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 #311 : Gre Subject Test: Chemistry

$Pb(OH)_{2}$ has a $K_{sp}$ equal to $1.2\times10^{-15}$ . What would be the numerical expression used to determine the molar solubility (S) of $Pb(OH)_{2}$ ?

Possible Answers:

$\sqrt[2]{4.0\times10^{-16}}$

$\sqrt[3]{6.0\times10^{-16}}$

$\sqrt[2]{1.2\times10^{-15}}$

$\sqrt[3]{4.0\times10^{-16}}$

Correct answer:

$\sqrt[3]{6.0\times10^{-16}}$

Explanation:

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

$Pb(OH)_2_{(s)}\rightleftharpoons\ Al^{2+}_{(aq)}\ +\ 2OH^{-}_{(aq)}$

The Ksp expression for the above chemical equation is:

$K_{sp}=S\times\2S^{2}=2S^{3}$

Rearranging this equation to solve for solubility (S) gives:

$\frac{K_{sp}}{2}=S^{3}$

$\sqrt[3]{\frac{K_{sp}}{2}}=S$

Plugging the value for Ksp into the equation gives:

$S=\sqrt[3]{\frac{1.2\times10^{-15}}{2}}=\sqrt[3]{6.0\times10^{-16}}$

Report an Error

Example Question #51 : Analytical Chemistry

Calculate the molar solubility of $Mn(OH)_{2}$ with a $K_{sp} = 1.6\times 10^{-13}$ if enough NaOH is added to raise the pH of the solution to pH 10.

Possible Answers:

$1.2\times10^{-7}\textup{ M}$

$3.2\textup{ M}$

$0.1992\textup{ M}$

$7.9\times10^{-3}\textup{ M}$

$1.6\times10^{-9}\textup{ M}$

Correct answer:

$1.6\times10^{-9}\textup{ M}$

Explanation:

The equation for the dissolution of $Mn(OH)_{2}$ in water is below: $Mn(OH)_{2(s)}\rightleftharpoons Mn^{2+}(aq)+2OH^{-}(aq)$

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

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

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

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

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

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

The solution was adjusted to pH 10 using NaOH , therefore:

pH +pOH=14

pOH=14-10=4

$[OH^{-}]=10^{-pOH}=10^{-4}=1.0\times10^{-4}$

Using an ICE table to process the data:

Screen shot 2015 11 13 at 4.35.54 pm

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

$Ksp=[X]\times [0.0001+2X^{2}]=1.6\times10^{-13}$

$2X^{2}<< 0.0001$

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

$Ksp=[X]\times [0.0001]=1.6\times10^{-13}$

$[X]=\frac{1.6\times10^{-13}}{[0.0001]}=1.6\times10^{-9}M$

Therefore, the solubility of $Mn(OH)_{2}$ in water is $1.6\times10^{-9}M$ .

Report an Error

Example Question #52 : Analytical Chemistry

Calculate the molar solubility of PbF_2 with $K_{sp} = 3.6\times 10^{-8}$ in a solution containing $0.01\textup{ M}$ NaF .

Possible Answers:

$2.1\times10^{-3}\textup{ M}$

$5.1\times10^{-8}\textup{ M}$

$5.8\times10^{-4}\textup{ M}$

$3.6\times10^{-6}\textup{ M}$

$1.0\times10^{-2}\textup{ M}$

Correct answer:

$3.6\times10^{-6}\textup{ M}$

Explanation:

The equation for the dissolution of PbF_2 in water is below: $PbF_{2(s)}\rightleftharpoons Pb^{2+}(aq)+2F^{-}(aq)$

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

$Ksp=[Pb^{2+}][F^{-}]^{2}=3.6\times10^{-8}$

Due to the solubility of PbF_2 in water, the concentration of $Pb^{2+}$ can be set to be:

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

Therefore F^- concentration will be double:

$F^{-}=2X$

Using an ICE table to process the data:

Screen shot 2015 11 09 at 9.29.39 pm

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

$Ksp=[X]\times [0.010+2X^{2}]=3.6\times10^{-8}$

$2X^{2}<< 0.010$

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

$Ksp=[X]\times [0.010]=3.6\times10^{-8}$

$[X]=\frac{3.6\times10^{-8}}{[0.010]}=3.6\times10^{-6}M$

Therefore, the solubility of PbF_2 in water is $3.6\times10^{-6}M$ .

Report an Error

Example Question #12 : Analyzing Solids

Calculate the molar solubility of PbF_2 with $K_{sp} = 3.6\times 10^{-8}$ in a solution containing $0.10\textup{ M}$ $Pb(NO_{3})_{2}$ .

Possible Answers:

$4.2\times10^{-4}\textup{ M}$

$0.111\textup{ M}$

$2.0\times10^{-3}\textup{ M}$

$3.1\times10^{-7}\textup{ M}$

$1.2\times10^{-2}\textup{ M}$

Correct answer:

$4.2\times10^{-4}\textup{ M}$

Explanation:

The equation for the dissolution of PbF_2 in water is below: $PbF_{2(s)}\rightleftharpoons Pb^{2+}(aq)+2F^{-}(aq)$

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

$Ksp=[Pb^{2+}][F^{-}]^{2}=3.6\times10^{-8}$

Due to the solubility of PbF_2 in water, taking into account the concentration of $Pb^{2+}$ from PbF_2 and $Pb(NO_{3})_{2}$ can be set to:

$[Pb^{2+}]=X+0.10$

The F^- concentration will be double the concentration of the $Pb^{2+}$ coming from PbF_2 :

$F^{-}=2X$

Using an ICE table to process the data:

Screen shot 2015 11 09 at 10.02.00 pm

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

$Ksp= [0.10+X]\times [2X^{2}]=3.6\times10^{-8}$

X<< 0.10

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

$Ksp=[0.10]\times [2X]^{2}=3.6\times10^{-8}$

$[2X]^{2}=\frac{3.6\times10^{-8}}{[0.10]}$

$[X]^{2}=\frac{3.6\times10^{-8}}{[0.10]\times 2}$

$[X]=\sqrt{\frac{3.6\times10^{-8}}{[0.10]\times 2}}=4.2\times10^{-4}\textup{ M}$

Therefore, the solubility of PbF_2 in water is $4.2\times10^{-4}\textup{ M}$ .

Report an Error

Example Question #13 : Analyzing Solids

$Al(OH)_{3}$ has a $K_{sp}$ equal to $1.3\times10^{-33}$ . What would be the numerical expression used to determine the molar solubility (S) of $Al(OH)_{3}$ ?

Possible Answers:

$\sqrt[3]{4.3\times10^{-33}}$

$\sqrt[3]{2.2\times10^{-33}}$

$\sqrt[4]{4.3\times10^{-34}}$

$\sqrt[4]{6.5\times10^{-34}}$

Correct answer:

$\sqrt[4]{4.3\times10^{-34}}$

Explanation:

The equation for the dissolution of $Al(OH)_{3}$ in water is below:

$Al(OH)_3_{(s)}\rightleftharpoons\ Al^{3+}_{(aq)}\ +\ 3OH^{-}_{(aq)}$

The Ksp expression for the above chemical equation is:

$K_{sp}=S\times\3S^{3}=3S^{4}$

Rearranging this equation to solve for solubility (S) gives:

$\frac{K_{sp}}{3}=S^{4}$

$\sqrt[4]{\frac{K_{sp}}{3}}=S$

Plugging the value for Ksp into the equation gives:

$S=\sqrt[4]{\frac{1.3\times10^{-33}}{3}}=\sqrt[4]{4.3\times10^{-34}}$

Report an Error

Example Question #56 : Analytical Chemistry

For the titration of $0.001 M\ Ag^{2+}$ with 0.150 L solution of $Cl^-_{(aq)}$ , the end point occurs upon addition of 0.200 L of the $Ag^{2+}$ solution. Determine the initial concentration of chloride that was present in the original solution.

Possible Answers:

$1.321\ M$

$0.063\ M$

$0.0026\ M$

$0.0013\ M$

Correct answer:

$0.0013\ M$

Explanation:

The equivalence point is when enough titrant has been added so that the number of moles of titrant equals the number of moles of analyte. We must first determine the number of moles of silver ions at the equivalence point.

At the equivalence point the following occurs in solution:

$moles\ of\ Ag^{2+}=moles\ of\ Cl^{-}$

The number of moles of silver ions at equivalence point is calculated as follows:

$0.200\ L\times \frac{0.001\ moles}{L}=2.0\times 10^{-4}\ moles\ Ag^{2+}$

$Molarity=\frac{moles\ of\ solute}{Liters\ of\ solution}$

$Liters\ of\ solution= Original\ volume\ of\ Cl^{-}\ solution$

Plugging these values into the equation gives:

$\frac{2\times10^{-4}\ moles}{0.150\ L}=0.0013\ M\ Cl^{-}$

Report an Error

Example Question #11 : Solubility And Solution Equilibrium

How many moles of fluoride ions are present in of a completely saturated solution of lead (II) fluoride?

$PbF_2(s)\rightarrow Pb^{2+}(aq)+2F^-(aq)$

$K_{sp}=3.7*10^{-8}$

Possible Answers:

$2.1*10^{-3}mol$

$1.5*10^{-7}mol$

$7.4*10^{-8}mol$

$4.2*10^{-3}mol$

$8.4*10^{-3}mol$

Correct answer:

$4.2*10^{-3}mol$

Explanation:

Recall that the solubility product constant is given by the equation below, for a reaction in the following format.

$aA\rightarrow cC+dD$

$K_{sp}=[C]^c[D]^d$

Using our balanced reaction, we can find the solubility product equation for the dissociation of lead (II) fluoride.

$PbF_2(s)\rightarrow Pb^{2+}(aq)+2F^-(aq)$

$K_{sp}=[Pb^{2+}][F^-]^2$

For each molecule of lead (II) fluoride that dissolves, it produces one lead ion and two fluoride ions. We can conclude that the concentration of fluoride ions in solution will be twice the concentration of lead ions.

$[Pb^{2+}] = x$

[F^-]=2x

Use these variables in the solubility product equation, along with the given value from the question.

$3.7*10^{-8}=(x) (2x)^2$

Now we can solve for the value of $\small x$ .

$3.7*10^{-8}=4x^3$

$x = \sqrt[3]{\frac{3.7*10^{-8}}{4}}=2.1*10^{-3}$

Remember that this value is equal to the concentration of lead ions in solution and half the concentration of fluoride ions in solution.

$[Pb^{2+}]=x=2.1*10^{-3}M$

$[F^-]=2x=2(2.1*10^{-3}M)=4.2*10^{-3}M$

Report an Error

Example Question #57 : Analytical Chemistry

Calculate the molar solubility of $Co(OH)_{2}$ which has a $Ksp = 1.3\times 10^{-15}$ .

Possible Answers:

$6.88\times10^{-6}M$

$3.6\times10^{-6}\textup{ M}$

$2.09\times10^{-3}\textup{ M}$

$8.66\times10^{-6}\textup{ M}$

$6.36\times10^{-6}\textup{ M}$

Correct answer:

$6.88\times10^{-6}M$

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 $Co^{2+}$ :

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

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

$Ksp=x\cdot (2x)^{2}$

Ksp=4x3

Now, we can solve for the x-variable by setting this value equal to the given $K_{sp}$ .

$4^{3}=1.3\times10^{-15}$

$\frac{1.3\times 10^{-15}}{4}=x^{3}$

$x=\sqrt[3]{\frac{1.3\times 10^{-15}}{4}}=6.87534\times10^{-6}$

Therefore, the solubility of $Co(OH)_{2}$ in water is $6.88\times10^{-6}M$ .

Report an Error

View Tutors

Yongmao
Certified Tutor

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

View Tutors

Gregory
Certified Tutor

Pennsylvania State University-Main Campus, Bachelor of Science, Chemical Engineering. Carnegie Mellon University, Doctor of P...

View Tutors

Austin
Certified Tutor

University of Arizona, Bachelor of Science, Physiology.

All GRE Subject Test: Chemistry Resources

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

Popular Subjects

Physics Tutors in Houston, SAT Tutors in Denver, French Tutors in Philadelphia, English Tutors in Washington DC, SSAT Tutors in Washington DC, SSAT Tutors in Philadelphia, Chemistry Tutors in Seattle, Chemistry Tutors in Miami, GMAT Tutors in Denver, Spanish Tutors in Chicago

Popular Courses & Classes

GMAT Courses & Classes in Denver, ISEE Courses & Classes in Seattle, SAT Courses & Classes in Boston, SSAT Courses & Classes in Boston, MCAT Courses & Classes in Seattle, GMAT Courses & Classes in New York City, SAT Courses & Classes in Philadelphia, MCAT Courses & Classes in Houston, Spanish Courses & Classes in Seattle, GRE Courses & Classes in Washington DC

Popular Test Prep

SAT Test Prep in Washington DC, SSAT Test Prep in San Francisco-Bay Area, ISEE Test Prep in Seattle, SSAT Test Prep in New York City, GMAT Test Prep in Philadelphia, ISEE Test Prep in Dallas Fort Worth, ACT Test Prep in San Diego, LSAT Test Prep in Seattle, LSAT Test Prep in New York City, GMAT Test Prep in New York City

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 #311 : Gre Subject Test: Chemistry

Example Question #51 : Analytical Chemistry

Example Question #52 : Analytical Chemistry

Example Question #12 : Analyzing Solids

Example Question #13 : Analyzing Solids

Example Question #56 : Analytical Chemistry

Example Question #11 : Solubility And Solution Equilibrium

Example Question #57 : Analytical 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: