Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 36

A \(0.28 \mathrm{~g}\) sample of an unknown monoprotic organic acid is dissolved in water and titrated with a 0.1 M sodium hydroxide solution. After the addition of \(17.5 \mathrm{ml}\) of base, a pH of \(5.0\) is recorded. The equivalence point is reached when a total of \(35.0 \mathrm{ml}\) of \(\mathrm{NaOH}\) is added. The molar mass of the organic acid is (a) 160 (b) 80 (c) 40 (d) 120

Short Answer

Expert verified
The molar mass of the organic acid is (b) 80 g/mol.

Step by step solution

01

Determine moles of NaOH at equivalence point

At the equivalence point, the amount of NaOH added is stoichiometrically equivalent to the amount of the unknown acid in the solution. First, calculate the moles of NaOH using the volume at the equivalence point and its molarity: \( Moles_{NaOH} = Molarity_{NaOH} \times Volume_{NaOH} \)\( Moles_{NaOH} = 0.1~M \times 0.0350~L = 0.0035~moles \)
02

Determine the molar mass of the acid

Since the acid is monoprotic, one mole of acid reacts with one mole of NaOH. The moles of acid is equal to the moles of NaOH at equivalence point. Using the mass of the acid sample, calculate its molar mass:\( Molar Mass_{acid} = \frac{Mass_{acid}}{Moles_{acid}} \)\( Molar Mass_{acid} = \frac{0.28~g}{0.0035~moles} = 80~g/mol \)
03

Choose the correct answer

With the calculated molar mass of 80 g/mol, select the appropriate answer among the choices given.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Monoprotic Acids
Monoprotic acids are a key concept in acid-base chemistry. They are defined as acids that donate one proton (hydrogen ion) per molecule during a chemical reaction. The proton donation typically occurs when the acid reacts with a base in a neutralization reaction. Water (H2O), being the universal solvent, often serves as the medium for these reactions.

For instance, the unknown organic acid in our exercise is monoprotic, which means for every molecule of the acid that dissolves in water, it releases one hydrogen ion. This is crucial for calculating stoichiometry in titration problems, as we will see next. In a titration with sodium hydroxide (NaOH), a strong base, each NaOH molecule will neutralize one hydrogen ion from the monoprotic acid, forming water and a salt.
Decoding the Equivalence Point
The equivalence point of a titration is a fundamental concept that denotes the exact point at which the amount of titrant (in this case, NaOH) added is stoichiometrically equivalent to the quantity of the substance being titrated (the unknown monoprotic acid).

At the equivalence point, the number of moles of base is equal to the number of moles of the monoprotic acid. This is vital information for determining the molar mass of the acid. In our exercise, the equivalence point is achieved when 35.0 ml of 0.1 M NaOH is added. The pH at this stage is not given or required since we know the stoichiometry of the reaction is one-to-one; every mole of acid reacts with a mole of base.
Calculating Molar Mass
Molar mass calculation is an integral part of solving titration problems when you need to determine the molecular weight of an unknown substance. The molar mass is the weight in grams of one mole of a substance.

After determining the moles of NaOH used at the equivalence point, we infer that this is also the moles of our unknown acid due to the one-to-one reaction ratio. In our problem, the sample weight of the acid (0.28 grams) divided by the moles (0.0035 moles) gives us the molar mass of the acid—80 grams per mole. This process allows us to discern the identity or purity of an unknown sample by comparing the calculated molar mass to known values.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\begin{array}{llll}\text { Solubility } & \text { products of } & \mathrm{Mg}(\mathrm{OH})_{2} \text { , }\end{array}\) \(\mathrm{Cd}(\mathrm{OH})_{2}, \mathrm{Al}(\mathrm{OH})_{3}\) and \(\mathrm{Zn}(\mathrm{OH})_{2}\) are \(4 \times 10^{-11}, 8 \times 10^{-6}, 8.5 \times 10^{-23}\) and \(1.8 \times 10^{-14}\), respectively. The cation that will precipitate first as hydroxide, on adding limited quantity of \(\mathrm{NH}_{4} \mathrm{OH}\) in a solution containing equimolar amount of metal cations, is (a) \(\mathrm{Al}^{3+}\) (b) \(\mathrm{Zn}^{2+}\) (c) \(\mathrm{Mg}^{2+}\) (d) \(\mathrm{Cd}^{2+}\)

The dissociation constant of a weak monoprotic acid is numerically equal to the dissociation constant of its conjugate base. What is \(\mathrm{pH}\) of \(0.1 \mathrm{M}\) solution of this acid? (a) \(7.0\) (b) \(6.0\) (c) \(8.0\) (d) \(4.0\)

An amount of \(0.10\) moles of \(\mathrm{AgCl}(\mathrm{s})\) is added to one litre of water. Next, the crystals of NaBr are added until \(75 \%\) of the \(\mathrm{AgCl}\) is converted to \(\mathrm{AgBr}(\mathrm{s})\), the less soluble silver halide. What is \(\mathrm{Br}^{-}\) at this point? \(K_{\mathrm{sp}}\) of \(\mathrm{AgCl}=2 \times 10^{-10}\) and \(K_{\mathrm{sp}}\) of \(\mathrm{AgBr}=4 \times 10^{-13}\) (a) \(0.075 \mathrm{M}\) (b) \(0.025 \mathrm{M}\) (c) \(1.5 \times 10^{-4} \mathrm{M}\) (d) \(0.027 \mathrm{M}\)

The buffer capacity \((\beta)\) for a weak acid (A) \(-\) conjugate base (B) buffer is defined as the number of moles of strong acid or base needed to change the \(\mathrm{pH}\) of \(1 \mathrm{~L}\) of solution by \(1 \mathrm{pH}\) unit, where \(\beta=\frac{2.303\left(C_{\mathrm{A}}+C_{\mathrm{B}}\right) K_{\mathrm{a}}\left[\mathrm{H}^{+}\right]}{\left(\left[\mathrm{H}^{+}\right]+K_{\mathrm{a}}\right)^{2}} .\) Under what condition will a buffer best resist a change in \(\mathrm{pH}\) ? (a) \(\mathrm{pH}=3 \mathrm{p} \mathrm{Ka}\) (b) \(2 \mathrm{pH}=\mathrm{p} \mathrm{Ka}\) (c) \(\mathrm{pH}=\mathrm{p} \mathrm{Ka}\) (d) \(\mathrm{pH}=2 \mathrm{p} \mathrm{Ka}\)

Solubility product constant \(\left(K_{\mathrm{sp}}\right)\) of salts of types \(\mathrm{MX}, \mathrm{MX}_{2}\) and \(\mathrm{M}_{3} \mathrm{X}\) at temperature, \(T\) are \(4.0 \times 10^{-8}, 3.2 \times 10^{-14}\) and \(2.7 \times 10^{-15}\), respectively. Solubilities (in \(\mathrm{M}\) ) of the salts at temperature, \(T\), are in the order (a) \(\mathrm{MX}>\mathrm{MX}_{2}>\mathrm{M}_{3} \mathrm{X}\) (b) \(\mathrm{M}_{3} \mathrm{X}>\mathrm{MX}_{2}>\mathrm{MX}\) (c) \(\mathrm{MX}_{2}>\mathrm{M}_{3} \mathrm{X}>\mathrm{MX}\) (d) \(\mathrm{MX}>\mathrm{M}_{3} \mathrm{X}>\mathrm{MX}_{2}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free