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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 #31 : Acid Base Chemistry

What is the pH of an aqueous solution with a $5.1\times10^{-6}\ M$ hydroxide ion concentration?

Possible Answers:

4.5

10.1

8.7

7.9

Correct answer:

8.7

Explanation:

We need to calculate the pH of a solution with a $5.1\times10^{-6}\ M$ hydroxide ion concentration.

$[OH^{-}]= 5.1\times10^{-6}\ M$

$pOH=-log(OH^{-})=-log(5.1\times10^{-6}\)=5.3$

The equation with the relationship between pH and pOH is below:

pH+pOH=14

We can calculate the pH by rearranging this equation:

pH=14.0-5.3=8.7

Another way of solving this problem is shown below. The equation with the relationship between $H_{3}O^{+}$

and $OH^{-}$ concentration is:

$K_{w}=[H_{3}O^{+}][OH^{-}]=1.0\times10^{-14}$

Rearranging this equation gives:

$[H_{3}O^{+}]=\frac{[K_{w}]}{[OH^{-}]}=\frac{1.0\times 10^{-14}}{5.1\times10^{-6}\ }=1.96\times10^{-9}$

We can calculate the pH of this solution using the equation below:

$pH=-log[H_{3}O^{+}]=-log(1.96\times10^{-9}\)=8.7$

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Example Question #1 : P H

Considering the $K_{a}$ for acetic acid $(CH_{3}COOH)$ is $1.8\times10^{-5}$ , what is the $K_{b}$ for acetate $(CH_{3}COO^{-})$ ?

Possible Answers:

$1.1\times10^{-8}$

$3.0\times10^{-3}$

$3.5\times10^{-4}$

$5.6\times10^{-10}$

Correct answer:

$5.6\times10^{-10}$

Explanation:

The equilibrium governing the dissolution of $CH_{3}COOH$ in water is:

$CH_{3}COOH_{(aq)}+H_{2}O_{(l)}\rightleftharpoons\ CH_{3}COO^{-}_{(aq)}+H_{3}O^{+}_{(aq)}$

$CH_{3}COOH$ is the conjugate acid of $CH_{3}COO^{-}$ . In other words, $CH_{3}COO^{-}$ is the conjugate base of $CH_{3}COOH$ .

Using the relationship, $K_{w}=K_{a}K_{b}$ , we can calculate the Kb.

By rearranging the equation we get:

$K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0\times 10^{-14}}{1.8\times 10^{-5}}=5.6\times10^{-10}$

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Example Question #31 : Acid Base Chemistry

Determine the pH of a $0.10\ M$ HCl solution.

Possible Answers:

Correct answer:

Explanation:

HCl is a strong acid and therefore completely ionizes in solution. Therefore, all the $H^{+}$ ions dissociate in the HCl molecules in solutions according to the below chemical equation:

$HCl(aq)\rightarrow\ H^{+}(aq)+Cl^{-}(aq)$

This means that the $[HCl]=[H^{+}]\ in\ solution$

Below is the equation to calculate pH:

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

Therefore,

pH=-log[0.1M]

pH=1

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Example Question #32 : Acid Base Chemistry

Determine the pH of a $0.02\ M$ $HNO_{3}$ solution.

Possible Answers:

4.3

5.5

7.0

1.7

Correct answer:

1.7

Explanation:

$HNO_{3}$ is a strong acid and therefore completely ionizes in solution. Therefore, all the $H^{+}$ ions dissociate in the $HNO_{3}$ molecules in solutions according to the below chemical equation:

$HNO_{3}(aq)\rightarrow\ H^{+}(aq)+NO_{3}^{-}(aq)$

This means that the $[HNO_{3}]=[H^{+}]\ in\ solution$

Below is the equation to calculate pH:

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

Therefore,

pH=-log[0.02M]

pH=1.7

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Example Question #41 : Acid Base Chemistry

Determine the pH of a $0.12\ M$ solution of KOH .

Possible Answers:

5.3

Correct answer:

Explanation:

KOH is a strong base and therefore completely ionizes in solution. Therefore, all the $OH^{-}$ ions dissociate in the KOH molecules in solutions according to the below chemical equation:

$KOH(aq)\rightarrow\ K^{+}(aq)+OH}^{-}(aq)$

This means that the $[KOH]=[OH^{-}]\ in\ solution$

NaOH solution. There is one mole of OH- in every mole

of NaOH, therefore:

$[OH^{-}]= 0.12\ M$

$pOH=-log(OH^{-})=-log(0.12)=0.92$

The equation with the relationship between pH and pOH is below:

pH+pOH=14

We can calculate the pH by rearranging this equation:

pH=14.0-0.92=13

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Example Question #42 : Acid Base Chemistry

Calculate the Ka for $HNO_{2}$ (pka=3.3) .

Possible Answers:

$2.5\times10^{-4}$

10.7

$5.0\times10^{-4}$

$5.6\times10^{-12}$

Correct answer:

$5.0\times10^{-4}$

Explanation:

The pka of $HNO_{2}$ is 3.3 . The pka is simply calculated based on the following equation:

pka=-log(ka)

Therefore, in order to determine the ka, simply use f the following equation:

$ka=10^{-pka}$

$ka=10^{-3.3}$

$ka=5.0\times10^{-4}$

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Example Question #43 : Acid Base Chemistry

Calculate the Ka for $CH_{3}COOH$ (pka=4.76) .

Possible Answers:

8.7

$1.7\times10^{-5}$

2.3

$1.4\times10^{-3}$

Correct answer:

$1.7\times10^{-5}$

Explanation:

The pka of $CH_{3}COOH$ is 4.76 . The pka is simply calculated based on the following equation:

pka=-log(ka)

Therefore, in order to determine the ka, simply use f the following equation:

$ka=10^{-pka}$

$ka=10^{-4.76}$

$ka=1.7\times10^{-5}$

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Example Question #44 : Acid Base Chemistry

Find the pH of $H_{2}O$ if its $K_{w}$ is $5.476\times10^{-14}$ at $50^{o}C$ .

Possible Answers:

8.32

6.63

6.11

7.23

Correct answer:

6.63

Explanation:

$K_{w}$ is a symbol for the ionization constant for water. This value is temperature dependent. At $50^{o}C$ , the $K_{w}$ for water is $5.476\times10^{-14}$ .

Water dissociates or ionizes according to the chemical equation below:

$H_{2}O(l)\rightleftharpoons\ H^{+}(aq)+OH^{-}(aq)$

Therefore, the ionization of water results in:

$[H^{+}]=[OH^{-}]$

$K_{w}=[H_{3}O^{+}][OH^{-}]$

Let us set $x=[H^{+}]=[OH^{-}]$

$K_{w}=x^{2}$

$\sqrt{K_{w}}=x$

$\sqrt{5.476\times10^{-14}}=x$

$x=2.340\times10^{-7}\ M$

$pH=-log[H^{+}]=-log(2.340\times10^{-7}\)=6.63$

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Example Question #45 : Acid Base Chemistry

Find the pH of $H_{2}O$ if its $K_{w}$ is $1.0\times10^{-14}$ at $25^{o}C$ .

Possible Answers:

5.7

Correct answer:

Explanation:

$K_{w}$ is a symbol for the ionization constant for water. This value is temperature dependent. At $25^{o}C$ , the $K_{w}$ for water is $8.8\times10^{-14}$ .

Water dissociates or ionizes according to the chemical equation below:

$H_{2}O(l)\rightleftharpoons\ H^{+}(aq)+OH^{-}(aq)$

Therefore, the ionization of water results in:

$[H^{+}]=[OH^{-}]$

$K_{w}=[H_{3}O^{+}][OH^{-}]$

Let us set $x=[H^{+}]=[OH^{-}]$

$K_{w}=x^{2}$

$\sqrt{K_{w}}=x$

$\sqrt{1.0\times 10^{-14}}=x$

$x=1.0\times10^{-7}\ M$

$pH=-log[H^{+}]=-log(1.0\times10^{-7}\)=7$

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Example Question #45 : Acid Base Chemistry

Considering the $K_{a}$ for HCN is $4.9\times10^{-10}$ , what is the $K_{b}$ for $CN^{-}$ ?

Possible Answers:

$2.0\times10^{-5}$

$3.5\times10^{-4}$

$4.3\times10^{-8}$

$1.7\times10^{-2}$

Correct answer:

$2.0\times10^{-5}$

Explanation:

The equilibrium governing the dissolution of HCN in water is:

$HCN_{(aq)}+H_{2}O_{(l)}\rightleftharpoons\ CN^{-}_{(aq)}+H_{3}O^{+}_{(aq)}$

HCN is the conjugate acid of $CN^{-}$ . In other words, $CN^{-}$ is the conjugate base of HCN .

Using the relationship, $K_{w}=K_{a}K_{b}$ , we can calculate the $K_{b}$ .

By rearranging the equation we get:

$K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0\times 10^{-14}}{4.9\times 10^{-10}}=2.0\times10^{-5}$

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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 #31 : Acid Base Chemistry

Example Question #1 : P H

Example Question #31 : Acid Base Chemistry

Example Question #32 : Acid Base Chemistry

Example Question #41 : Acid Base Chemistry

Example Question #42 : Acid Base Chemistry

Example Question #43 : Acid Base Chemistry

Example Question #44 : Acid Base Chemistry

Example Question #45 : Acid Base Chemistry

Example Question #45 : Acid Base Chemistry

All GRE Subject Test: Chemistry Resources

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