GRE Subject Test: Chemistry - Acid-Base Reactions

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.

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.

Which of the following acids is considered polyprotic?

$H_{2}CO_3$

All of these

Explanation

Carbonic acid ( $H_{2}CO_{3}$ ) is considered a polyprotic acid because it has two ionizable protons ( atoms) in its molecular formula. The protons dissociate in an aqueous solution according to the acid-base equilibria below:

$H_{2}CO_{3} \leftrightarrow H^{+} + HCO_{3}^{-}$

$HCO_{3}^{-} \leftrightarrow H^{+} + CO_{3}^{2-}$

The other acids are all monoprotic acids.

Which of the following acids is polyprotic?

$H_{2}SO_{4}$

$CH_{3}CH_{2}COOH$

$HNO_{3}$

None of these

Explanation

Sulfuric acid ( $H_{2}SO_{4}$ ) is considered a polyprotic acid because it has two ionizable protons in its molecular formula. The protons dissociate in an aqueous solution according to the acid-base equilibria below:

$H_{2}SO_{4} \leftrightarrow H^{+} + HSO_{4}^{-}$

$HSO_{4}^{-} \leftrightarrow H^{+} + SO_{4}^{2-}$

All the other acids listed in the answer choices are monoprotic.

$H_{2}SO_{4}_{(aq)}+2NaOH_{(aq)}\rightarrow\ 2H_{2}O_{(l)}+Na_{2}SO_{4}_{(aq)}$

Based on the above balanced equation for a neutralization reaction, what is the concentration of a solution if of $H_{2}SO_{4}$ is needed to neutralize a solution of ?

Explanation

We need to convert milliliters to liters:

$25.0\ mL \times\frac{1L}{1000mL}=0.0250\ L$

We need to determine the moles of $H_{2}SO_{4}$ using dimensional analysis the concentration as a conversion factor:

$Moles\ of\ H_{2}SO_{4}=0.0250\ L \times \frac{0.100\ moles}{L}=0.00250\ moles\ H_{2}SO_{4}$

Based on the chemical equation given:

$1\ mole\ of H_{2}SO_{4}=2\ moles\ of\ NaOH$

We can used the relationship of moles of and moles of $H_{2}SO_{4}$ as a conversion factor to determine the moles of :

$Moles\ of\ NaOH=0.00250\ moles\ H_{2}SO_{4} \times\frac{2\ moles\ NaOH}{1\ mole\ H_{2}SO_{4}}=0.00500\ moles\ NaOH$

Concentration in molarity can be calculated using the following formula:

$Molarity\ of NaOH=\frac{moles\ of\ NaOH}{Liters\ of\ solution}$

Let's convert liters of to the mL:

$Liters\ of\ NaOH\ solution=15.0\ mL \times\frac{1L}{1000mL}=0.0150\ L$

Therefore, the concentration of is:

$Molarity\ of NaOH=\frac{0.00500\ moles\ NaOH}{0.0150\ L}=0.333\ M$

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Determine the acid dissociation constant for a $0.1\ M$ monoprotic acid $\left ( HA\right )$ dissolved in the $H_{2}O$ in which a pH meter for read pH for the solution.

$2.51\times10^{-6}$

$1.52\times10^{-3}$

$3.3\times10^{-2}$

$7.9\times10^{-12}$

Explanation

The acid dissociation constant expression for this reaction is: $K_{a}=\frac{[H_{3}O^{+}][A^{-}]}{[HA]}$

Due to the dissociation of :

$x=[H_{3}O^{+}]=[A^{-}]$

Using the following equation, we can calculate the $H_{3}O^{+}$ concentration:

$[H_{3}O^{+}]=10^{-pH}$

$[H_{3}O^{+}]=10^{-3.3}$

$[H_{3}O^{+}]=5.01\times10^{-4}M$

$[HA]=original\ HA\ concentration-x=0.1M-x\approx0.1M$

Plug the values obtained in to the acid dissociation constant expression:

$K_{a}=\frac{[H_{3}O^{+}][A^{-}]}{[HA]}=\frac{x\times x}{0.1\ M}=\frac{(5.01\times10^{-4})^{2}M}{0.1\ M}=2.51\times10^{-6}$

Considering the Ka for is $1.1*10^{-2}$ , what is the Kb for $ClO_2^{-}$ ?

$9.1*10^{-13}$

$2.1*10^{-10}$

$5*10^{-3}$

$6.3*10^{-12}$

$4.8*10^{-11}$

Explanation

The equilibrium governing the dissolution of in water is:

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

is the conjugate acid of $ClO_2^{-}$ . In other words, $ClO_2^{-}$ is the conjugate base of .

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

Rearrange the equation and solve:

$K_{b}=\frac{K_{w}}{K_{a}}=\frac{1.0*10^{-14}}{1.1*10^{-2}}=9.1*10^{-13}$

Which of the following is the weakest acid?

$HClO_{4}$

Explanation

(hydrofluoric acid) is the weakest acid. Fluoride ion is the most electronegative ion. Among the other halogens, its atomic radius is smaller, and therefore bonds more strongly with hydrogen and therefore does not completely dissociate in solution as compared to , , and . Perchloric acid is a strong acid and dissociates completely in solution.

$HCl_{(aq)}+KOH_{(aq)}\rightarrow\ H_{2}O_{(l)}+KCl_{(aq)}$

What is the type of reaction given?

Neutralization Reaction

Sublimation Reaction

Synthesis Reaction

Catalytic reaction

Redox Reaction

Explanation

The reaction given is called a neutralization reaction because the acid and base components react to counterbalance each other making them chemically neutral. This type of reaction occurs between an acid and base to form a water and a salt. A neutralization reaction between a strong acid and a strong base react to form a neutral solution of pH 7.

Acid-Base Reactions

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