/*! 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 185 Consider 1000 . mL of a \(1.00 \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 185

Consider 1000 . mL of a \(1.00 \times 10^{-4}-M\) solution of a certain acid HA that has a \(K_{\text { a value equal to } 1.00 \times 10^{-4} . \text { How much }}\) water was added or removed (by evaporation) so that a solution remains in which 25.0\(\%\) of \(\mathrm{HA}\) is dissociated at equilibrium? Assume that HA is nonvolatile.

Short Answer

Expert verified
To achieve 25% dissociation of HA at equilibrium, 2.00 L (or 2000 mL) of water was added to the solution.

Step by step solution

01

Determine the initial concentration of HA

Given that we have a 1000 mL solution with a molarity of \(1.00 \times 10^{-4} M\), we can calculate the number of moles of HA as follows: \[\text{Moles of HA} = \text{Molarity} \times \text{Volume} = (1.00 \times 10^{-4} M)(1000\,mL)\] Note that 1000 mL is equal to 1 L. \[\text{Moles of HA} = (1.00 \times 10^{-4} M)(1\,L) = 1.00 \times 10^{-4}\,mol\]
02

Calculate the moles of HA dissociated when it reaches 25% dissociation

Since we know the amount of HA required to have a 25% equilibrium dissociation, we can calculate the amount of HA dissociated at equilibrium: Amount of dissociated HA = \(25 \% \times\) moles of HA Amount of dissociated HA = \(\frac{25}{100} \times 1.00 \times 10^{-4} mol = 2.50 \times 10^{-5} mol\)
03

Use the equilibrium constant to find the concentrations of dissociated HA, H+ and A- ions

The equilibrium constant is given as \(K_a = 1.00 \times 10^{-4}\), and we can use this value to establish the relationship between the equilibrium concentrations of H+, A-, and HA. Assuming x as the moles of HA dissociated at equilibrium. Since the amount of dissociated HA is given: \(x = 2.50 \times 10^{-5}\) Now, the dissociation equation is as follows: \(HA \rightleftharpoons H^+ + A^-\) At equilibrium: \([H^+] = x\), \([A^-] = x\), \([HA] = \text{moles of HA - x}\) We can now use the equilibrium constant expression: \[K_a = \frac{[H^+][A^-]}{[HA]} = \frac{x^2}{(\text{moles of HA} - x)}\]
04

Solve for x

Put the values of Ka, and HA into the equation: \(1.00 \times 10^{-4} = \frac{(2.50 \times 10^{-5})^2}{(1.00 \times 10^{-4} - 2.50 \times 10^{-5})}\) Solve the equation to find the value of x: \(x = 2.50 \times 10^{-5}\,mol\)
05

Calculate the change in volume

Now we have the new concentration of the dissociated HA, which is: \[\text{New concentration} = \frac{\text{Amount of dissociated HA}}{\text{Amount of HA at equilibrium}} = \frac{2.50 \times 10^{-5}\,mol}{1.00 \times 10^{-4} - 2.50 \times 10^{-5}\,mol}\] To find the new volume, rearrange the equation for volume: \[\text{New volume} = \frac{\text{Amount of HA at equilibrium}}{\text{New concentration}}\] Calculate the new volume: \[\text{New volume} = \frac{1.00 \times 10^{-4} - 2.50 \times 10^{-5}\,mol}{2.50 \times 10^{-5}} = 3.00\,L\]
06

Calculate the amount of water added or removed

Subtract the initial volume (1L) from the new volume (3L) to find the amount of water added or removed: \[\text{Change in volume} = \text{New volume} - \text{Initial volume} = 3.00\,L - 1.00\,L = 2.00\,L\] This means that 2.00 L (or 2000 mL) of water was added to the solution to achieve 25% dissociation of HA at equilibrium.

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.

Equilibrium Constant
The equilibrium constant, denoted as \( K_a \) in this context, is a crucial concept in understanding how acids dissociate in solution. It reveals the ratio of the concentration of the products (ions) to the reactants (undissociated acid) at equilibrium. In mathematical terms, it's expressed as:
  • \[ K_a = \frac{[H^+][A^-]}{[HA]} \]
This formula emphasizes the balance between the dissociated ions (\([H^+]\) and \([A^-]\)) and the undissociated form (\([HA]\)). The greater the \( K_a \), the stronger the acid, indicating a higher tendency to dissociate. Understanding how to use \( K_a \) helps predict acid strength and behavior in different conditions.
Molarity
Molarity is a measure of the concentration of a solute in a solution. It is expressed as moles of solute per liter of solution, denoted as \( M \). In the given problem, the initial molarity of the acid HA is \(1.00 \times 10^{-4} M\). Calculating molarity involves:
  • Defining the volume of the solution.
  • Calculating the number of moles of solute.
The formula for molarity is:
  • \[ ext{Molarity} = \frac{ ext{Moles of solute}}{ ext{Volume of solution in liters}} \]
This concept is essential for determining how much of a substance is present and plays a critical role in calculating changes in volume and concentration during chemical reactions.
Chemical Equilibrium
Chemical equilibrium occurs when a reversible reaction's forward and reverse rates balance out. For an acid like HA, equilibrium is reached when the rate of dissociation into \([H^+]\) and \([A^-]\) equals the rate of recombination. This concept is key because it establishes the conditions under which the amounts of reactants and products remain constant in absence of external changes. At equilibrium, the concentrations satisfy the equilibrium constant expression.
  • The reaction: \( HA \rightleftharpoons H^+ + A^- \)
  • At equilibrium: \([H^+] = [A^-] = x\), \([HA] = ext{Initial moles} - x\)
Understanding equilibrium in processes is crucial for predicting how changes in conditions like pressure, temperature, or concentration can affect the outcome.
Dissociation Percentage
The dissociation percentage measures how much of an acid dissociates into ions in a given solution. In this problem, 25% of HA dissociates. This tells us that only a quarter of the acid molecules separate into \([H^+]\) and \([A^-]\), helping to determine the remaining concentration of \([HA]\) and ensuring proper calculations using \( K_a \).
  • Calculation example: \( ext{Dissociated HA} = 25\% \times ext{Initial moles of HA} \)
  • \[ ext{Percentage} = \frac{ ext{Moles dissociated}}{ ext{Total initial moles}} \times 100\% \]
Understanding dissociation percentages is vital for predicting a solution's pH and the behavior of weak acids in a variety of chemical environments.

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

Rank the following 0.10\(M\) solutions in order of increasing \(\mathrm{pH.}\) a. \(\mathrm{NH}_{3}\) b. KOH c. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) d. \(\mathrm{KCl}\) e. HCl

When determining the pH of a weak acid solution, sometimes the 5\(\%\) rule can be applied to simplify the math. At what \(K_{\mathrm{a}}\) values will a \(1.0-M\) solution of a weak acid follow the 5\(\%\) rule?

Calculate the percent dissociation of the acid in each of the following solutions. a. 0.50\(M\) acetic acid b. 0.050\(M\) acetic acid c. 0.0050\(M\) acetic acid d. Use Le Ch芒telier's principle to explain why percent dissociation increases as the concentration of a weak acid decreases. e. Even though the percent dissociation increases from solutions a to \(c,\) the \(\left[\mathrm{H}^{+}\right]\) decreases. Explain.

Quinine \(\left(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{N}_{2} \mathrm{O}_{2}\right)\) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine, \(\mathrm{p} K_{\mathrm{b}_{1}}=5.1\) and \(\mathrm{p} K_{\mathrm{b}_{2}}=9.7\left(\mathrm{p} K_{\mathrm{b}}=-\log K_{\mathrm{b}}\right) .\) Only 1 g quinine will dissolve in 1900.0 \(\mathrm{mL}\) of solution. Calculate the pH of a saturated aqueous solution of quinine. Consider only the reaction \(\mathrm{Q}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{QH}^{+}+\mathrm{OH}^{-}\) described by \(\mathrm{p} K_{\mathrm{b}_{1}},\) where \(\mathrm{Q}=\) quinine.

A solution of formic acid (HCOOH, \(K_{\mathrm{a}}=1.8 \times 10^{-4} )\) has a \(\mathrm{pH}\) of 2.70 . Calculate the initial concentration of formic acid in this solution.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

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.