Call Now to Set Up Tutoring:

(888) 888-0446

MCAT Physical : Reactions and Titrations

Study concepts, example questions & explanations for MCAT Physical

Create An Account

All MCAT Physical Resources

8 Diagnostic Tests 303 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 Next →

Example Question #1 : Acid Base Chemistry

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.

$HA \leftrightharpoons H^{+} + A^{-}$

All of the bases proceed in a similar fashion.

$BOH \leftrightharpoons B^{+} + OH^{-}$

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.

Which of the following expressions most closely approximates the equilibrium constant of pure water?

Possible Answers:

$K_{w} = [}H^{+}}][H_{2}O]$

$K_{w} = [H_{2}O][OH^{-}]$

$K_w=\frac{[H^+]}{[OH^-]}$

$K_{w} = [}H^{+}}][OH^{-}]$

$K_{w} = [}H^{+}}]$

Correct answer:

$K_{w} = [}H^{+}}][OH^{-}]$

Explanation:

The autionization of water proceeds as follows:

$H_{2}O \leftrightharpoons H^{+} + OH^{-}$

The equilibrium constant is calculated by the following formula.

$K_{eq}=\frac{[product]}{[reactant]}$

The equilibrium expression is $K_{w} = [}H^{+}}][OH^{-}]$ , with the pure water reactant excluded. Pure solids and liquids are not included in the equilibrium calculation.

Report an Error

Example Question #2 : Acid Base Chemistry

What is the hydroxide concentration in an aqueous solution with a pH of 2?

Possible Answers:

There are no hydroxide ions in the solution

$1*10^{-12}M$

$1*10^{-2}M$

$1*10^{-4}M$

Correct answer:

$1*10^{-12}M$

Explanation:

The hydroxide concentration can be determined by considering the autoionization of water in a solution. At 25^oC , water has the equilibrium constant of $K_{w} = 1*10^{-14}$ . This value is based on the autoionization of water.

$H_2O\rightleftharpoons H^++OH^-$

$K_w=[H^+][OH^-]=1*10^{-14}$

Since the reaction is in terms of proton and hydroxide ion concentrations, we can set the expression equal to this value in order to determine the concentration of hydroxide ions at the given pH. We start by determining the concentration of protons in the solution by using the following equation:

pH=-log[H^+]

$[H^{+}] = 10^{-pH}$

$[H^{+}] = 10^{-2} = 1*10^{-2}M$

Now, we can solve for the hydroxide concentration.

$K_w=[H^+][OH^-]=1*10^{-14}$

$[OH^{-}][1*10^{-2}] = 1*10^{-14}$

$[OH^{-}] = 1*10^{-12}M$

This problem can also be solved by calculating the pOH of the solution and using this value to find the hydroxide ion concentration.

Report an Error

Example Question #1 : Reactions And Titrations

What is the pH of a solution which has a hydroxide ion concentration of 5 * 10^-4M?

Possible Answers:

$\small 5*10^{10}$

$\small 10.70$

$\small 3.30$

$\small 13.30$

$\small 2*10^{-11}$

Correct answer:

$\small 10.70$

Explanation:

First convert concentration of OH^- into concentration of H⁺. Remember that K_w is 1*10^-14.

$\small [H^{+}]=\frac{1*10^{-14}}{[OH^{-}]} = \frac{1*10^{-14}}{5*10^{-4}}=2*10^{-11}$

Then, convert concentration of H⁺ into pH.

$\small pH = -log[H^{+}]=-log(2*10^{-11})\approx10.70$

Alternatively, you can use the hydroxide concentration to solve for pOH and convert to pH.

Report an Error

Example Question #1 : Acid Base Equilibrium

Diffusion can be defined as the net transfer of molecules down a gradient of differing concentrations. This is a passive and spontaneous process and relies on the random movement of molecules and Brownian motion. Diffusion is an important biological process, especially in the respiratory system where oxygen diffuses from alveoli, the basic unit of lung mechanics, to red blood cells in the capillaries.

Capture

Figure 1 depicts this process, showing an alveoli separated from neighboring cells by a capillary with red blood cells. The partial pressures of oxygen and carbon dioxide are given. One such equation used in determining gas exchange is Fick's law, given by:

ΔV = (Area/Thickness) · D_gas · (P_{1 –}P₂)

Where ΔV is flow rate and area and thickness refer to the permeable membrane through which the gas passes, in this case, the wall of the avlveoli. P₁and P₂refer to the partial pressures upstream and downstream, respectively. Further, D_gas, the diffusion constant of the gas, is defined as:

D_gas= Solubility / (Molecular Weight)^(1/2)

Which following pair gives the correct equation, and change in pH, when carbon dioxide diffuses into blood?

Possible Answers:

CO₂ + H₂O HCO₃^–| pH decreases

CO₂ + H₂O H₂CO₃| pH increases

O₂ + CO₂+ H₂O H₂O₂ + CO₃^2–| pH decreases

CO₂ + H₂O H₂CO₃| pH decreases

Correct answer:

CO₂ + H₂O H₂CO₃| pH decreases

Explanation:

This is essentially a two-part question. The first asks you to generate the production of carbonic acid; and the second, requires you to infer that an acid will lower blood pH. Only one answer choice is both balanced correctly and gives an accurate change in pH. Although this knowledge is not required for the MCAT, your body constantly monitors blood pH to determine if concentrations of CO₂are too high, or alternatively, if O₂concentrations are too low.

Report an Error

Example Question #1 : Acid Base Equilibrium

As the value of K_a increases, __________

Possible Answers:

the strength of the acid increases

all of these are true

pKa decreases

the strength of the conjugate base decreases

the reaction HA → H⁺+ A^– favors the products

Correct answer:

all of these are true

Explanation:

K_a 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 K_a increases the strength of the acid increases, and the strength of the conjugate base decreases. pK_a is defined as the –log(K_a), so as K_a increases pK_a decreases. All of these answer choices are correct.

Report an Error

Example Question #1 : Acid Base Chemistry

What is the pK_a of acetic acid? (K_a= 1.8 * 10^–5)

Possible Answers:

5.3

2.1

4.7

4.2

8.6

Correct answer:

4.7

Explanation:

We know that pK_a is equal to –log(K_a). Thus, pK_a 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 pK_a 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.

Report an Error

Example Question #7 : Acid Base Chemistry

HCN dissociates based on the following reaction.

$HCN + H_{2}O \rightarrow CN^{-}+ H_{3}O^{+}$

The K_a for hydrogen cyanide is $\small 6.2*10^{-10}$ .

What is the K_bfor CN^-?

Possible Answers:

$\small 6.2*10^{4}$

$\small 1.3*10^{-4}$

$\small 1.6*10^{-5}$

$\small 6.2*10^{-24}$

Correct answer:

$\small 1.6*10^{-5}$

Explanation:

Remember that the K_a for the acid and the K_b for the conjugate base, when multiplied will equal the autoionization of water constant (K_w).

$K_{a} K_{b} = K_{w}=\small 1*10^{-14}$

$\small K_{b} = \frac{K_w}{K_a}=\frac{1*10^{-14}}{6.2*10^{-10}}$

$\small K_{b}= 1.6*10^{-5}$

Report an Error

Example Question #1 : Reactions And Titrations

Compound	Base strength, K_b
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 4

Compound 2

Compound 3

Compound 1

Correct answer:

Compound 4

Explanation:

The strongest acid of the group will also have the smallest K_b value. As the weakest base (smallest K_b), 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 K_b value.

Report an Error

Example Question #9 : Acid Base Chemistry

$HA \leftrightharpoons H^{+} + A^{-}$

All of the bases proceed in a similar fashion.

$BOH \leftrightharpoons B^{+} + OH^{-}$

In aqueous conditions the equilibrium constant for a Brønsted-Lowry base, $A^{-}$ , can be expressed as which of the following?

Possible Answers:

$K_b=\frac{[A^-][OH^-]}{[HA]}$

$K_b=\frac{[OH^-]}{[A^-]}$

$K_b=\frac{[HA][OH^-]}{[A^-]}$

$K_b=\frac{[HA][OH^-]}{[H_2O]}$

$K_b=\frac{[HA]}{[A^-]}$

Correct answer:

$K_b=\frac{[HA][OH^-]}{[A^-]}$

Explanation:

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

$A^{-} +H_{2}O \leftrightharpoons HA + OH^{-}$

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

$K_{eq}=\frac{[product]}{[reactant]}$

$K_b=\frac{[HA][OH^-]}{[A^-]}$

Report an Error

Example Question #1 : Acid Base Equilibrium

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

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

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

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

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

Possible Answers:

None of the above.

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

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

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

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

Correct answer:

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

Explanation:

The acid dissociation constant, K_a, 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 K_a of these acids, making it the strongest, and H₂CO₃ has the smallest K_a, making it the weakest.

Report an Error

← Previous 1 2 3 4 Next →

View MCAT Chemical and Physical Foundations of Biological Systems Tutors

KAITLYN
Certified Tutor

Fairfield University, Bachelor of Science, Biology, General. NUI Galway Ireland, Master of Science, Neuroscience.

View MCAT Chemical and Physical Foundations of Biological Systems Tutors

Yongmao
Certified Tutor

Nankai University, Bachelor of Science, Chemistry. Washington University in St Louis, Doctor of Philosophy, Organic Chemistry.

View MCAT Chemical and Physical Foundations of Biological Systems Tutors

Erik Weijie
Certified Tutor

University of Florida, Bachelor of Science, Chemical Engineering.

All MCAT Physical Resources

8 Diagnostic Tests 303 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

English Tutors in Philadelphia, Computer Science Tutors in Dallas Fort Worth, GMAT Tutors in Washington DC, GRE Tutors in Boston, SSAT Tutors in San Diego, Calculus Tutors in Washington DC, MCAT Tutors in Dallas Fort Worth, Math Tutors in Los Angeles, Statistics Tutors in Atlanta, Biology Tutors in New York City

Popular Courses & Classes

SAT Courses & Classes in Dallas Fort Worth, SAT Courses & Classes in Seattle, GMAT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Chicago, SAT Courses & Classes in Philadelphia, GRE Courses & Classes in Los Angeles, SAT Courses & Classes in Los Angeles, Spanish Courses & Classes in Houston, LSAT Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in San Diego

Popular Test Prep

SSAT Test Prep in Dallas Fort Worth, GRE Test Prep in Phoenix, GMAT Test Prep in New York City, SSAT Test Prep in Denver, GMAT Test Prep in Philadelphia, ISEE Test Prep in Seattle, ISEE Test Prep in New York City, LSAT Test Prep in Los Angeles, SSAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Miami

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

MCAT Physical : Reactions and Titrations

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #1 : Acid Base Chemistry

Example Question #2 : Acid Base Chemistry

Example Question #1 : Reactions And Titrations

Example Question #1 : Acid Base Equilibrium

Example Question #1 : Acid Base Equilibrium

Example Question #1 : Acid Base Chemistry

Example Question #7 : Acid Base Chemistry

Example Question #1 : Reactions And Titrations

Example Question #9 : Acid Base Chemistry

Example Question #1 : Acid Base Equilibrium

All MCAT Physical Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: