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GRE Subject Test: Chemistry : pH

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

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Example Questions

Example Question #1 : P H

What is the molarity of a HCl solution that has a pH of 5.0 ?

Possible Answers:

3.50 M

$1 * 10^{-5} M$

0.5M

1.50 M

$5.5 * 10^{-1} M$

Correct answer:

$1 * 10^{-5} M$

Explanation:

The pH of the solution is , therefore the concentration would be $1 * 10^{-5} M$ . This solution is based on the equation, $[H^{+}] = 10^{-pH}$ and because hydrochloric is a strong acid, it can be assumed to completely dissociate in solution.

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Example Question #112 : Gre Subject Test: Chemistry

What is the pH of a $2.0*10^{-4} M$ solution of NaOH ?

Possible Answers:

10.3

7.8

Correct answer:

10.3

Explanation:

We need to calculate the pH of a $2*10^{-4} M$ NaOH solution. There is one mole of OH^- in every mole of NaOH , therefore:

$[OH^{-}]=2*10^{-4}\ M$

$pOH=-log(OH^{-})=-log(2*10^{-4})=3.7$

The equation with the relationship between pH and pOH is:

pH+pOH=14

We can calculate the pH by rearranging this equation:

pH= 14.00-3.7=10.3

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

$K_{w}=[H_{3}O^{+}][OH^{-}]=1.0*10^{-14}$

Rearrange this equation:

$[H_{3}O^{+}]=\frac{[K_{w}]}{[OH^{-}]}=\frac{1.0*10^{-14}}{2*10^{-4}}=5*10^{-11}$

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

$pH=-log(H_{3}O^{+})=-log(5.0*10^{-11})=10.3$

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Example Question #113 : Gre Subject Test: Chemistry

What is the pH of a $5.0*10^{-4} M$ solution of $Ca(OH)_{2}$ ?

Possible Answers:

Correct answer:

Explanation:

First we need to calculate the [OH^-] of the $5*10^{-4} M$ $Ca(OH)_{2}$ solution. For every mole of $Ca(OH)_{2}$ , there is double the number of moles of hydroxide ions:

$[OH^{-}]=2*(5 *10^{-4} M)=0.001M$

The pOH can be calculated using the below equation:

$pOH=-log(OH^{-})=-log(0.001)=3$

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

pH+pOH=14

We can calculate the pH by rearranging this equation:

pH= 14.00-3.00=11

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

$K_{w}=[H_{3}O^{+}][OH^{-}]=1.0*10^{-14}$

Rearrange this equation:

$[H_{3}O^{+}]=\frac{[K_{w}]}{[OH^{-}]}=\frac{1.0*10^{-14}}{0.001}=1*10^{-11}$

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

$pH=-log(H_{3}O^{+})=-log(1.0*10^{-11})=11$

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Example Question #2 : P H

Considering the K_a of (hydrofluoric acid) is $6.8\times10^{-4}$ , what is the pK_b of the base F^- ?

Possible Answers:

8.4

6.4

10.8

5.5

7.5

Correct answer:

10.8

Explanation:

The relationship between K_a and K_b is:

$K_{a}*\ K_{b}=K_{w}$

$K_{w}=1.0\times10^{-14}$

Rearranging this equation gives:

$K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0\times 10^{-14}}{6.8\times 10^{-4}}=1.5\times10^{-11}$

In order to calculate the pK_b , we must use this relationship:

$pK_{b}=-log(K_{b})=-log(1.5\times 10^{-11})=10.8$

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Example Question #31 : Acid Base Chemistry

What is the pH of a $0.05 \textup{ M}$ solution of HCl ?

Possible Answers:

11.0

0.05

5.5

4.1

1.3

Correct answer:

1.3

Explanation:

Below is the equilibria of HCl in an aqueous solution:

$HCl(l) +H_{2}O(l)\rightleftharpoons\ Cl^{-}(aq)+H_{3}O^{+}(aq)$

HCl is a strong acid so it completely ionizes in solution. It has a high $K_{eq}$ of ^.

There is 100% dissociation of HCl , therefore the resulting $H_{3}O^+$ in solution equals to the concentration of original HCl .

$[H_{3}O^{+}]=0.05M$

$pH=-log[H_{3}O^{+}]=-log[0.05]=1.3$

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Example Question #31 : Acid Base Chemistry

What is the pH of a $0.10 \textup{ M}$ solution of NaOH ?

Possible Answers:

5.0

8.1

1.0

0.1

Correct answer:

Explanation:

Below is the equilibria of NaOH in an aqueous solution:

$NaOH(s)\rightleftharpoons\ Na^{+}(aq)+OH^{-}(aq)$

NaOH is a strong base so it completely ionizes in solution. It has a high solubility product constant.

There is 100% dissociation of NaOH , therefore the resulting $OH^{-}$ in solution equals to the concentration of original NaOH .

$[OH^{-}]=0.10M$

$pOH=-log[OH^{-}]=-log[0.10]=1$

pH+pOH=14

pH=14-pOH=14-1=13

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Example Question #1 : P H

If $3.11\times10^{-4}\ moles$ of NaOH is reacted with $5.71\times10^{-4}\ moles$ of HCl , what is the pH if the total volume of the solution is 1 liter ?

Possible Answers:

2.1

3.6

8.4

6.6

Correct answer:

3.6

Explanation:

Based on the chemical equation, HCl and NaOH react in a 1:1 mole ratio. Therefore, in order to find the pH of this solution you must first determine the difference in moles of the two reactants and the limiting reactant. The reactant in excess will determine the pH of the solution. $5.71\times 10^{-4}\ moles\ HCl-3.11\times 10^{-4}\ moles\ NaOH=2.59\times 10^{-4}\ moles\ HCl$

There HCl is in excess and the pH of the solution will be based on the moles of the excess HCl .

$moles\ of\ H^{+}=2.59\times 10^{-4}}\ moles$

$Molarity\ [H^+]=\frac{moles\ of\ solute}{Liters\ of\ solution}=\frac{2.6\ moles}{1L}=2.6M$

$pH=-log[H^{+}]=-log[2.59\times 10^{-4}]=3.6$

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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 #1 : P H

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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All GRE Subject Test: Chemistry Resources

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GRE Subject Test: Chemistry : pH

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

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #1 : P H

Example Question #112 : Gre Subject Test: Chemistry

Example Question #113 : Gre Subject Test: Chemistry

Example Question #2 : P H

Example Question #31 : Acid Base Chemistry

Example Question #31 : Acid Base Chemistry

Example Question #1 : P H

Example Question #31 : Acid Base Chemistry

Example Question #1 : P H

Example Question #1 : P H

All GRE Subject Test: Chemistry Resources

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