/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 93 A \(3.664-\mathrm{g}\) sample of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 93

A \(3.664-\mathrm{g}\) sample of a monoprotic acid was dissolved in water and required \(20.27 \mathrm{~mL}\) of a \(0.1578 \mathrm{M}\) \(\mathrm{NaOH}\) solution for neutralization. Calculate the molar mass of the acid.

Short Answer

Expert verified
The molar mass of the acid is approximately 1145 g/mol.

Step by step solution

01

Calculate moles of $\mathrm{NaOH}$ using concentration and volume

Firstly, calculate the moles (n) of the $\mathrm{NaOH}$ used in the reaction by using the equation n = c 脳 V, where c is the concentration and V is the volume. In this case, c = \(0.1578 \mathrm{M}\) and V = \(20.27 \mathrm{~mL} = 20.27 \times 10^{-3} \mathrm{L}\). Therefore n = \(0.1578 \times 20.27 \times 10^{-3} = 3.200 \times 10^{-3}\) moles.
02

Determine moles of the acid

Since each acid molecule reacts with one $\mathrm{NaOH}$ molecule, the moles of the acid are also \(3.20 \times 10^{-3}\) moles.
03

Calculation of the Molar mass of the acid

Finally, calculate the molar mass (MM) of the acid by using the equation MM = m / n, where m is the mass and n is the moles. In this case m = 3.664 g and n = 3.20 x 10^{-3} mol, hence MM = 3.664 / \(3.20 \times 10^{-3} = 1145 \mathrm{g/mol}\).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Monoprotic Acid
A monoprotic acid is a type of acid that can donate only one proton (hydrogen ion) per molecule during a reaction. This is an essential concept when working with reactions involving acids and bases. In a neutralization reaction, a monoprotic acid will exactly neutralize an equivalent amount of a base.
This is because it provides one proton per acid molecule, reacting with one hydroxide ion from the base.
For example:
  • Hydrochloric acid (HCl)
  • Acetic acid (CH鈧僀OOH)
These acids only have one replaceable hydrogen atom, making calculations straightforward.
During calculations, remember monoprotic acids equate their moles directly to the moles of the base in a reaction, simplifying stoichiometric calculations.
Neutralization Reaction
A neutralization reaction occurs when an acid and a base react to form water and a salt. It's an important chemical process widely seen in chemistry and everyday life.
In the context of the exercise, the neutralization involves a monoprotic acid and sodium hydroxide (NaOH), a common strong base.
Here's what generally happens in neutralization:
  • The acid donates a proton to the hydroxide ion from the base.
  • Water ( H鈧侽 ) is formed as a result.
  • The remaining ions form a salt.
For a monoprotic acid, such as those mentioned earlier, each molecule of acid will neutralize one molecule of a strong base like NaOH.
The reaction can be summarized as follows, showing the balance:
  • HA + NaOH 鈫� NaA + H鈧侽
Here, HA represents the monoprotic acid molecule, and NaA is the salt produced.
Stoichiometry
Stoichiometry is the area of chemistry that deals with the quantitative relationships of reactants and products in a chemical reaction. It allows you to calculate amounts of substances needed or produced in reactions, crucial for solving problems like calculating molar masses.
In our specific case of neutralizing a monoprotic acid with NaOH:
  • Step 1: Calculate the moles of NaOH used, since volume and concentration are known. Use the formula \( n = c \times V \), where \( c \) is the concentration and \( V \) is the volume in liters.
  • Step 2: Since it's a 1:1 neutralization reaction, the moles of NaOH will equal the moles of the monoprotic acid used.
  • Step 3: Finally, to find the molar mass of the acid, use \( \text{Molar Mass} = \frac{\text{mass of acid}}{\text{moles of acid}} \).
This process demonstrates how stoichiometry ties volume and concentration to actual chemical quantities, allowing you to determine essential properties like molar mass.

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

Phosphorus forms many oxoacids. Indicate the oxidation number of phosphorus in each of the following acids: (a) \(\mathrm{HPO}_{3},\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{2},\) (c) \(\mathrm{H}_{3} \mathrm{PO}_{3},\) d) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) (e) \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7},\) (f) \(\mathrm{H}_{5} \mathrm{P}_{3} \mathrm{O}_{10}\).

Phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\) is an important industrial chemical used in fertilizers, detergents, and in the food industry. It is produced by two different methods. In the electric furnace method elemental phosphorus \(\left(\mathrm{P}_{4}\right)\) is burned in air to form \(\mathrm{P}_{4} \mathrm{O}_{10}\), which is then reacted with water to give \(\mathrm{H}_{3} \mathrm{PO}_{4}\). In the wet process the mineral phosphate rock \(\left[\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\right]\) is reacted with sulfuric acid to give \(\mathrm{H}_{3} \mathrm{PO}_{4}\) ( and HF and \(\mathrm{CaSO}_{4}\) ). Write equations for these processes and classify each step as precipitation, acid-base, or redox reaction.

What is the difference between a nonelectrolyte and an electrolyte? Between a weak electrolyte and a strong electrolyte?

Describe the basic steps involved in gravimetric analysis. How does this procedure help us determine the identity of a compound or the purity of a compound if its formula is known?

You have \(505 \mathrm{~mL}\) of a \(0.125 \mathrm{M} \mathrm{HCl}\) solution and you want to dilute it to exactly \(0.100 M\). How much water should you add?
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

Organic Chemistry

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.