$ is: $Total\ oxygen\ oxidation\ number=(-2) \times 4=-8$ . Potassium,, is a group 1 metal and has an oxidation number of . The sum of all oxidation number in a neutral compound such as $KMnO_{4}$ is zero. We will give the oxidation number of equal to :

$((-2) \times 4)+(+1)+x=0$

Simplifying the above equation:

(-8)+(+1)+x=0

(-7)+x=0

Rearranging to solve for :

x=+7

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Example Question #1 : Oxidation Reduction Analysis

What is the oxidation state of manganese in $K_{3}MnO_{4}$ ?

Possible Answers:

Correct answer:

Explanation:

In this compound oxygen has an oxidation number of . There are 4 oxygens, therefore the total oxidation number for the oxygens in $K_{3}MnO_{4}$ is: $Total\ O\ oxidation\ number=(-2) \times 4=-8$ . Potassium, , is a group 1 metal and has an oxidation number of . There are 3 potassium ions present in $K_{3}MnO_{4}$ , therefore the total oxidation number for the potassium ions is:

$Total\ K\ oxidation\ number=(+1) \times 3=+3$ .The sum of all oxidation number in a neutral compound such as $K_{3}MnO_{4}$ is zero. We will give the oxidation number of equal to :

(-8)+(+3)+x=0

Simplifying the above equation:

(-5)+x=0

Rearranging to solve for :

x=+5

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All GRE Subject Test: Chemistry Resources

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GRE Subject Test: Chemistry : Analyzing Reactions

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

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #21 : Analytical Chemistry

Example Question #291 : Gre Subject Test: Chemistry

Example Question #32 : Analytical Chemistry

Example Question #32 : Analytical Chemistry

Example Question #292 : Gre Subject Test: Chemistry

Example Question #1 : Oxidation Reduction Analysis

Example Question #1 : Oxidation Reduction Analysis

Example Question #2 : Oxidation Reduction Analysis

Example Question #4 : Oxidation Reduction Analysis

Example Question #1 : Oxidation Reduction Analysis

All GRE Subject Test: Chemistry Resources

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