GRE Subject Test: Chemistry - Acid-Base Chemistry

What volume of 0.375M H2SO4 is needed to fully neutralize 0.5L of 0.125M NaOH?

83.3mL

1.5L

167mL

41.7mL

0.5L

Explanation

This question requires use of the simple titration equation M1V1 = M2V2. The key is to identify that sulfuric acid has two equivalents of acidic hydrogens while NaOH has only one hydroxide equivalent. All wrong answer choices result from making this mistake or other calculation errors.

What volume of 0.375M H2SO4 is needed to fully neutralize 0.5L of 0.125M NaOH?

83.3mL

1.5L

167mL

41.7mL

0.5L

Explanation

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

$5.6\times10^{-10}$

$3.5\times10^{-4}$

$1.1\times10^{-8}$

$3.0\times10^{-3}$

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:

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

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

$5.6\times10^{-10}$

$3.5\times10^{-4}$

$1.1\times10^{-8}$

$3.0\times10^{-3}$

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:

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

$NaOH+CH_3COOH\rightarrow CH_3COONa +H_2O$

Considering the given chemical reaction, determine the number of moles of $CH_{3}COOH$ in a 20mL solution if it takes 19.00mL of a 0.0500M solution to reach the endpoint of a titration.

$9.5\times10^{-4}$

$1.0\times 10^{-4}$

Explanation

$CH_{3}COOH$ and react in a 1:1 mole ratio. Therefore the number of moles of at the end point of the reaction equals to the number of moles of $CH_{3}COOH$ . We can use the concentration as a conversion factor to determine the number of moles reacted.

$0.01900L \times \frac{0.0500moles}{L}=9.5\times10^{-4}\ moles\ NaOH=moles\ of\ CH_{3}COOH$

$NaOH+CH_3COOH\rightarrow CH_3COONa +H_2O$

Considering the given chemical reaction, determine the number of moles of $CH_{3}COOH$ in a 20mL solution if it takes 19.00mL of a 0.0500M solution to reach the endpoint of a titration.

$9.5\times10^{-4}$

$1.0\times 10^{-4}$

Explanation

$0.01900L \times \frac{0.0500moles}{L}=9.5\times10^{-4}\ moles\ NaOH=moles\ of\ CH_{3}COOH$

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

Explanation

The relationship between and 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 , we must use this relationship:

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

Considering the Ka for is $2.5*10^{-9}$ , what is the Kb for ?

$4*10^{-6}$

$2*10^{-8}$

$1*10^{-9}$

$3*10^{-1}$

Explanation

The equilibrium governing the dissolution of in water is:

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

is the conjugate acid of . In other words, is the conjugate base of .

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

Rearrange the equation and solve:

$K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0*10^{-14}}{2.5*10^{-9}}=4*10^{-6}$

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Based on the equilibrium shown, what does $Cl^{-}$ act as?

A base

An acid

A radical

A cation

A catalyst

Explanation

A base is a substance that can accept a proton. The conjugate base of an acid is formed when the acid donates a proton. In this case, $Cl^{-}$ is the conjugate base to the acid . This is because donates a hydrogen ion to the organic molecule to form $Cl^{-}$ , the conjugate base.