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GRE Subject Test: Chemistry : Analytical Chemistry

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Example Questions

Example Question #31 : Analytical Chemistry

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

Possible Answers:

radical

a base

an anion

a catalyst

an acid

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 #32 : Analytical Chemistry

Which molecule would be considered a Lewis acid?

Possible Answers:

Cu(s)

$NH_{3}$

$OH^{-}$

$O_{2}$

$Zn^{2+}$

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 #33 : Analytical Chemistry

Which of the following would be considered a Lewis base?

Possible Answers:

$CH_{2}Cl_{2}$

$Mn^{2+}$

$H^{+}$

$Br^{-}$

Zn(s)

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 #34 : Analytical Chemistry

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

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

Possible Answers:

Combination

Double replacement

Redox reaction

Combustion

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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Example Question #11 : Reactions And Titrations

250mL of 2N HClO_4 is added to 100mL of 5N NH_3 . An indicator in the solution is known to be yellow at any pH greater than 8.3 and green at any pH less than 8.3. Which of the following best describes the solution once it reaches equilibrium?

Possible Answers:

Neutral and green

Basic and green

Acidic and green

Basic and yellow

Acidic and yellow

Correct answer:

Acidic and green

Explanation:

Each of these compounds requires one equivalent of H⁺ is added to 100mL of 5N NH₃mol, so for HClO₄, 2N = 2M, and for NH₃, 5N = 5M.

Using the concentrations and volumes, we can find the moles, finding $\dpi{100} \small 0.25\times 2=0.5mol$ HClO₄ and $\dpi{100} \small 0.10\times 5=0.5mol$ NH₃.

In this case, equivalents of acid are equal to the equivalents of base, meaning that we are at the equivalence point in a titration. HClO₄ is a strong acid, and NH₃ is a weak base.

Thus, the acid will fully dissociate, while the base will not, resulting in a greater concentration of H⁺ than OH^– in the solution. This means the resulting solution will be acidic. We know that the indicator changes from yellow to green at 8.3, which is a basic pH. Our initial solution is basic, and we must pass through the pH of 8.3 to reach our final acidic solution, with pH < 8.3, meaning that the indicator must change from yellow to green during the reaction. This gives out final answer that the solution will be acidic and green.

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GRE Subject Test: Chemistry : Analytical Chemistry

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

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #31 : Analytical Chemistry

Example Question #32 : Analytical Chemistry

Example Question #32 : Analytical Chemistry

Example Question #292 : Gre Subject Test: Chemistry

Example Question #32 : Analytical Chemistry

Example Question #33 : Analytical Chemistry

Example Question #34 : Analytical Chemistry

Example Question #4 : Oxidation Reduction Analysis

Example Question #1 : Oxidation Reduction Analysis

Example Question #11 : Reactions And Titrations

All GRE Subject Test: Chemistry Resources

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