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GRE Subject Test: Chemistry : Acid-Base Chemistry

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Example Questions

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:

$5.6\times10^{-12}$

10.7

$5.0\times10^{-4}$

$2.5\times10^{-4}$

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:

$1.4\times10^{-3}$

$1.7\times10^{-5}$

8.7

2.3

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:

7.23

6.11

8.32

6.63

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 #16 : P H

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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Example Question #18 : P H

What is the pH of a $7.1\times10^{-4}\ M$ solution of NaOH ?

Possible Answers:

8.4

10.9

5.5

3.2

Correct answer:

10.9

Explanation:

We need to calculate the pH of a $7.1\times10^{-4}\ M$ NaOH solution. There is one mole of $OH^{-}$ in every mole

of NaOH , therefore:

$[OH^{-}]= 7.1\times10^{-4}\ M$

$pOH=-log(OH^{-})=-log(7.1\times10^{-4}\)=3.1$

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

pH+pOH=14

We can calculate the pH by rearranging this equation:

pH=14.0-3.1=10.9

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:

$[H_{3}O^{+}]=\frac{[K_{w}]}{[OH^{-}]}=\frac{1.0\times 10^{-14}}{7.1\times 10^{-4}}=1.4\times10^{-11}$

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

$pH=-log[H_{3}O^{+}]=-log(1.4\times10^{-11}\)=10.9$

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Example Question #131 : General Chemistry

What is the pH of a solution in which $H^{+}$ concentration is $1.2\times10^{-8}M$ ?

Possible Answers:

7.9

Correct answer:

7.9

Explanation:

The pH of a solution is related to the number of hydrogen ions in solutions. The equation to determine the pH of a solution is below:

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

Plugging the concentration given into the equation above gives:

$pH=-log[1.2\times 10^{-8}M]$

Therefore, the pH of the solution is:

pH=7.9

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Example Question #132 : General Chemistry

What is the $H^{+}$ concentration of a solution with pH ?

Possible Answers:

$1.4\times10^{-6}M$

14M

$7.0\times10^{-1}M$

$1.0\times10^{-7}M$

Correct answer:

$1.0\times10^{-7}M$

Explanation:

The pH of a solution is related to the number of hydrogen ions in solutions. The equation to determine the pH of a solution is below:

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

The above equation can be converted into the following form in order to determine the hydrogen ion concentration:

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

Therefore, the hydrogen ion concentration of a solution of pH 7 is:

$[H^{+}]=10^{-7}=1.0\times10^{-7}M$

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Example Question #133 : General Chemistry

What is the pH of a solution with a pOH of ?

Possible Answers:

4.2

Correct answer:

Explanation:

The equation that expresses the relationship between pH and pOH is:

14=pOH+pH

Plugging the pOH given into the equation above gives:

14=13+pH

Rearranging this expression gives:

14-13=pH

Therefore, the pH of a solution with a pOH of is:

1=pH

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GRE Subject Test: Chemistry : Acid-Base Chemistry

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

All GRE Subject Test: Chemistry Resources

Example Questions

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 #16 : P H

Example Question #45 : Acid Base Chemistry

Example Question #18 : P H

Example Question #131 : General Chemistry

Example Question #132 : General Chemistry

Example Question #133 : General Chemistry

All GRE Subject Test: Chemistry Resources

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