MCAT Physical : MCAT Physical Sciences

Study concepts, example questions & explanations for MCAT Physical

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Example Questions

Example Question #1 : Reactions And Titrations

As the value of Ka increases, __________

Possible Answers:

the reaction HA → H+ + A favors the products

all of these are true

the strength of the conjugate base decreases

the strength of the acid increases

pKa decreases

Correct answer:

all of these are true

Explanation:

Ka represents the equilibrium constant for the acid dissociation in water, HA → H+ A, so it is a measure of the products divided by the reactants. As this value increases, the reaction favors the products (the ions) more, meaning that the acid dissociates more. The definition of a strong acid is one that fully dissociates in water, so as Ka increases the strength of the acid increases, and the strength of the conjugate base decreases. pKa is defined as the –log(Ka), so as Ka increases pKa decreases. All of these answer choices are correct.

Example Question #1 : Acid Base Equilibrium

What is the pKa of acetic acid? (K= 1.8 * 10–5)

Possible Answers:

8.6

2.1

4.7

4.2

5.3

Correct answer:

4.7

Explanation:

We know that pKa is equal to –log(Ka). Thus, pKa of acetic acid is –log(1.8 * 10–5). This is not an easy problem to solve in your head, but there is a trick.

We know that 1 * 10–4 > 1.8 * 10–5 > 1 * 10–5, and we know that –log(1 * 10–4) = 4 and –log(1 * 10–5) = 5. Now we can conclude that our pKa is somewhere between 4 and 5.

Two answer choices fall in this range: 4.2 and 4.7. 

1.8 * 10–5 is closer to 1 * 10–5 than it is to 1 * 10–4, so we can pick the answer closer to 5 than to 4 : 4.7.

Example Question #4 : Acid Base Chemistry

HCN dissociates based on the following reaction.

The Ka for hydrogen cyanide is .

What is the Kfor CN-?

Possible Answers:

Correct answer:

Explanation:

Remember that the Ka for the acid and the Kb for the conjugate base, when multiplied will equal the autoionization of water constant (Kw).

Example Question #5 : Acid Base Chemistry

Compound

Base strength, Kb

1

1012

2

 105

3

 101

4

 10-8

Students in a chemistry class are given one of four unknown samples in a laboratory. A student is told that his compound is the strongest acid of the four compounds. Based on the information in the above table, which compound was the student given?

Possible Answers:

Compound 3

Compound 4

Compound 2

Compound 1

Correct answer:

Compound 4

Explanation:

The strongest acid of the group will also have the smallest Kb value. As the weakest base (smallest Kb), compound 4 will only partially dissociate in solution because it has a fairly strong conjugate acid. Were this question asking which base was the strongest, compound 1 would be the answer, due to its large Kb value.

Example Question #3 : Acid Base Equilibrium

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 aqueous conditions the equilibrium constant for a Brønsted-Lowry base, , can be expressed as which of the following?

Possible Answers:

Correct answer:

Explanation:

The base reaction will essentially be the reverse of the acid reaction. In the question, the aqueous conditions mean that the base, , reacts with water to give the following reaction:

Following normal equilibrium convention, we omit water from the equation because it is a pure liquid.

Example Question #7 : Acid Base Chemistry

H_{2}CO_{3} \left ( K_{a} = 4.5 * 10^{-7} \right )

HCHO_{2} \left ( K_{a} = 1.8 * 10^{-4} \right )

HC_{2}H_{3}O_{2} \left ( K_{a} = 1.8 * 10^{-5} \right )

HF \left ( K_{a} = 6.3 * 10^{-4} \right )

Given the above values of Ka, place the acids in order from strongest to weakest.

Possible Answers:

HF > H_{2}CO_{3} > HC_{2}H_{3}O_{2} > HCHO_{2}

None of the above.

H_{2}CO_{3} > HC_{2}H_{3}O_{2} > HCHO_{2} > HF

HF > HCHO_{2} > HC_{2}H_{3}O_{2} > H_{2}CO_{3}

HCHO_{2} > HF > HC_{2}H_{3}O_{2} > H_{2}CO_{3}

Correct answer:

HF > HCHO_{2} > HC_{2}H_{3}O_{2} > H_{2}CO_{3}

Explanation:

The acid dissociation constant, Ka, describes how strongly an acid tends to break apart into hydrogen ions (H+) and its conjugate base (A-). The higher the dissociation constant, the stronger the acid. HF has the largest Ka of these acids, making it the strongest, and H2CO3 has the smallest Ka, making it the weakest.

Example Question #11 : Reactions And Titrations

A solution of hydrofluoric acid has a concentration of 

The  for  is .

What is the pH of the solution?

Possible Answers:

Correct answer:

Explanation:

Since hydrofluoric acid is a weak acid, an ICE table needs to be set up in order to determine the hydronium ion concentration. Since both fluoride ion and hydronium ion concentrations will increase by , while the acid concentration will decrease by , the equilibrium expression comes out to be:

Note that the  in the denominator will have a negligible effect and can be ignored.

Since  is equal to the hydronium ion concentration, we can calculate the pH by taking the negative log of the concentration:

Example Question #12 : Acid Base Chemistry

A solution of hydrofluoric acid has a concentration of .

The  for  is .

If sodium hydroxide is slowly added to the acid solution, what will the pH be at the equivalence point? 

Assume the concentration of the acid is not changed with the addition.

Possible Answers:

Correct answer:

Explanation:

At the equivalence point, there are equimolar amounts of acid and base. This means that all weak acid has been neutralized, and only the conjugate base remains. Since the conjugate base of a weak acid will affect the pH, we need to use an ICE table in order to find the pH. First, we start by finding the base dissociation constant of the conjugate base, using the equation:

The balanced equation for the conjugate base dissociation is:

As the hydroxide ion and acid concentrations increase by , the fluoride ion concentration will decrease by . This makes the equilibrium expression:

Since this is the hydroxide concentration, we can find the pH by taking the negative log of this value, then subtracting from 14:

 

 

Example Question #12 : Acid Base Equilibrium

Based on the above information, it is expected that __________.

Possible Answers:

The concentration of the cation is 

The concentration of the hydroxide ion is 

The product of the anion and cation concentrations is 

There are an equal number of water molecules, hydronium ions, and hydroxide ions

Correct answer:

The concentration of the cation is 

Explanation:

Since the product of the cation and anion is , the only true statement is that the concentration of the cation  is the square root of this number:

Example Question #1 : Electrolytic Cells

In an electrolytic cell, oxidation takes place at the __________.

Possible Answers:

anode

Oxidation does not take place in an electrolytic cell.

cathode

salt bridge

Correct answer:

anode

Explanation:

Electrolytic cells have a negative electromotive force and require an outside energy source to power a non-spontaneous reaction. Galvanic cells, in contrast, have positive potentials and facilitate spontaneous reactions without the need of a power source.

Regardless of cell type, however, oxidation always takes place at the anode and reduction always takes place at the cathode. The flow of electrons is always from the anode to cathode.

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