We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

AP Chemistry : AP Chemistry

Study concepts, example questions & explanations for AP Chemistry

Create An Account

All AP Chemistry Resources

6 Diagnostic Tests 225 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Solubility

Pb(OH)₂ has a K_sp of 1.43 * 10^-20.

What is its solubility?

Possible Answers:

1.53 * 10^-7M

1.20 * 10^-10M

8.42 * 10^-12M

2.43 * 10^-7M

Correct answer:

1.53 * 10^-7M

Explanation:

Lead (II) hydroxide dissociates according to this reaction: $Pb(OH)_2\rightleftharpoons Pb^{2+}+2OH^-$ . Solubility is equal to the number of moles that will dissociate at equilibrium, and can be found using this reaction and the value of K_sp.

K_sp = [Pb²⁺][OH^-]²

If [Pb²⁺] = x, then [OH^-] = 2x. They exist in a 1:2 ratio.

K_sp = 1.43 * 10^-20 = x(2x)² = 4x³

x = 1.53 x 10^-7M

Report an Error

Example Question #3 : Solubility

Which of the following factors will generally lead a solute to dissolve more easily into a solution (i.e. increase solubility)?

Possible Answers:

Decreasing the temperature of the solution

The common ion effect

Increasing the temperature of the solution

None of these choices will increase solubility

Correct answer:

Increasing the temperature of the solution

Explanation:

Increasing the temperature of a solution will generally increase its solubility, while decreasing temperature will have the opposite effect. The common ion effect refers to the phenomenon when two compunds in a solution dissociate to produce at least one similar ion, and will make it harder for additional amounts of that ion in the solution to dissociate, decreasing solubility.

Report an Error

Example Question #4 : Solubility

$NaCl (s) \rightarrow Na^{+} (aq) + Cl^{-} (aq)$

If the K_sp of the above dissociation is found to be 2.70 * 10^-4, what is the solubility of NaCl?

Possible Answers:

1.44 * 10^-2M

1.89 * 10^-2M

1.64 * 10^-2M

7.29 * 10^-8M

Correct answer:

1.64 * 10^-2M

Explanation:

First, set up a balanced reaction and enter X for the amount of each solute that is going to be dissolved. As NaCl dissociates, each ion amount will increase by some amount, X.

$NaCl (s) \rightarrow Na^{+}(aq) + Cl^{-}(aq)$

$\wr$

Then set up the equation for the value of K_sp and enter in the value of X for each solute. Use this equation to solve for X, the solubility (in mol/L). Remember that NaCl si not included in the equation, as it is a pure solid.

$K_{sp}= [Na^{+}][Cl^{-}] = (X)(X) = X^{2}$

$2.70 * 10^{-4} = X^{2}$

$X= 1.64 * 10^{-2}M$

Report an Error

Example Question #5 : Solubility

$CaF_{2}(s) \rightarrow Ca^{2+}(aq) + 2F^{-}(aq)$

If the solubility of CaF₂ in the dissociation reaction above was determined to be 1.54 * 10^-3M, what is the K_sp value for CaF₂?

Possible Answers:

2.24 * 10^-6

1.98 * 10^-8

1.42 * 10^-6

1.46 * 10^-8

Correct answer:

1.46 * 10^-8

Explanation:

Use the balanced reaction of the dissociation to assign values of X for the solubility of each ion. As CaF₂ dissociates, each ion amount will increase by some amount; calcium will increase by X and flouride will increase by 2X.

$CaF_{2}(s) \rightarrow Ca^{2+}(aq) + 2F^{-}(aq)$

$\wr$ +2X

$\wr$

Then plug in the equation to solve K_sp, using the X values assigned for each solute. Use the given value for solubility to find K_sp. Remember that NaCl si not included in the equation, as it is a pure solid.

$K_{sp}= [Ca^{2+}][F^-]^2=(X)(2X)^{2} = 4X^{3}$

$K_{sp}= 4(1.9 * 10^{-2})^{3} = 1.46 * 10^{-8}$

Report an Error

Example Question #73 : Solutions

Consider the following balanced equation for the solubility of barium hydroxide in an aqueous solution.

$\small Ba(OH)_{2}_{(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2OH^{-}_{(aq)}$

$\small K_{sp} = 5*10^{-3}$

What is the solubility of barium hydroxide?

Possible Answers:

$\small 0.108M$

$\small 0.035M$

$\small 0.05M$

$\small 0.071M$

Correct answer:

$\small 0.108M$

Explanation:

In order to solve for the solubility of barium hydroxide, we need to establish an ICE table for the reaction.

I. Initially, there are no barium ions or hydroxide ions in the solution, so we can call their concentrations zero at the beginning of the reaction.

C. For every molecule of dissolved barium hydroxide, there will be one ion of barium and two ions of hydroxide. As a result, the increases in each ion can be designated $\small +x$ and $\small +2x$ , respectively.

E. Finally, we plug these values into the equilibrium expression and set it equal to the solubility product constant.

$\begin{matrix} Ba(OH)_2 & Ba^{2+} &OH^- \\ /& 0 & 0\\ / & x & 2x \end{matrix}$

$K_{sp}=[Ba^{2+}][OH^-]^2$

$\small 5*10^{-3} = [x][2x]^{2}$

$\small x = 0.108M$

Report an Error

Example Question #7 : Solubility

Four different insoluble salts are added to an aqueous solution in equal amounts. Assume that the salts do not interact with one another in any way. The following list shows the salts and their solubility product constants.

$\small BaSO_{4}$ $\small K_{sp} = 1.1*10^{-10}$

$\small CaCrO_{4}$ $\small K_{sp} = \small 7.1*10^{-4}$

$\small MgF_{2}$ $\small K_{sp } = 3.7*10^{-8}$

$\small NiCO_{3}$ $\small K_{sp} = 6.6*10^{-9}$

Which of the following ions will have the highest concentration in the solution?

Possible Answers:

$\small SO_{4}^{2-}$

$\small CO_{3}^{2-}$

$\small \small CrO_{4}^{2-}$

$\small F-$

Correct answer:

$\small \small CrO_{4}^{2-}$

Explanation:

The solubility product constants help us envision which ions will have the largest concentrations after dissolving to equilibrium. $\small CaCrO_{4}$ has the largest solubility product constant, meaning it will dissolve the most out of the four options.

Upon the dissolving of $\small MgF_{2}$ , twice the number of fluoride ions will be released as magnesium ions; however, the solubility constant for magnesium fluoride is so much smaller, compared to the solubility constant of $\small CaCrO_{4}$ , that it will not become the ion with the largest concentration.

$2(3.7*10^{-8})<7.1*10^{-4}$

$[F^-]<[CrO_4^{2-}]$

Report an Error

Example Question #2 : Solubility

Consider the following balanced equation for the dissociation of barium hydroxide in an aqueous solution.

$\small Ba(OH)_{2}_{(s)} \rightleftharpoons Ba^{2+}_{(aq)} + 2OH^{-}_{(aq)}$

$\small K_{sp} = 5*10^{-3}$

A 1M solution of barium nitrate is added to the solution. What is the solubility of barium hydroxide after this addition?

Possible Answers:

$\small 0.003M$

$\small 0.050M$

$\small 0.035M$

$\small 0.108M$

0.118M

Correct answer:

$\small 0.035M$

Explanation:

Before we solve for the solubility, remember Le Chatelier's principle. Barium nitrate is a soluble salt that will completely dissolve in solution. This means that there will be a one molar concentration of barium ions before the salt dissolves. Since we have added to the products side of the reaction, we can predict that the equilibrium will be shifted to the left, and the barium hydroxide will be less soluble. This change in equilibrium is referred to as the common ion effect.

We will use an ICE table to determine the solubility of the salt.

I. Before the salt begins to dissolve, there is a one molar concentration of barium ions (from the barium nitrate) with no hydroxide ions in the solution.

C. Since every dissolved molecule of barium hydroxide will yield one barium ion and two hydroxide ions, the unknown increases in each ion can be designated $\small 1+x$ and $\small 2x$ respectively. Notice how the barium ion increase will be added to the previously added one molar concentration in the solution.

E. Now, we set the solubility expression equal to the solubility product constant.

$\begin{matrix} Ba(OH)_2 & Ba^{2+} & OH^-\\ / & 1M & 0M\\ /& 1+x & x \end{matrix}$

$K_{sp}=[Ba^{2+}][OH^-]^2$

$\small 5*10^{-3} = [1+x][2x]^{2}$

Since 1 is such a larger number than the value we will get for $\small x$ , we can consider the value in the barium ion expression to be negligible. This allows us to simplify the equation to the following setup.

$\small 5*10^{-3} = [1][2x]^{2}$

$\small x = 0.035M$

The solubility of barium hydroxide without the common ion effect is 0.108M. Notice how the solubility for the salt has decreased, as was predicted in the beginning by Le Chatelier's principle.

Report an Error

Example Question #9 : Solubility

A chemist has a fully saturated solution of chemical X, with solid X resting on the bottom. If she adds a chemical that will react with the aqueous X, what will happen to the solid?

Possible Answers:

A layer of sediment will cover the solid

More solid will precipitate out of solution

It will start to dissolve into the solution

The solid will change color

Nothing will change

Correct answer:

It will start to dissolve into the solution

Explanation:

A fully saturated solution is at equilibrium between the solid and aqueous states, but it is a dynamic equilibrium. As the reacting chemical reacts with aqueous X, more and more of the solid will dissolve into solution.

Report an Error

Example Question #3 : Solubility

The partial pressure of an arbitrary gas, $X_{2\hspace{1 mm}(g)}$ , in an enclosed chanber with a pool of water is 7.2atm at 298K.

The Henry's Law constant for $X_{2\hspace{1 mm}(g)}$ is $5.6*10^{-2}\frac{mol}{L\cdot atm}$ .

What is the solubility of $X_{2\hspace{1 mm}(g)}$ ?

Possible Answers:

$4.0\frac{mol}{L}$

$4.0*10^{-1}\frac{mol}{L\cdot atm}$

$4.0*10^{-1}\frac{mol}{L}$

$129\frac{mol}{L}$

None of the available answers

Correct answer:

$4.0*10^{-1}\frac{mol}{L}$

Explanation:

Henry's law can be expressed:

$S_{gas}=k_H\times P_{gas}$

Where $k_H$ is the Henry's Law constant, and $P_{gas}$ is the partial pressure of the gas above the solution. We can substitute our givens into this equation:

$S_{gas}=\frac{5.6\times 10^{-2}\hspace{1 mm}mol}{L\cdot atm}\times 7.2\hspace{1 mm}atm=4.0\times 10^{-1}\hspace{1 mm}\frac{mol}{L}$

Report an Error

Example Question #161 : Solutions And States Of Matter

What is the formula for the dissociation of K_2CO_3 ?

Possible Answers:

$K_2CO_3\rightarrow K^+ + CO_3^-$

$2K_2CO_3\rightarrow 2K^+ + C_2 + 3O_2^{2-}$

None of the available answers

$K_2CO_3 \rightarrow K^{2+}+CO_3^{2-}$

$K_2CO_3\rightarrow 2K^+ + CO_3^{2-}$

Correct answer:

$K_2CO_3\rightarrow 2K^+ + CO_3^{2-}$

Explanation:

K_2CO_3 is composed of two types of ions, K^+ and $CO_3^{2-}$ . When it dissociates, it forms:

$K_2CO_3 \rightarrow 2K^+ +\hspace{1 mm}CO_3^{2-}$

Report an Error

View AP Chemistry Tutors

James
Certified Tutor

University of Georgia, Bachelor in Arts, Political Science and Government.

View AP Chemistry Tutors

Leonard
Certified Tutor

Florida State University, Bachelor of Science, Environmental Science.

View AP Chemistry Tutors

Alexandra
Certified Tutor

Rice University, Bachelor in Arts, Biochemistry and Molecular Biology. University of North Texas Health Science Center, Maste...

All AP Chemistry Resources

6 Diagnostic Tests 225 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

ISEE Tutors in Boston, GRE Tutors in Boston, SSAT Tutors in San Francisco-Bay Area, GRE Tutors in Seattle, ACT Tutors in Philadelphia, Statistics Tutors in Dallas Fort Worth, Reading Tutors in Seattle, LSAT Tutors in Los Angeles, Algebra Tutors in New York City, MCAT Tutors in Philadelphia

Popular Courses & Classes

ACT Courses & Classes in Phoenix, SAT Courses & Classes in Chicago, ISEE Courses & Classes in New York City, Spanish Courses & Classes in Denver, GRE Courses & Classes in Atlanta, SSAT Courses & Classes in Philadelphia, Spanish Courses & Classes in Atlanta, GRE Courses & Classes in Chicago, LSAT Courses & Classes in Seattle, SSAT Courses & Classes in Miami

Popular Test Prep

MCAT Test Prep in Houston, ACT Test Prep in Philadelphia, GMAT Test Prep in Dallas Fort Worth, GRE Test Prep in San Francisco-Bay Area, SAT Test Prep in Chicago, SAT Test Prep in New York City, LSAT Test Prep in Chicago, MCAT Test Prep in Denver, GMAT Test Prep in Houston, MCAT Test Prep in Philadelphia

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

AP Chemistry : AP Chemistry

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #2 : Solubility

Example Question #3 : Solubility

Example Question #4 : Solubility

Example Question #5 : Solubility

Example Question #73 : Solutions

Example Question #7 : Solubility

Example Question #2 : Solubility

Example Question #9 : Solubility

Example Question #3 : Solubility

Example Question #161 : Solutions And States Of Matter

All AP Chemistry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: