We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

GRE Subject Test: Chemistry : GRE Subject Test: Chemistry

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

Create An Account

All GRE Subject Test: Chemistry Resources

1 Diagnostic Test 152 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Analyzing Reactions

Calculate the concentration of hydrogen ions in the following acetic acid solution.

pKa = 4.76

$\frac{[CH_3COOH]}{[CH_3COO^-]} = 0.5$

Possible Answers:

8.69*10^-^6 M

3.26*10^-^6M

7.89*10^-^5M

4.23*10^-^7M

Correct answer:

8.69*10^-^6 M

Explanation:

To answer this question you need to use the Henderson-Hasselbalch equation:

$pH = pKa +log\frac{[conjugate\: base]}{[acid]}$

The ratio given in the question is $\frac{[CH_3COOH]}{[CH_3COO^-]}$ , or $\frac{[acid]}{[conjugate \: base]}$ .

To use the correct ratio for the Henderson-Hasselbalch equation, we need to convert this ratio to its reciprocal:

$\frac{CH_3COO^-}{CH_3COOH} = \frac{1}{0.5} = 2$

Plugging the given values into the equation gives us:

pH = 4.76 + log(2) = 5.06

The question is asking for the concentration of hydrogen ions. To solve for this we have to use the definition of pH.

pH = -log[H^+]

Solving for the concentration of hydrogen ions gives us:

[H+] = 10^-^p^H

[H^+] = 10^-^5^.^0^6

[H+] = 8.69 * 10^-^6M

Report an Error

Example Question #1 : Henderson Hasselbalch Equation

Which of the following is true regarding the Henderson-Hasselbalch equation?

I. The pH of the solution is always greater than the pKa of the solution

II. As the ratio of conjugate base to acid increases, the pH increases

III. The hydrogen ion concentration can never equal the acid dissociation constant

Possible Answers:

I only

I and II

II only

II and III

Correct answer:

II only

Explanation:

The Henderson-Hasselbalch equation is a tool that allows us to calculate the pH of an acid solution using the pKa of the acid and the relative concentrations of the acid and its conjugate base. It is defined as:

$pH = pKa + log\frac{[A-]}{[HA]}$

By looking at the equation we can determine that if the ratio inside the logarithm is greater than 1, then the pH of the solution will be greater than the pKa; however, if the ratio is less than 1 (meaning, if the concentration of the acid is greater than the concentration of conjugate base), then the pH will be less than the pKa. Statement I is false.

Increasing the ratio of [A^-] to [HA] will increase the logarithm, and subsequently the pH of the solution. This makes sense because you will have more conjugate base than acid, thereby making the solution more alkaline and increasing the pH. Statement II is true.

pH and pKa are defined as follows:

pH = -log[H^+]

pKa = -log(-Ka)

If we have the same concentration of hydrogen ions as the acid dissociation constant ( K_a ), then the pH will equal the pKa. According to the Henderson-Hasselbalch equation, the pH equals the pKa if the concentration of the conjugate base equals the concentration of acid; therefore, it is possible for the hydrogen ion concentration to equal the acid dissociation constant. Statement III is false.

Report an Error

Example Question #1 : Henderson Hasselbalch Equation

A researcher prepares two solutions. Solution A contains an unknown acid, HA, and solution B contains an unknown acid, HB. The researcher performs several tests and collects the following data.

1. Both solutions contain weak acids

2. $\frac{[A^-]}{[HA]} = 1000$

3. pH_A=5.60

4. $\frac{[B^-]}{[HB]} = 100$

5. pH_B=4.60

What can you conclude about these two solutions?

Possible Answers:

Acid HB is more acidic than acid HA

The acids, HA and HB, are identical

The acid dissociation constant for HA is greater than that of HB

The hydrogen ion concentration of solution A is greater than that of solution B

Correct answer:

The acids, HA and HB, are identical

Explanation:

The Henderson-Hasselbalch equation states that:

$pH = pKa + log\frac{[A^-]}{[HA]}$

The question gives us information regarding the ratio of conjugate base to acid AND the pH for each acidic solution. Using this information, we can solve for the pKa values of both solutions.

$pKa = pH - log\frac{[A^-]}{[HA]}$

$\text{pKa of solution A }= 5.60 - log (1000) = 2.6$

$\text{pKa of solution B }= 4.60 - log (100) = 2.6$

The pKa values of both solutions are the same. This means that both solution contains the same acid; therefore, the identity of HA is the same as the identity of HB.

The hydrogen ion concentration of solution A is lower than that of solution B because the pH of solution A is greater. Acidity, or strength, of an acid is determined by the pKa. Since we have the same pKa for both acids, HA and HB will have the same acidity. Acid dissociation constant, Ka, is defined as:

Ka = 10^-^p^K^a

Acid dissociation constant only depends on the pKa; therefore, the Ka for both acids is the same.

Report an Error

Example Question #1 : Acid Base Analysis

47.0g of nitrous acid, HNO₂, is added to 4L of water. What is the resulting pH? $\dpi{100} \small \left (K_{a}=4.1\times 10^{-4} \right )$

Possible Answers:

3.0

3.2

2.5

3.5

2.0

Correct answer:

2.0

Explanation:

HNO₂ is a weak acid; it will not fully dissociate, so we need to use the HA → H⁺+ A^– reaction, with $\dpi{100} \small K_{a}=\frac{\left [ products \right ]}{\left [ reactants \right ]}=\left \frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=4.1\times 10^{-4}$ .

47.0g HNO₂ is equal to 1mol. 1mol into 4L gives a concentration of 0.25M when the acid is first dissolved; however, we want the pH at equilibrium, not at the initial state. As the acid dissolves, we know [HNO₂] will decrease to become ions, but we don't know by how much so we indicate the decrease as "x". As HNO₂ dissolves by a factor of x, the ion concentrations will increase by x.

HNO₂ → H⁺+ NO₂^–

Initial 0.25M 0 0

Equilibrium 0.25 – x x x

Now, we can fill in our equation: $\dpi{100} \small K_{a}=\left \frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=\frac{\left ( x \right )\left ( x \right )}{0.25-x}$ .

Since x is very small, we can ignore it in the denominator: $\dpi{100} \small K_{a}=\left \frac{\left ( x \right )\left ( x \right )}{\left (0.25 \right )}=4.1\times 10^{-4}$

(they expect you to do this on the MCAT; you will never have to solve with x in the denominator on the exam!)

Solve for x, and you find $x=1\times10^{-2}$ . Looking at our table, we know that $\dpi{100} \small x=\left [ H^{+} \right ]$

Now we can solve for pH: $pH=-log[H^+]=-log(1\times10^{-2})=2.0$

Report an Error

Example Question #1 : Acid Base Analysis

What is the [H^+] of a 0.12M solution of HCN that has a pH of 5.1?

Possible Answers:

$7.94*10^{-6}M$

$1.30*10^{-6} M$

1.36 M

$1.01*10^{-5} M$

4.22M

Correct answer:

$7.94*10^{-6}M$

Explanation:

We can calculate the hydrogen ion concentration by using the following equation:

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

Plug in the given value for pH and solve:

$[H^{+}]=10^{-5.1}=7.94*10^{-6} M$

Report an Error

Example Question #291 : Gre Subject Test: Chemistry

Based on the equilibrium above, what does $NH_{4}^{+}$ act as?

Possible Answers:

radical

a base

an acid

an anion

a catalyst

Correct answer:

an acid

Explanation:

An acid is a substance that can donates a proton. The conjugate acid of a base is formed when the base accepts a proton. In this case, $NH_{4}^{+}$ is the conjugate acid to the base $NH_{3}$ . This is because $NH_{3}$ accepts a hydrogen ion from the water molecule to form $NH_{4}^{+}$ , the conjugate acid.

Report an Error

Example Question #292 : Gre Subject Test: Chemistry

Given the following equilibria, what is the hydroxide ion concentration of a solution containing $0.06\textup{ M}$ $NH_{3}$ ?

$NH_{3}+H_{2}O\rightleftharpoons\ NH_{4}^{+}+OH^{-}$

The $K_{b}$ for $NH_{3}$ is $1.75\times10^{-5}$

Possible Answers:

$2.13\textup{ M}$

$0.0010 \textup{ M}$

$0.06\textup{ M}$

$0.106\textup{ M}$

$0.344\textup{ M}$

Correct answer:

$0.0010 \textup{ M}$

Explanation:

Below is the acid-base equilibria of $NH_{3}$ in an aqueous solution:

$NH_{3}+H_{2}O\rightleftharpoons\ NH_{4}^{+}+OH^{-}$

A solution of ammonia ( $NH_{3}$ ) is basic based on the chemical equation given. The $K_{b}$ for this reaction is:

$Kb= \frac{[NH_{}4]^{+}][OH^{-}]}{[NH_{3}]}=1.75\times10^{-5}$

Based on the chemical equation, $[OH^{-}]=[NH_{4}^{+}]$ .

We can make the concentration of these species equal to :

$[OH^{-}]=[NH_{4}^{+}]=X$

The $NH_{3}$ concentration is equal to:

$NH_{3}=0.06-[OH^{-}]$

$NH_{3}$ has a low $K_{b}$ so we can assume the following:

$NH_{3}=0.06-[OH^{-}]\approx0.06$

Plugging the values into the base dissociation constant equation gives:

$Kb= \frac{[NH_{}4]^{+}][OH^{-}]}{[NH_{3}]}=\frac{X^{2}}{0.06}=1.75\times10^{-5}$

Solve for :

$X^{2}= (0.06)\times(1.75\times10^{-5})=1.05\times10^{-6}$

$X=\sqrt{1.05\times 10^{-6}}=0.0010\ M$

Report an Error

Example Question #32 : Analytical Chemistry

Given the following equilibria, what is the hydrogen ion concentration of a solution containing $0.01\textup{ M}$ $C_{6}H_{5}COOH$ (benzoic acid)?

$C_{6}H_{5}COOH(aq)\rightleftharpoons\ C_{6}H_{5}COO^{-}(aq)+H^{+}(aq)$

The $K_{a}$ for $NH_{3}$ is $6.3\times10^{-5}$ .

Possible Answers:

$6.9\times10^{-9}\textup{ M}$

$0.01\textup{ M}$

$7.9\times10^{-4}\textup{ M}$

$1.5\times10^{-7}\textup{ M}$

$3.6\times10^{-4}\textup{ M}$

Correct answer:

$7.9\times10^{-4}\textup{ M}$

Explanation:

Below is the acid-base equilibria of $C_{6}H_{5}COOH$ in an aqueous solution:

$C_{6}H_{5}COOH(aq)\rightleftharpoons\ C_{6}H_{5}COO^{-}(aq)+H^{+}(aq)$

A solution of benzoic acid ( $C_{6}H_{5}COOH$ ) is acidic based on the chemical equation given. The $K_{a}$ for this reaction is:

$K_{a}= \frac{[C_{6}H_{5}COO^{-}][H^{+}]}{[C_{6}H_{5}COOH]}=6.3\times10^{-5}$

Based on the chemical equation, $[H^{+}]=[C_{6}H_{5}COO^{-}]$ .

We can make the concentration of these species equal to :

$[H^{+}]=[C_{6}H_{5}COO^{-}]=X$

The $C_{6}H_{5}COOH$ concentration is equal to:

$[C_{6}H_{5}COOH]=0.01-[H^{+}]$

$C_{6}H_{5}COOH$ has a low $K_{a}$ so we can assume the following:

$C_{6}H_{5}COOH=0.01-[H^{+}]\approx0.01$

Plugging the values into the base dissociation constant equation gives:

$Kb= \frac{[C_{6}H_{5}COO^{-}][H^{+}]}{[C_{6}H_{5}COOH]}=\frac{X^{2}}{0.01}=6.3\times10^{-5}$

Solve for :

$X^{2}= (0.01)\times(6.3\times10^{-5})=6.3\times10^{-7}$

$X=\sqrt{6.3\times10^{-7}}=7.9\times10^{-4}\textup{ M}$

Report an Error

Example Question #292 : Gre Subject Test: Chemistry

Given the following equilibria, what is the hydronium ion concentration of a solution containing $0.025\textup{ M}$ HOCl ?

$HOCl(aq)+H_{2}O\rightleftharpoons\ OCl^{-}(aq)+H_{3}O^{+}(aq)$

The $K_{a}$ for HOCl is $3.0\times10^{-8}$ .

Possible Answers:

$7.9\times10^{-4}\textup{ M}$

$2.7\times10^{-5}\textup{ M}$

$5.5\times10^{-2}\textup{ M}$

$8.4\times10^{-9}\textup{ M}$

$0.332\textup{ M}$

Correct answer:

$2.7\times10^{-5}\textup{ M}$

Explanation:

Below is the acid-base equilibria of HOCl in an aqueous solution:

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

A solution of HOCl is acidic based on the chemical equation given. The Ka for this reaction is:

$K_{a}= \frac{[OCl^{-}][H_{3}O^{+}]}{[HOCl]}=3.0\times10^{-8}$

Based on the chemical equation, $[H_{3}O^{+}]=[OCl^{-}]$ .

We can make the concentration of these species equal to :

$[H_{3}O^{+}]=[OCl^{-}]=X$

The HOCl concentration is equal to:

$[HOCl]=0.025-[H^{+}]$

HOCl has a low $K_{a}$ so we can assume the following:

$HOCl=0.025-[H_{3}O^{+}]\approx0.025$

Plugging the values into the base dissociation constant equation gives:

$K_{a}= \frac{[OCl^{-}][H_{3}O^{+}]}{[HOCl]}=\frac{X^{2}}{0.01}=3.0\times10^{-8}$

Solve for :

$X^{2}= (0.025)\times(3.0\times10^{-8})=7.5\times10^{-10}$

$X=\sqrt{7.5\times10^{-10}}=2.7\times10^{-5}\textup{ M}$

Report an Error

Example Question #291 : Gre Subject Test: Chemistry

Which molecule would be considered a Lewis acid?

Possible Answers:

$Zn^{2+}$

$O_{2}$

Cu(s)

$NH_{3}$

$OH^{-}$

Correct answer:

$Zn^{2+}$

Explanation:

A lewis acid is an electron pair acceptor. $Zn^{2+}$ would be considered a lewis acid based on the definition. Because $Zn^{2+}$ is electron-deficient based on its oxidation number, it is able to accept an electron pair.

Report an Error

View Tutors

Nate
Certified Tutor

University of Wisconsin-Oshkosh, non degree, Prepharmacy. University of Wisconsin-Madison, Pharmaceutical Chemist, Pharmacy.

View Tutors

Ekta
Certified Tutor

Punjab Agricultural University, India, Doctorate (PhD), Chemistry.

View Tutors

Kimberly
Certified Tutor

University of Connecticut, Bachelor of Chemistry, Chemistry. University of Connecticut, Doctor of Philosophy, Organic Chemist...

All GRE Subject Test: Chemistry Resources

1 Diagnostic Test 152 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

French Tutors in New York City, Math Tutors in San Diego, ISEE Tutors in Miami, English Tutors in Houston, ACT Tutors in Boston, Algebra Tutors in Philadelphia, Reading Tutors in San Francisco-Bay Area, French Tutors in Boston, French Tutors in Phoenix, Spanish Tutors in Phoenix

Popular Courses & Classes

ISEE Courses & Classes in Boston, GRE Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in New York City, ACT Courses & Classes in Phoenix, GMAT Courses & Classes in New York City, MCAT Courses & Classes in Atlanta, ACT Courses & Classes in Chicago, SSAT Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in Philadelphia, LSAT Courses & Classes in Seattle

Popular Test Prep

ACT Test Prep in Miami, GMAT Test Prep in New York City, SSAT Test Prep in Boston, GMAT Test Prep in Chicago, GMAT Test Prep in Atlanta, MCAT Test Prep in Phoenix, SSAT Test Prep in Washington DC, GMAT Test Prep in Houston, SSAT Test Prep in Houston, GMAT Test Prep in Phoenix

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

GRE Subject Test: Chemistry : GRE Subject Test: Chemistry

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

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #1 : Analyzing Reactions

Example Question #1 : Henderson Hasselbalch Equation

Example Question #1 : Henderson Hasselbalch Equation

Example Question #1 : Acid Base Analysis

Example Question #1 : Acid Base Analysis

Example Question #291 : Gre Subject Test: Chemistry

Example Question #292 : Gre Subject Test: Chemistry

Example Question #32 : Analytical Chemistry

Example Question #292 : Gre Subject Test: Chemistry

Example Question #291 : Gre Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: