/*! 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 57 Calculate the hydronium ion conc... [FREE SOLUTION] | 91影视

91影视

Calculate the hydronium ion concentration and \(\mathrm{pH}\) in a \(0.20 \mathrm{M}\) solution of ammonium chloride, \(\mathrm{NH}_{4} \mathrm{Cl}\).

Short Answer

Expert verified
[H鈧僌鈦篯 for 0.20 M NH鈧凜l is approximately calculated using K鈧 and then pH is found from \(-\log[H鈧僌鈦篯\).

Step by step solution

01

Understand the Problem

We have a 0.20 M solution of ammonium chloride (NH鈧凜l), which is a salt formed from a weak base (NH鈧凮H) and a strong acid (HCl). This solution will act as an acidic solution because NH鈧勨伜 can donate a proton to water, forming hydronium ions, which we need to calculate.
02

Write the Reaction Equation

NH鈧凜l dissociates in water to produce NH鈧勨伜 ions. The ammonium ions then react with water to form hydronium ions: NH鈧勨伜 + H鈧侽 鈬 NH鈧 + H鈧僌鈦. This equilibrium reaction will help us determine the H鈧僌鈦 concentration.
03

Use the Equilibrium Expression

The equilibrium constant for the reaction is the ionization constant of NH鈧勨伜, represented as K鈧. First, find K鈧 using the relation K鈧 = \( \dfrac{K_w}{K_b} \), where K_w is the ion-product constant for water \(1.0 \times 10^{-14} \) and K_b is the base ionization constant for NH鈧凮H. Use given or standard tables to find these constants.
04

Set Up the Equilibrium Table

Create a table showing initial concentrations (initially, [NH鈧勨伜] = 0.20 M and [H鈧僌鈦篯 = [NH鈧僝 = 0), the change in concentrations, and equilibrium concentrations. Let 'x' be the change in concentration to form H鈧僌鈦 and NH鈧. Therefore, equilibrium concentrations are [NH鈧勨伜] = 0.20 - x, [H鈧僌鈦篯 = x, and [NH鈧僝 = x.
05

Solve the Equilibrium Equation

Using the K鈧 expression for NH鈧勨伜: \( K鈧 = \dfrac{[NH鈧僝[H鈧僌鈦篯}{[NH鈧勨伜]} \) Substitute the equilibrium concentrations into the expression: \( K鈧 = \dfrac{x^2}{0.20-x} \). Assuming 0.20-x 鈮 0.20 (because x is small), solve for x to find [H鈧僌鈦篯.
06

Calculate pH

Once the [H鈧僌鈦篯 is determined, use the formula for pH: \( ext{pH} = -\log[H鈧僌鈦篯 \). Evaluate the logarithm to find the pH of the solution.

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.

Hydronium Ion Concentration
When working with solutions like ammonium chloride, it is crucial to calculate the hydronium ion concentration. Ammonium chloride, with a concentration of 0.20 M, dissociates fully in water to form ammonium ions (NH鈧勨伜). These ammonium ions can interact with water molecules to form hydronium ions (H鈧僌鈦). This reaction can be represented as:
  • NH鈧勨伜 + H鈧侽 鈬 NH鈧 + H鈧僌鈦
The hydronium ion concentration is directly involved in determining the acidity of the solution. In most cases, the higher the H鈧僌鈦 concentration, the more acidic the solution. By establishing the equilibrium conditions of this reaction, we can find the concentration of hydronium ions to further calculate the pH.
Equilibrium Expression
An equilibrium expression helps quantify the relationship between the concentrations of reactants and products in a chemical reaction at equilibrium. For the reaction involving NH鈧勨伜 and water:
  • NH鈧勨伜 + H鈧侽 鈬 NH鈧 + H鈧僌鈦
The equilibrium constant K鈧 is used to express this balance and is given by:
  • \[ K鈧 = \dfrac{[NH鈧僝[H鈧僌鈦篯}{[NH鈧勨伜]} \]
Knowing the initial concentrations and the changes occurring during the reaction assists in setting up an equilibrium table. This table estimates how concentrations shift, enabling us to solve for the exact concentration of hydronium ions at equilibrium. Understanding and manipulating the equilibrium expression is key in solving for unknown concentrations, especially in establishing pH.
Ammonium Chloride Solution
Ammonium chloride (NH鈧凜l) is a white crystalline solid that is highly soluble in water. When dissolved, it dissociates into ammonium ions (NH鈧勨伜) and chloride ions (Cl鈦). Ammonium chloride solutions tend to be acidic due to the weak base (NH鈧凮H) and strong acid (HCl) from which it originates. The balanced dissociation equation in water is:
  • NH鈧凜l 鈫 NH鈧勨伜 + Cl鈦
The solution then influences the pH by producing hydronium ions through the action of ammonium ions interacting with water. Understanding the nature and behavior of an ammonium chloride solution allows us to predict if a solution will be acidic, neutral, or basic.
Ionization Constant
The ionization constant, or K鈧, is crucial when dealing with weak acids or bases and their salts like ammonium chloride. It represents the strength of the acid or base in a solution, showing how readily it donates protons. For NH鈧勨伜, you can find K鈧 using the formula:
  • \[ K鈧 = \dfrac{K_w}{K_b} \]
Where K鈧 is the dissociation constant for NH鈧勨伜, K_w is the ion-product constant for water ( \(1.0 \times 10^{-14} \)), and K_b is the base ionization constant for NH鈧凮H. This relationship translates the basicity to acidity for effective calculations in acidic solutions. Calculating and understanding the ionization constant allows chemists to anticipate how acidic or basic a solution will be, based on its components.

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

An aqueous solution has a pH of \(3.75 .\) What is the hydronium ion concentration of the solution? Is it acidic or basic?

In each of the following acid-base reactions, identify the Bronsted acid and base on the left and their conjugate partners on the right. (a) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) \rightleftarrows\) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq})\) (b) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{aq})+\mathrm{HSO}_{4}^{-}(\mathrm{aq}) \rightleftarrows\) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\) (c) \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightleftarrows\) \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)\)

What is the \(\mathrm{pH}\) of a \(0.0015 \mathrm{M}\) solution of \(\mathrm{Ba}(\mathrm{OH})_{2} ?\)

A 0.025 M solution of hydroxylamine has a pH of 9.11. What is the value of \(K_{\mathrm{b}}\) for this weak base? $$\mathrm{H}_{2} \mathrm{NOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightleftarrows \mathrm{H}_{3} \mathrm{NOH}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$$

Decide whether each of the following substances should be classified as a Lewis acid or a Lewis base. (a) \(\mathrm{BCl}_{3}\) (Hint: Draw the electron dot structure.) (b) \(\mathrm{H}_{2} \mathrm{NNH}_{2},\) hydrazine (Hint: Draw the electron dot structure.) (c) the reactants in the reaction $$\mathrm{Ag}^{+}(\mathrm{aq})+2 \mathrm{NH}_{3}(\mathrm{aq}) \rightleftarrows\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(\mathrm{aq})$$

See all solutions

Recommended explanations on Chemistry Textbooks

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.