Call Now to Set Up Tutoring:

(888) 888-0446

College Chemistry : Titrations

Study concepts, example questions & explanations for College Chemistry

Create An Account

All College Chemistry Resources

1 Diagnostic Test 100 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Titrations

The titration of a $15.0\text{mL}$ sample of an $\text{HCl}$ solution at an unknown concentration requires $17.08\text{mL}$ of a $0.200\text{M NaOH}$ solution. What is the concentration of the unknown hydrochloric acid solution in ?

Possible Answers:

0.228

0.237

0.189

0.245

Correct answer:

0.228

Explanation:

Start by writing the chemical equation for this acid-base reaction.

$\text{HCl}(aq)+\text{NaOH}(aq)\rightarrow\text{H}_2\text{O}(l)+\text{NaCl}(aq)$

Next, find the number of moles of $\text{NaOH}$ that was added.

$17.08\text{mL NaOH}\cdot\frac{1\text{L}}{1000\text{mL}}\cdot\frac{0.200\text{ moles NaOH}}{\text{L}}=0.003416\text{ moles of NaOH}$

At the equivalence point of a titration, the moles of acid and the moles of base are the same.

$\text{Moles of HCl}=\text{Moles of NaOH}=0.003416$

Finally, find the molarity of the solution.

$\text{Molarity}=\frac{\text{moles of solute}}{\text{liters of solution}}=\frac{0.003416}{15.0\times10^{-3}}=0.228M$

Make sure to round the answer to three significant figures.

Report an Error

Example Question #1 : Titrations

A $75.0\text{mL}$ sample of 0.100M sodium hydroxide is titrated with 0.150M hydrochloric acid. Find the pH of the solution after adding $45.0\text{mL}$ of hydrochloric acid.

Possible Answers:

12.2

11.2

11.8

10.4

Correct answer:

11.8

Explanation:

Start by finding the initial number of moles of sodium hydroxide by using the given volume and molarity. Since sodium hydroxide is a strong base, it will dissociate completely into $\text{OH}^-$ .

$\text{Initial moles of NaOH}=\text{Initial moles of OH}^-=0.075\text{L}(\frac{0.100\text{ mol}}{\text{L}})=0.0075\text{ moles}$

Next, find the number of moles of the hydrochloric acid added by using the given volume and molarity. Since hydrochloric acid is a strong acid, all of the $\text{H}_3\text{O}^+$ will come from the acid.

$\text{Moles of HCl added}=\text{Moles of H}_3\text{O}^+=0.045\text{L}(\frac{0.150\text{ moles}}{\text{L}})=0.00675\text{ moles}$

Now, as the acid is added, it will neutralize some of the base as shown by the following equation:

$\text{OH}^-(aq)+\text{H}_3\text{O}^+(aq)\rightarrow 2\text{H}_2\text{O}(l)$

Since the reactants are found in a 1:1 ratio, subtract the number of moles of acid added from the number of initial moles of base to get the amount of remaining base.

$\text{Remaining moles of NaOH}=0.0075-0.00675=7.5\times10^{-4}\text{ moles of NaOH}$

Now, divide this number of moles by the total volume of the solution to find the concentration of the base.

$[\text{OH}]^-=\frac{7.5\times 10^{-4}\text{ moles}}{0.075\text{L}+0.045\text{L}}=0.00625\text{ M}$

Next, find the $\text{pOH}$ value.

$\text{pOH}=-\log(0.00625)=2.204$

Finally, find the $\text{pH}$ value.

$\text{pH}=14-\text{pOH}$

$\text{pH}=14-2.204=11.80$

Your answer should have significant figures.

$\text{pH}=11.8$

Report an Error

Example Question #3 : Acid Base Reactions

Screen shot 2016 03 24 at 9.10.57 pm

Determine the volume in mL of HCl solution it would take to neutralize $35.0\ mL$ of a $0.0025\ M$ NaOH with a $0.012\ M$ HCl solution during a titration.

Possible Answers:

$24.0\ mL$

$23.0\ mL$

$131.0\ mL$

$72.0\ mL$

Correct answer:

$72.0\ mL$

Explanation:

Determine the moles of NaOH used:

$0.035\ L\times \frac{0.025\ moles}{L}=8.75\times10^{-4}\ moles\ NaOH$

At the equivalence point of the titration:

$moles\ of\ NaOH=moles\ of\ HCl=8.75\times10^{-4}\ moles$

Convert the moles of HCl to liters:

$\frac{L}{0.012\ moles}\times8.75\times10^{-4}\ moles=0.072\ L\ of HCl$

Convert the liters to milliliters:

$\frac{1\ mL}{10^{-3}L}\times0.072\ L=72.0\ mL\ HCl$

Report an Error

Example Question #1 : Titrations

Screen shot 2016 03 24 at 9.10.57 pm

Determine the volume in mL of HCl solution it would take to neutralize $42.0\ mL$ of a $0.0021\ M$ NaOH with a $0.014\ M$ HCl solution during a titration.

Possible Answers:

$4.2\ mL$

$6.3\ mL$

$1.2\ mL$

$40.5\ mL$

Correct answer:

$6.3\ mL$

Explanation:

Determine the moles of NaOH used:

$0.042\ L\times \frac{0.0021\ moles}{L}=8.82\times10^{-5}\ moles\ NaOH$

At the equivalence point of the titration:

$moles\ of\ NaOH=moles\ of\ HCl=8.82\times10^{-5}\ moles$

Convert the moles of HCl to liters:

$\frac{L}{0.014\ moles}\times8.82\times10^{-5}\ moles=0.0063\ L\ of HCl$

Convert the liters to milliliters:

$\frac{1\ mL}{10^{-3}L}\times0.0063\ L=6.3\ mL\ HCl$

Report an Error

Example Question #2 : Titrations

NaOH reacts with $CH_{3}COOH$ in aqueous solution according to the chemical equation provided. If it took $32.0\ mL$ of a $0.1\ M$ NaOH solution to titrate $25.0\ mL$ of an $CH_{3}COOH$ solution, what was the concentration of original $CH_{3}COOH$ solution?

Possible Answers:

$0.0032\ M$

$0.25\ M$

$0.13\ M$

$1.4\ M$

Correct answer:

$0.13\ M$

Explanation:

Using the concentration of NaOH as a conversion factor, we can convert the volume of NaOH to moles by dimensional analysis:

$\frac{0.1\ moles}{L}\times0.032\ L=0.0032\ moles\ NaOH$

At the equivalence point of the titration:

$moles\ of\ NaOH=moles\ of\ CH_{3}COOH=0.0032\ moles$

$Molarity=\frac{0.0032\ moles}{0.025\ L}=0.13\ M$

Report an Error

Example Question #3 : Titrations

Household vinegar contains the organic compound acetic acid with chemical formula, $CH_{3}COOH$ . Acetic acid reacts with NaOH in aqueous solution according to the chemical equation provided. Determine the percentage (by volume) of vinegar in a $40.0\ mL$ sample that reacted with $15.0\ mL$ of $0.1\ M$ NaOH during a titration to reach the equivalence. (Density of $CH_{3}COOH$ = $1.05\frac{g}{mL}$ )

Possible Answers:

$45\%$

$10\%$

$0.21\%$

$3.5\%$

Correct answer:

$0.21\%$

Explanation:

$\frac{0.1\ moles}{L}\times0.015L=0.0015\ moles$

At the equivalence point of the titration:

$moles\ of\ NaOH=moles\ of\ CH_{3}COOH=0.0015\ moles$

Convert the moles of CH_3COOH to grams:

$0.0015\ moles\ CH_{3}COOH\times \frac{60.05\ g}{moles}=0.090\ g\ CH_{3}COOH$

Using the density of CH_3COOH given, convert the grams to mL:

$0.09\ g \times \frac{mL}{1.05g}=0.086\ mL\ CH_{3}COOH$

Determine the percentage of CH_3COOH by dividing the volume of by the total volume of vinegar:

$\frac{0.086\ mL}{40.0\ mL}=0.21\%\ CH_{3}COOH\ in\ vinegar$

Report an Error

Example Question #1 : Titrations

NaOH reacts with $CH_{3}COOH$ in aqueous solution according to the chemical equation provided. If it took $26.7\ mL$ of a $0.1\ M$ NaOH solution to titrate $20.8\ mL$ of an $CH_{3}COOH$ solution, what was the concentration of original $CH_{3}COOH$ solution?

Possible Answers:

$0.309\ M$

$0.005\ M$

$0.154\ M$

$1.4\ M$

Correct answer:

$0.154\ M$

Explanation:

Using the concentration of NaOH as a conversion factor, we can convert the volume of NaOH to moles by dimensional analysis:

$\frac{0.1\ moles}{L}\times0.0267\ L=0.00267\ moles\ NaOH$

At the equivalence point of the titration:

$moles\ of\ NaOH=moles\ of\ CH_{3}COOH=0.00267\ moles$

$Molarity=\frac{0.0032\ moles}{0.0208\ L}=0.154\ M$

Report an Error

Example Question #1 : Titrations

Household vinegar contains the organic compound acetic acid with chemical formula, $CH_{3}COOH$ . Acetic acid reacts with NaOH in aqueous solution according to the chemical equation provided. Determine the percentage (by volume) of acetic acid in a $100.0\ mL$ sample of vinegar that reacted with $30.0\ mL$ of $2.0\ M$ NaOH during a titration to reach the equivalence. (Density of $CH_{3}COOH$ = $1.05\frac{g}{mL}$ )

Possible Answers:

$7.0\%$

$24\%$

$1.5\%$

$3.4\%$

Correct answer:

$3.4\%$

Explanation:

Convert the liters of NaOH to moles:

$\frac{2.00\ moles}{L}\times0.030L=0.06\ moles\ NaOH$

At the equivalence point of the titration:

$moles\ of\ NaOH=moles\ of\ CH_{3}COOH=0.06\ moles$

Convert the moles of CH_3COOH to grams:

$0.06\ moles\ CH_{3}COOH\times \frac{60.05\ g}{moles}=3.6\ g\ CH_{3}COOH$

Using the density of CH_3COOH given, convert the grams to mL:

$3.6\ g \times \frac{mL}{1.05g}=3.4\ mL\ CH_{3}COOH$

Determine the percentage of CH_3COOH by dividing the volume of by the total volume of vinegar:

$\frac{3.4\ mL}{100.0\ mL}=3.4\%\ CH_{3}COOH\ in\ vinegar$

Report an Error

View Tutors

Sara
Certified Tutor

Michigan State University, Bachelor of Science, Environmental Studies.

View Tutors

Kimberly
Certified Tutor

Columbia University in the City of New York, Bachelor in Arts, American Literature. Columbia University in the City of New Yo...

View Tutors

Alejandro
Certified Tutor

Kennesaw State University, Bachelor of Science, Computer Software Engineering.

All College Chemistry Resources

1 Diagnostic Test 100 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

French Tutors in Miami, SAT Tutors in Miami, SAT Tutors in Los Angeles, Computer Science Tutors in Los Angeles, ACT Tutors in Houston, Computer Science Tutors in Washington DC, Chemistry Tutors in San Francisco-Bay Area, Biology Tutors in New York City, LSAT Tutors in Seattle, Chemistry Tutors in Atlanta

Popular Courses & Classes

LSAT Courses & Classes in Washington DC, SAT Courses & Classes in Houston, SAT Courses & Classes in Washington DC, SSAT Courses & Classes in San Diego, LSAT Courses & Classes in Phoenix, ISEE Courses & Classes in Dallas Fort Worth, LSAT Courses & Classes in Atlanta, GMAT Courses & Classes in Phoenix, GMAT Courses & Classes in New York City, Spanish Courses & Classes in Miami

Popular Test Prep

GRE Test Prep in Washington DC, LSAT Test Prep in Phoenix, SSAT Test Prep in Miami, ISEE Test Prep in New York City, MCAT Test Prep in Philadelphia, ACT Test Prep in Washington DC, ACT Test Prep in Denver, ISEE Test Prep in Denver, ISEE Test Prep in Boston, 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

College Chemistry : Titrations

Study concepts, example questions & explanations for College Chemistry

All College Chemistry Resources

Example Questions

Example Question #1 : Titrations

Example Question #1 : Titrations

Example Question #3 : Acid Base Reactions

Example Question #1 : Titrations

Example Question #2 : Titrations

Example Question #3 : Titrations

Example Question #1 : Titrations

Example Question #1 : Titrations

All College Chemistry Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: