GRE Subject Test: Chemistry : Acid-Base Chemistry

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

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Example Questions

Example Question #1 : Acid Base Chemistry

What is the definition of a Brønsted-Lowry base?

Possible Answers:

A compound that donates protons in solution

A compound that accepts an electron pair in solution

A compound that donates an electron pair in solution

A compound that accepts protons in solution

Correct answer:

A compound that accepts protons in solution

Explanation:

A Brønsted-Lowry base is any compound that accepts protons in solution. Lewis acids and bases refer to the accepting or donating of an electron pair, respectively.

Example Question #13 : Defining/Classifying Acids And Bases

Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.

All of the bases proceed in a similar fashion.

 

The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A- in the above reaction. In the reverse reaction, A- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.

In the reverse reaction of , the proton is acting as a(n) __________, and is thus a __________.

Possible Answers:

electron acceptor . . . Lewis base

electron donor . . . Lewis acid

electron acceptor . . . Lewis acid

electron donor . . . Brønsted-Lowry base

electron donor . . . Lewis base

Correct answer:

electron acceptor . . . Lewis acid

Explanation:

In terms of the passage, the lone proton can be considered a proton donor and would, therefore, be a Brønsted-Lowry acid. This is not an answer choice.

The third acid-base definition is the Lewis definition, which states that acids are electron acceptors and bases are electron donors. The negative charge on the signifies that it is a Lewis base with available electrons to donate. The proton is accepting these electrons from , and is thus acting as a Lewis acid.

Example Question #1 : Lewis, Brønsted Lowry, And Arrhenius Definitions

Which of the following molecules or ions have the greatest ability to act like a Lewis acid?

Possible Answers:

Correct answer:

Explanation:

Lewis acids are electron pair acceptors. Molecules and ions that have a full octet cannot act a Lewis acid, therefore  and  are not lewis acids.  is very stable and insoluble and cannot accept an electron pair.  is a well known base and has extremely weak acidity.  is a transition metal ion. Transition metal are known to be Lewis acids because of their positive charge which gives them the ability to accept electron pairs.

Example Question #3 : Lewis, Brønsted Lowry, And Arrhenius Definitions

Which of the following is not a strong electrophile?

Possible Answers:

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All of these are strong electrophiles.

Correct answer:

Explanation:

One of the easiest ways of determining if a molecule is an electrophile is by the presence of a positive charge. Electrophiles are in need of electrons, therefore they are electron deficient and can be attacked by nucleophiles (compounds that are electron rich). A nucleophile is a compound that provides a pair of electrons to form a new covalent bond. Nucleophiles are electron rich and one of the easiest types of nucleophiles to recognize are ones carrying a negative charge.  is the only option given that contains a negative charge and therefore is not an electrophile.

Example Question #1 : Acid Base Chemistry

100mL of an unknown solution of NaOH is titrated with 3M HCl until neutralized. The resulting solution is evaporated, and 3.0g of white crystal are recovered. What was the concentration of the NaOH solution?

Possible Answers:

0.3M

1.0M

1.25M

0.7M

0.5M

Correct answer:

0.5M

Explanation:

In the neutralization reaction between NaOH and HCl, NaCl salt is formed. When the solution is evaporated, this salt is left behind.

3.0g of NaCl is equivalent to 0.05mol NaCl. Since the titration is between a strong acid and a strong base, all of the NaOH in the original solution is converted to NaCl in a one-to-one ratio, meaning that mol NaCl = mol NaOH.

We now know that there was 0.05mol NaOH in the 100mL solution, so the concentration must have been \dpi{100} \small 0.5M\left ( \frac{0.05}{0.1}=0.5 \right ).

Example Question #1 : Acid Base Chemistry

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

Possible Answers:

41.7mL

83.3mL

1.5L

167mL

0.5L

Correct answer:

83.3mL

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.

Example Question #31 : Acid Base Chemistry

Consider the following reaction:

 Which of the following changes will increase the pH of this solution? 

Possible Answers:

Increasing the pKa of 

Adding sodium acetate

Increasing the acetic acid concentration

Decreasing the volume of 

Correct answer:

Adding sodium acetate

Explanation:

To answer this question you need to use Le Chatelier’s principle. Adding sodium acetate to the solution will cause it to dissociate as follows:

The dissociation reaction will produce more acetate ions. According to Le Chatelier’s principle, the increase in acetate ions will shift the equilibrium of the reaction (given in the question) to the left. This means that and  will be utilized to form . This will cause a decrease in the amount of hydronium ions in solution. Recall that pH is increased when the concentration of hydrogen ions (or hydronium ions, ) is decreased; therefore, adding sodium acetate will increase the pH of the solution.

Increasing acetic acid concentration will shift the equilibrium to the right and produce more hydronium ions, thereby decreasing the pH. Recall that you can never change the pKa of an acid. The pKa of acetic acid is around 4.75, and it cannot be altered. Le Chatelier’s principle only applies when there is a change in amount of aqueous or gaseous substances; liquid and solid substances will not shift the equilibrium. Changing the volume of liquid water will not change the concentration of hydronium ions.

Example Question #1 : Acid Base Reactions

Which of the following acids is polyprotic?

Possible Answers:

None of these

Correct answer:

Explanation:

Sulfuric acid () 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:

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

Example Question #2 : Acid Base Reactions

Which of the following acids is considered polyprotic?

Possible Answers:

All of these

Correct answer:

Explanation:

Carbonic acid () 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:

The other acids are all monoprotic acids.

Example Question #91 : General Chemistry

Considering the Ka for  is , what is the Kb for ?

Possible Answers:

Correct answer:

Explanation:

The equilibrium governing the dissolution of  in water is:

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

Using the relationship, , we can calculate the Kb.

Rearrange the equation and solve:

 

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