/*! 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 46 \(0.005 \mathrm{M}\) acid soluti... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 46

\(0.005 \mathrm{M}\) acid solution has \(5 \mathrm{pH}\). The percentage ionization of acid is a. \(0.8 \%\) b. \(0.6 \%\) c. \(0.4 \%\) d. \(0.2 \%\)

Short Answer

Expert verified
The percentage ionization of the acid is 0.2%.

Step by step solution

01

Understanding pH

The pH of a solution is given as 5. This means the concentration of hydrogen ions, \([ ext{H}^+]\), in the solution is \(10^{-5}\) M.
02

Use Formula for Percentage Ionization

The formula for percentage ionization is \(\text{Percentage Ionization} = \left( \frac{[ ext{H}^+]}{[ ext{HA}]} \right) \times 100\). Here, \([ ext{H}^+] = 10^{-5}\) M and \([ ext{HA}] = 0.005\) M.
03

Calculate Percentage Ionization

Substitute the values into the formula: \[\text{Percentage Ionization} = \left( \frac{10^{-5}}{0.005} \right) \times 100 = \left( \frac{10^{-5}}{5 \times 10^{-3}} \right) \times 100= \left( \frac{1}{500} \right) \times 100 \approx 0.2\%\].
04

Conclusion Based on Calculation

The calculated percentage ionization is closest to the value given in option d. Hence, the percentage ionization of the acid is \(0.2\%\).

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.

Understanding Percentage Ionization
Percentage ionization is the term used to describe the proportion of a weak acid that dissociates into ions in a solution. This percentage indicates how much of the acid breaks up into hydrogen ions (H鈦�) and its conjugate base. Understanding this concept is crucial:
  • It's primarily used for weak acids because strong acids are almost completely ionized in water.
  • It gives insight into the strength of an acid鈥攖he higher the percentage ionization, the stronger the acid in a given concentration.
To calculate percentage ionization, we use the formula:\[\text{Percentage Ionization} = \left( \frac{[\text{H}^+]}{[\text{HA}]} \right) \times 100\]Where \([\text{H}^+]\) is the concentration of hydrogen ions, and \([\text{HA}]\) is the initial concentration of the acid.
Understanding Acid Concentration
Acid concentration refers to the amount of acid present in a solution. In solutions, concentration is often expressed in terms of molarity, which is moles per liter (M). For example, a 0.005 M solution means that there are 0.005 moles of the acid per liter of solution.
  • Acid concentration gives a direct measure of how many acid molecules are available to ionize.
  • It is a crucial factor in calculating both the pH of the solution and the percentage ionization, as seen in the formula for percentage ionization.
In more practical terms, if you know the molarity and the dissociation constant of an acid, you can predict how it behaves in water.
Hydrogen Ion Concentration and Its Importance
Hydrogen ion concentration is vital in determining the acidity of a solution. It is symbolized as \([\text{H}^+]\) and directly correlates with pH measurement.
The pH scale, which ranges from 0 to 14, measures acidity or basicity. The concentration of hydrogen ions determines this scale:
  • Neutral solutions have a hydrogen ion concentration of \(10^{-7}\) M, corresponding to a pH of 7.
  • Acidic solutions have a higher concentration of hydrogen ions and a lower pH.
  • Basic solutions have a lower concentration of hydrogen ions and a higher pH.
For example, a solution with pH 5 has a hydrogen ion concentration of \(10^{-5}\) M.
Solution Molarity in Context
Molarity is a way of expressing the concentration of a solution. It is defined as the number of moles of solute per liter of solution, denoted as M.
For preparing solutions in a laboratory or industrial setting, molarity is an immensely handy concept:
  • When calculating pH or reactions, knowing the molarity allows chemists to control and predict the behavior of substances.
  • It serves as a bridge for calculations, such as those needed to determine percentage ionization.
By knowing the molarity of an acid solution, as in our example with 0.005 M, we can derive more significant information such as percentage ionization by using equations and known pH levels.

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

\(\mathrm{N}_{2}+3 \mathrm{H}_{2} \leftrightarrow 2 \mathrm{NH}_{3}\) Which is correct statement if \(\mathrm{N}_{2}\) is added at equilibrium condition? a. The equilibrium will shift to forward direction because according to 2 nd law of thermodynamics the entropy must increases in the direction of spontaneous reaction b. The condition for equilibrium is \(\mathrm{GN}_{2}+3 \mathrm{GH}_{2}\) \(=2 \mathrm{GNH}_{3}\), where \(\mathrm{G}\) is Gibbs free energy per mole of the gaseous species measured at the partial pressure. The condition of equilibrium in unaffected by the use of catalyst,which increases the rate of both the forward and backward reactions to the same extent. c. The catalyst will increase the rate of forward reaction by \(\alpha\) and that of backward reaction by \(\beta\). d. Catalyst will not alter the rate of either of the reaction.

Cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\) undergoes a molecular rearrangement in the presence of \(\mathrm{AlCl}_{3}\) to form methylcyclopentane \(\left(\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right)\), according to the equation \(\mathrm{C}_{6} \mathrm{H}_{12} \rightleftharpoons \mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\) If \(\mathrm{K}_{\mathrm{C}}=0.143\) at \(25^{\circ} \mathrm{C}\) for this reaction, find the equilibrium concentrations of \(\mathrm{C}_{6} \mathrm{H}_{12}\) and \(\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\) if the initial concentrations are \(0.200 \mathrm{M}\) and \(0.100\) M respectively a. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.286\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.016 \mathrm{M}\) b. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.262\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.038 \mathrm{M}\) c. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.186\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.162 \mathrm{M}\) d. \(\left[\mathrm{C}_{6} \mathrm{H}_{12}\right]=0.164\) and \(\left[\mathrm{CH}_{3} \mathrm{C}_{5} \mathrm{H}_{9}\right]=0.621 \mathrm{M}\)

An equilibrium mixture for the reaction \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \leftrightarrow 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{S}_{2}(\mathrm{~g})\) had 1 mole of \(\mathrm{H}_{2} \mathrm{~S}, 0.2\) mole of \(\mathrm{H}_{2}\) and \(0.8\) mole of \(\mathrm{S}_{2}\) in a 2 litre flask. The value of \(\mathrm{K} \mathrm{c}\) in \(\mathrm{mol} \mathrm{L}^{-1}\) is a. \(0.08\) b. \(0.016\) c. \(0.004\) d. \(0.160\)

The reactions in which the yield of the products can not be increased by the applications of high pressure is/are a. \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) b. \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) c. \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\) d. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\)

Match the following: Column I Column II A. \(0.5 \mathrm{M} \mathrm{NO}_{3}+0.1 \mathrm{M} \quad\) (p) 7 \(\mathrm{NH}_{4} \mathrm{OH}\) B. \(0.1 \mathrm{M} \mathrm{KCl}+0.1 \mathrm{M} \quad\) (q) greater then 7 \(\mathrm{KNO}_{3}\) C. \(0.2 \mathrm{M} \mathrm{NaOH}+0.5 \mathrm{M}(\mathrm{r})\) between 1 to 7 \(\mathrm{HCOOH}\) D. \(0.1 \mathrm{NH}_{4} \mathrm{Cl}+0.1 \mathrm{M}(\mathrm{s}) 0.7\) \(\mathrm{KOH}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

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.