/*! 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 84 An acid \(\mathrm{HX}\) is 25\(\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 84

An acid \(\mathrm{HX}\) is 25\(\%\) dissociated in water. If the equilibrium concentration of \(\mathrm{HX}\) is \(0.30 \mathrm{M},\) calculate the \(K_{\mathrm{a}}\) value for \(\mathrm{HX}\) .

Short Answer

Expert verified
The Ka value for the acid HX is 0.025.

Step by step solution

01

Write the dissociation reaction for the acid

For a generic acid, HX, the dissociation reaction in water is : HX (aq) <=> H+ (aq) + X- (aq)
02

Determine the initial, change, and equilibrium concentrations of all species in the reaction using the percent dissociation and ICE table

Given that the acid is 25% dissociated, and the total concentration of HX is 0.3 M, we can determine the concentrations of all species in the reaction: Percent dissociation = (dissociated HX / initial HX) * 100 25% = (dissociated HX / 0.3 M) * 100 We can determine the dissociated HX: dissociated HX = 0.25 * 0.3 M = 0.075 M Since HX dissociates into H+ and X-, each dissociated HX contributes to an equal amount of both ions. Therefore, the initial concentrations of H+ and X- are 0 and 0, respectively. Now we can set up the ICE table: ``` Initial Change Equilibrium HX 0.3 M -0.075 M 0.225 M H+ 0 +0.075 M 0.075 M X- 0 +0.075 M 0.075 M ```
03

Write the Ka expression and solve for Ka

We can now write the Ka expression for the reaction. Ka is the equilibrium constant for acids and is defined as the product of the concentrations of the products divided by the concentration of the reactants: \(K_a = \dfrac{[H^+][X^-]}{[HX]}\) Plugging in the equilibrium concentrations from the ICE table: \(K_a = \dfrac{(0.075)(0.075)}{0.225}\) Finally, calculate Ka: \(K_a = \dfrac{0.005625}{0.225} = 0.025 \) So, the Ka value for the acid HX is 0.025.

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 Concentration
Understanding equilibrium concentration in a chemical reaction is crucial because it helps us decode the balance of chemical species involved.
For the given problem, we start with
  • HX at an initial concentration of 0.30 M.
  • At equilibrium, after considering the 25\(\%\) dissociation, we find HX is now 0.225 M.
This change occurs as some of the HX dissociates into H+ and X-. When we talk about equilibrium, it means that all chemical reactions have reached a state where the concentrations of reactants and products remain constant over time.
This does not mean the reactions stop, but the rate of conversion between products and reactants is equal, allowing stable concentrations of each species involved.
Percent Dissociation
Percent dissociation is a measure of how much of the acid molecule has dissociated into ions in solution. It is a way to understand how strong or weak an acid is.
For HX, we know it is 25\(\%\) dissociated in water. This tells us that under the given conditions, 25\(\%\) of the initial acid molecules
  • have released hydrogen ions (H+).
  • formed their corresponding conjugate base (X-).
This percentage helps us calculate how much of the acid has turned into ions.
In our exercise, the calculation showed that 0.075 M of the initial 0.30 M dissociates. Percent dissociation is vital for determining the strength and reactivity of the acid, as a higher percentage often indicates a stronger acid.
ICE Table
An ICE table, representing Initial, Change, and Equilibrium stages, is a simple yet powerful tool for analyzing reactions. For our example, the ICE table is used to break down concentration changes as the reaction progresses:
  • Initial State: Start with initial concentrations: HX = 0.3 M, H+ = 0, X- = 0
  • Change: Due to dissociation, HX decreases by 0.075 M, while H+ and X- increase by the same amount as seen in the table.
  • Equilibrium State: Showing concentrations after reaction reaches balance: HX = 0.225 M, H+ = 0.075 M, X- = 0.075 M

This systematic approach helps visualize what happens to each species involved in a reaction right from the start to the point where no further changes occur. Once known, these values feed into the equilibrium constant expression for calculating \( K_a \), confirming the dissociation characteristics of the acid.

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

Calculate the mass of sodium hydroxide that must be added to 1.00 \(\mathrm{L}\) of \(1.00-M \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) to double the pH of the solution (assume that the added NaOH does not change the volume of the solution).

For the following, mix equal volumes of one solution from Group I with one solution from Group II to achieve the indicated pH. Calculate the pH of each solution. $$\begin{aligned} \text { Group I: } & 0.20 M \mathrm{NH}_{4} \mathrm{Cl}, 0.20 \mathrm{MCl}, 0.20 M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl} \\ & 0.20 M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NHCl} \end{aligned}$$ $$\begin{aligned} \text { Group II: } 0.20 \quad M\quad \mathrm{KOI}, 0.20 \quad\mathrm{M} \quad\mathrm{NaCN}, 0.20\quad \mathrm{M}\quad \mathrm{KOCl}, 0.20 \\ \mathrm{M}\quad \mathrm{NaNO}_{2} \end{aligned}$$ a. the solution with the lowest pH b. the solution with the highest pH c. the solution with the pH closest to 7.00

Calculate the concentration of all species present and the pH of a \(0.020-M\) HF solution.

Calculate the pH of a \(0.10-M \mathrm{CoCl}_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.0 \times 10^{-5} .\)

Hemoglobin (abbreviated Hb) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is pH- dependent. The relevant equilibrium reaction is $$ \mathrm{HbH}_{4}^{4+}(a q)+4 O_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q) $$ Use Le Ch芒telier's principle to answer the following. a. What form of hemoglobin, HbH \(_{4}^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4},\) is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygen-binding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise \(146 .\) ) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary? (Hint: CO, blood levels increase during cardiac arrest.)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic 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.