We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

GRE Subject Test: Chemistry : Solubility and Precipitation

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 #2 : Solubility

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:

$\small 0.05M$

$\small 0.071M$

$\small 0.035M$

$\small 0.108M$

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 #304 : Gre Subject Test: Chemistry

Write the solubility product, $K_{sp}$ , for $BaSO_4_{(s)}$ .

Possible Answers:

$K_{sp} = [Ba][SO_{4}]^{2-}$

$K_{sp} = [Ba][SO_{4}]$

$K_{sp} = [Ba^{2+}][SO_{4^}^{}^{2-}]$

$K_{sp} = [Ba]^{4+}[SO_{4}]^{2-}$

$K_{sp} = [Ba]^{2+}[SO_{4}]^{2-}$

Correct answer:

$K_{sp} = [Ba^{2+}][SO_{4^}^{}^{2-}]$

Explanation:

The solubility product constant ( $K_{sp}$ ) is an expression that describes the extent to which a compound is soluble in an aqueous solution. It describes the equilibrium between a solid and its constituent ions in a solution. The equilibrium constant for $BaSO_4_{(s)}$ can be written as:

$K_{eq} = \frac{[Ba^{2+}][SO_{4^}^{}^{2-}]}{[BaSO_{4}(s)]}$

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

$K_{eq}*[BaSO_{4}]=[Ba^{2+}][SO_{4^}^{}^{2-}]$

The product, $K_{eq}*[BaSO_{4}]$ , can be considered a constant expression called the solubility product constant ( $K_{sp}$ ) and the equation can be written in the form:

$K_{sp} = [Ba^{2+}][SO_{4^}^{}^{2-}]$

Report an Error

Example Question #2 : Solubility And Precipitation

Write the solubility product, $K_{sp}$ , for $AgCN_{(s)}$ .

Possible Answers:

$K_{sp} = [Ag^{2+}][CN^{2-}]$

$K_{sp} = [Ag][CN^{-}]$

$K_{sp} = [Ag][CN]$

$K_{sp} = [Ag^{+}][CN^{-}]$

$K_{sp} = [Ag]^{+}[CN]^{-}$

Correct answer:

$K_{sp} = [Ag^{+}][CN^{-}]$

Explanation:

$K_{eq} = \frac{[Ag^{+}][CN^{-}]}{[AgCN(s)]}$

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

$K_{eq}*[AgCN(s)]=[Ag^{+}][CN^{-}]$

The product, $K_{eq}*[AgCN]$ , can be considered a constant expression called the solubility product constant ( $K_{sp}$ ) and the equation can be written in the form:

$K_{sp} = [Ag^{+}][CN^{-}]$

Report an Error

Example Question #1 : Analyzing Solids

Write the $K_{sp}$ expression for the following equilibria:

$Mg(OH)_{2(s)} \leftrightarrow Mg^{2+(aq)} + 2 OH^{-}_{(aq)}$

Possible Answers:

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

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

$Ksp = [Mg^{2+}][OH^{-}]^{2}$

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

None of these

Correct answer:

$Ksp = [Mg^{2+}][OH^{-}]^{2}$

Explanation:

The solubility product constant ( $K_{sp}$ ) is an expression that describes the extent to which a compound is soluble in an aqueous solution. It describes the equilibrium between a solid and its constituent ions in a solution.The equilibrium between undissolved $Mg(OH)_{2(s)}$ and its ions dissolved in solution is:

$Mg(OH)_{2(s)} \leftrightarrow Mg^{2+(aq)} + 2 OH^{-}_{(aq)}$

The equilibrium constant for $Mg(OH)_{2(s)}$ can be written as:

$K_{eq} = \frac{[Mg^{2+}][OH^{-}]^{2}}{[Mg(OH)_{2}(s)]}$

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

$K_{eq}[Mg(OH)_{2}](s)=[Mg^{2+}][OH^{-}]^{2}$

The product, $K_{eq}[Mg(OH)_{2}]$ , can be considered a constant expression called the solubility product constant ( $K_{sp}$ ) and the equation can be written in the form:

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

Report an Error

Example Question #2 : Analyzing Solids

Considering that $K_{sp} = 1.8 * 10^{-11}$ for the equilibrium reaction:

$Mg(OH)_{2(s)} \leftrightarrow Mg^{2+(aq)} + 2 OH^{-}_{(aq)}$

What would be the $Mg^{2+}$ concentration for a solution buffered at a pH of 8.5?

Possible Answers:

11 M

5 M

16 M

1.8 M

Correct answer:

1.8 M

Explanation:

The pH of the solution is 8.5 , therefore the pOH would be 5.5 considering the relationship:

14 = pH + pOH

Therefore $[OH]=3.2x10^{-6}$ based on the equation $[OH] = 10^{-pOH}$

The solubility product expression for $Mg(OH)_{2(s)}$ is : $K_{sp} = [Mg^{2+}][OH^{-}]^{2}$

By inserting the knowns in to the expression, gives:

$1.8 * 10^{-11} = [Mg^{2+}][3.2 * 10^{-6}]^{2}$

Rearrange this equation

$[Mg^{2+}] = \frac{1.8 * 10^{-11}}{(3.2 * 10^{-6})^{2}} = 1.8M$

Report an Error

Example Question #2 : Analyzing Solids

The K_sp of BaSO_4 is $5.0*10^{-9}$ . What is the molar solubility of BaSO_4 in water?

Possible Answers:

$3*10^{-3}$

$7.1*10^{-5}$

$8.5*10^{-6}$

$5.3*10^{-8}$

$2.3*10^{-3}$

Correct answer:

$7.1*10^{-5}$

Explanation:

The equation for the dissolution of BaSO_4 in water is below:

$BaSO_{4}(s)\rightleftharpoons Ba^{2+}(aq)\ +\ SO_{4}^{2-}(aq)$

The K_spfor the above equation is:

$Ksp=[Ba^{2+}][SO_{4}^{2-}]=5.0*10^{-9}$

Due to the solubility of BaSO_4 of in water, the concentration of $Ba^{2+}$ and $SO_4^{2-}$ ^- should be equal:

$[Ba^{2+}]\ =\ [SO_{4}^{2-}]=X$

Plug into the K_sp equation:

$Ksp=X*X=X^{2}=5.0*10^{-9}$

Therefore:

$X=\sqrt{5.0\ *10^{-9}}=7.1*10^{-5}$

The solubility of BaSO_4 in water is $7.1*10^{-5}$

Report an Error

Example Question #311 : Gre Subject Test: Chemistry

The K_sp of CuBr is $5.3*10^{-9}$ . What is the molar solubility of CuBr inwater?

Possible Answers:

$5.6*10^{-9}$

$2.2*10^{-3}$

$7.3*10^{-5}$

$7.7*10^{-3}$

$3.4*10^{-5}$

Correct answer:

$7.3*10^{-5}$

Explanation:

The equation for the dissolution of CuBr in water is below:

$CuBr(s)\rightleftharpoons Cu^{+}(aq)\ +\ Br^{-}(aq)$

The K_sp for the above equation is:

$Ksp=[Cu^{+}][Br^{-}]=5.3*10^{-9}$

Due to the solubility of CuBr of in water, the concentration of Cu^+ and Br^- should be equal:

$[Cu^{+}]\ =\ [Br^{-}]=X$

Plugging X into the K_sp equation gives:

$Ksp=X*X=X^{2}=5.3*10^{-9}$

Therefore:

$X=\sqrt{5.3\ *10^{-9}}=7.3*10^{-5}$

The solubility of CuBr in water is $7.3*10^{-5}$

Report an Error

Example Question #2 : Solubility And Precipitation

Which salt is insoluble in water?

Possible Answers:

$CuSO_{4}$

$Al(OH)_{3}$

AgCl

$NaNO_{3}$

$CuCl_{2}$

Correct answer:

AgCl

Explanation:

Solubility rules must be followed for substances in aqueous media. Below are some of the solubility rules and they must be followed in the order given. For example, rule 1 should have precedence over rule 2.

1. All alkali metal $(Li, Na, K, Rb, \text{and}\ Cs)$ and NH_4^+ compounds are soluble.

2. Nitrate (NO_3^-) , perchlorate (ClO_4^-) , chlorate, (ClO_3^3) , and acetate (CH_3COO^-) salts are soluble.

3. Silver, lead, and mercury salts are insoluble.

Thus, we see that silver chloride is insoluble due to rule 3.

Report an Error

Example Question #2 : Solubility And Precipitation

The $K_{sp}$ of CdCO_3 is $1.8*10^{-14}$ . What is the molar solubility of CdCO_3 in water?

Possible Answers:

$0.011\textup{ M}$

$3.1*10^{-2} \textup{ M}$

$4.3*10^{-5}\textup{ M}$

$1.3*10^{-7} \textup{ M}$

$3.22\textup{ M}$

Correct answer:

$1.3*10^{-7} \textup{ M}$

Explanation:

The equation for the dissolution of CdCO_3 in water is:

$CdCO_{3(s)}\rightleftharpoons Cd^{2+}_{(aq)}+CO_{3}^{2-}_{(aq)}$

The $K_{sp}$ for the above equation is:

$K_{sp}=[Cd^{2+}][CO_{3}^{2-}]=1.8 *10^{-14}$

Due to the molar ratios of the species of CdCO_3 , the concentration of $Cd^{2+}$ and $CO_3^{2-}$ should be equal when dissolved:

$[Cd^{2+}]=[CO_{3}^{2-}]=X$

Plugging into the $K_{sp}$ equation gives:

$K_{sp}=X* X=X^{2}=1.8*10^{-14}$

$X=\sqrt{1.8* 10^{-14}}=1.3*10^{-7}$

Therefore, the solubility of CdCO_3 in water is $1.3*10^{-7}\textup{ M}$

Report an Error

Example Question #3 : Solubility And Precipitation

Calculate the molar solubility of $Co(OH)_{2}$ with $Ksp = 1.3\times 10^{-15}$ if enough NaOH is added to raise the pH of the solution to pH 12.

Possible Answers:

$0.5421\textup{ M}$

$4.11\textup{ M}$

$1.3\times10^{-13}\textup{ M}$

$3.1\times10^{-3}\textup{ M}$

$2.3\times10^{-3}\textup{ M}$

Correct answer:

$1.3\times10^{-13}\textup{ M}$

Explanation:

The equation for the dissolution of $Co(OH)_{2}$ in water is:

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

The $K_{sp}$ for the above equation is:

$Ksp=[Co^{2+}][OH^{-}]^{2}=1.3\times10^{-15}$

Due to the solubility of $Co(OH)_{2}$ in water, the concentration of $Co^{2+}$ can be set to:

$[Co^{2+}]=X$

Therefore the concentration of $OH^{-}$ should be double that of $Co^{2+}$ :

$[OH^{-}]=2X$

The solution was adjusted to pH 12 using NaOH , therefore:

pH +pOH=14

pOH=14-12=2

$[OH^{-}]=10^{-pOH}=10^{-2}=1.0\times10^{-2}$

Using an ICE table to process the data:

Screen shot 2015 11 13 at 4.39.13 pm

Plugging these variables into the $K_{sp}$ equation gives:

$Ksp=[X]\times [0.01+2X^{2}]=1.3\times10^{-15}$

$2X^{2}<< 0.01$

Therefore, the $K_{sp}$ equation can be approximated to:

$Ksp=[X]\times [0.01]=1.3\times10^{-15}$

$[X]=\frac{1.3\times10^{-15}}{[0.01]}=1.3\times10^{-13}M$

Therefore, the solubility of $Co(OH)_{2}$ in water is $1.3\times10^{-13}M$ .

Report an Error

View Tutors

Ekta
Certified Tutor

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

View Tutors

Nate
Certified Tutor

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

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

SAT Tutors in Boston, GMAT Tutors in Seattle, GMAT Tutors in Houston, GMAT Tutors in San Francisco-Bay Area, Biology Tutors in Miami, MCAT Tutors in Dallas Fort Worth, French Tutors in New York City, LSAT Tutors in Houston, Math Tutors in Phoenix, English Tutors in New York City

Popular Courses & Classes

ISEE Courses & Classes in Atlanta, LSAT Courses & Classes in Houston, MCAT Courses & Classes in San Francisco-Bay Area, LSAT Courses & Classes in Phoenix, ACT Courses & Classes in Phoenix, ISEE Courses & Classes in Phoenix, LSAT Courses & Classes in Philadelphia, SAT Courses & Classes in Chicago, SSAT Courses & Classes in Boston, GMAT Courses & Classes in New York City

Popular Test Prep

ACT Test Prep in Phoenix, SAT Test Prep in Boston, MCAT Test Prep in San Diego, SAT Test Prep in Dallas Fort Worth, GRE Test Prep in Boston, SSAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in Denver, MCAT Test Prep in San Francisco-Bay Area, ISEE Test Prep in Los Angeles, ISEE Test Prep in Dallas Fort Worth

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 : Solubility and Precipitation

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

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #2 : Solubility

Example Question #304 : Gre Subject Test: Chemistry

Example Question #2 : Solubility And Precipitation

Example Question #1 : Analyzing Solids

Example Question #2 : Analyzing Solids

Example Question #2 : Analyzing Solids

Example Question #311 : Gre Subject Test: Chemistry

Example Question #2 : Solubility And Precipitation

Example Question #2 : Solubility And Precipitation

Example Question #3 : Solubility And Precipitation

All GRE Subject Test: Chemistry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: