/*! 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 81 Calculate the normality of each ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 81

Calculate the normality of each of the following solutions. a. \(0.250 M \mathrm{HCl}\) b. \(0.105 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) c. \(5.3 \times 10^{-2} M \mathrm{H}_{3} \mathrm{PO}_{4}\)

Short Answer

Expert verified
The normality of the solutions are: a. 0.250 N for 0.250 M HCl. b. 0.210 N for 0.105 M H2SO4. c. \(15.9 \times 10^{-2}\) N for \(5.3 \times 10^{-2}\) M H3PO4.

Step by step solution

01

Identify the number of acidic hydrogens present in the given acids.

Before calculating the normality of the solutions, we need to recognize the number of acidic hydrogens in the given acids: a. HCl has 1 acidic hydrogen. b. H2SO4 has 2 acidic hydrogens. c. H3PO4 has 3 acidic hydrogens.
02

Calculate the normality of the given solutions.

Now using the formula N = M 脳 n, let's calculate the normality of the following solutions: a. For 0.250 M HCl, N = M 脳 n = 0.250 M 脳 1 N = 0.250 N b. For 0.105 M H2SO4, N = M 脳 n = 0.105 M 脳 2 N = 0.210 N c. For \(5.3 \times 10^{-2}\) M H3PO4, N = M 脳 n = \(5.3 \times 10^{-2}\) M 脳 3 N = \(15.9 \times 10^{-2}\) N Now, we have calculated the normality of the given solutions. The normality of 0.250 M HCl is 0.250 N. The normality of 0.105 M H2SO4 is 0.210 N. The normality of \(5.3 \times 10^{-2}\) M H3PO4 is \(15.9 \times 10^{-2}\) N.

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.

Acidic Hydrogens
Acidic hydrogens are the hydrogen atoms in a compound that can be released as protons ( H鈦� ) when dissolved in water. It's important to identify the number of acidic hydrogens to calculate the normality of a solution since normality relates to the equivalent concentration of the solution.

For different acids, the number of acidic hydrogens can vary:
  • In HCl , there is 1 acidic hydrogen.
  • In H鈧係O鈧� , there are 2 acidic hydrogens.
  • In H鈧働O鈧� , there are 3 acidic hydrogens.
Knowing the number of acidic hydrogens helps us determine how many moles of protons are released per mole of acid, which is crucial for calculating normality.
Molarity
Molarity is a measure of the concentration of a solute in a solution, specifically the number of moles of solute per liter of solution. It is denoted by the letter M and is usually expressed in units of mol/L .

When discussing acids or bases, understanding molarity is essential as it forms the basis for calculating other concepts like normality. For instance:
  • 0.250 M HCl means there are 0.250 moles of HCl in each liter of solution.
  • 0.105 M H鈧係O鈧� indicates that each liter contains 0.105 moles of H鈧係O鈧� .
  • 5.3 脳 10^{-2} M H鈧働O鈧� signifies that there are 5.3 脳 10^{-2} moles of H鈧働O鈧� per liter.
Molarity is crucial in stoichiometric calculations and when preparing solutions for reactions.
Chemical Solutions
Chemical solutions are homogeneous mixtures composed of two or more substances. In a solution, the solute is the substance dissolved, while the solvent is the substance that does the dissolving, typically in greater quantity.

Chemical solutions are fundamental in various scientific experiments and real-world applications, with water being a universal solvent. For instance:
  • HCl dissolved in water to create hydrochloric acid solution.
  • H鈧係O鈧� in water forms sulfuric acid solution.
  • H鈧働O鈧� in water results in phosphoric acid solution.
Understanding the properties of chemical solutions is essential for tasks such as titration, where precise concentrations are needed for accurate results.
Acid-Base Chemistry
Acid-base chemistry revolves around the behavior of acids and bases, particularly their ability to donate or accept protons ( H鈦� ). In this field, the concepts of molarity and normality are highly relevant.

An acid is a substance that donates protons, while a base accepts them, leading to reactions that can neutralize each other. The normality of an acid or a base gives a measure of its reactive capacity.
  • In acid-base reactions, normality reveals how many equivalents of reactants actually participate.
  • For instance, HCl contributes one equivalent per mole, while H鈧係O鈧� and H鈧働O鈧� contribute two and three equivalents, respectively.
Understanding these principles helps in balancing chemical equations and predicting the outcome of reactions, particularly in titration processes.

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

What \(mass\) of the indicated solute does each of the following solutions contain? a. \(17.8 \mathrm{~mL}\) of \(0.119 \mathrm{M} \mathrm{CaCl}_{2}\) b. \(27.6 \mathrm{~mL}\) of \(0.288 M \mathrm{KCl}\) c. \(35.4 \mathrm{~mL}\) of \(0.399 \mathrm{M} \mathrm{FeCl}_{3}\) d. \(46.1 \mathrm{~mL}\) of \(0.559 \mathrm{M} \mathrm{KNO}_{3}\)

Calculate the normality of each of the following solutions. a. \(0.134 M \mathrm{NaOH}\) b. \(0.00521 M \mathrm{Ca}(\mathrm{OH})_{2}\) c. \(4.42 M \mathrm{H}_{3} \mathrm{PO}_{4}\)

A solution is \(0.1 \%\) by mass calcium chloride. Therefore, \(100 .\) g of the solution contains ______________ g of calcium chloride.

Calculate the number of moles of each ion present in each of the following solutions. a. \(10.2 \mathrm{~mL}\) of \(0.451 \mathrm{M} \mathrm{AlCl}_{3}\) solution b. \(5.51 \mathrm{~L}\) of \(0.103 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution c. \(1.75 \mathrm{~mL}\) of \(1.25 \mathrm{M} \mathrm{CuCl}_{2}\) solution d. \(25.2 \mathrm{~mL}\) of \(0.00157 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) solution

Baking soda (sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3}\) ) is often used to neutralize spills of acids on the benchtop in the laboratory. What mass of \(\mathrm{NaHCO}_{3}\) would be needed to neutralize a spill consisting of \(25.2 \mathrm{~mL}\) of \(6.01 \mathrm{M}\) hydrochloric acid solution?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Kinetics

Read Explanation

Nuclear 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.