/*! 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 151 Solubility of \(\mathrm{AgCl}\) ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 151

Solubility of \(\mathrm{AgCl}\) in \(\mathrm{H}_{2} \mathrm{O}, 0.02 \mathrm{M} \mathrm{MgCl}_{2}, 0.02 \mathrm{M}\) \(\mathrm{NaCl}\), and \(0.05 \mathrm{M} \mathrm{AgNO}_{3}\) are \(\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}^{2}\), and \(\mathrm{S}_{4}\) respectively then: (a) \(\mathrm{S}_{1}>\mathrm{S}_{3}>\mathrm{S}_{5}>\mathrm{S}_{4}\) (b) \(\mathrm{S}_{1}>\mathrm{S}_{3}>\mathrm{S}_{2}>\mathrm{S}_{4}\) (c) \(S_{1}>S_{2}=S_{3}>S_{4}\) (d) \(\mathrm{S}_{4}>\mathrm{S}_{3}>\mathrm{S}_{1}>\mathrm{S}_{2}\)

Short Answer

Expert verified
Option (b) is correct: \( S_1 > S_3 > S_2 > S_4 \).

Step by step solution

01

Understand the Problem

We need to compare the solubility of AgCl in different solutions: pure water (S_1), 0.02 M MgCl2 (S_2), 0.02 M NaCl (S_3), and 0.05 M AgNO3 (S_4). The solubility product (Ksp) for AgCl can be used to calculate solubility in these solutions.
02

Establish the Ksp for AgCl

The solubility product constant (Ksp) for AgCl is given by the equation:\[ K_{sp} = [Ag^+][Cl^-] = 1.8 \times 10^{-10} \] This value is used to calculate the solubility (S) in different conditions by determining the concentration of ions in solution.
03

Calculate Solubility in Pure Water (S1)

In pure water, AgCl dissociates into Ag+ and Cl-. The solubility, S_1, is represented by:\[ S_1 = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} \]Solving this gives:\[ S_1 = 1.34 \times 10^{-5} \text{ M} \]
04

Solubility in 0.02 M NaCl (S3)

In NaCl solution, Cl- from NaCl contributes to the ionic concentration, shifting the dissolution equilibrium and reducing solubility. The chloride ion concentration is 0.02 M. Here,\[ K_{sp} = [Ag^+](0.02 + [Cl^-]) \approx 0.02[Ag^+] \]Solving for [Ag+]:\[ [Ag^+] = \frac{K_{sp}}{0.02} = \frac{1.8 \times 10^{-10}}{0.02} = 9 \times 10^{-9} \text{ M} \]Thus, \( S_3 = 9 \times 10^{-9} \text{ M} \).
05

Solubility in 0.02 M MgCl2 (S2)

MgCl2 dissociates to provide two Cl- ions for every formula unit, increasing the chloride concentration more than NaCl does. This provides 0.04 M Cl-, so:\[ K_{sp} = [Ag^+](0.04 + [Cl^-]) \approx 0.04[Ag^+] \]Solving for [Ag+]:\[ [Ag^+] = \frac{K_{sp}}{0.04} = \frac{1.8 \times 10^{-10}}{0.04} = 4.5 \times 10^{-9} \text{ M} \]Thus, \( S_2 = 4.5 \times 10^{-9} \text{ M} \).
06

Solubility in 0.05 M AgNO3 (S4)

AgNO3 provides excess Ag+ ions, decreasing the amount of Cl- needed to reach equilibrium:\[ 1.8 \times 10^{-10} = (0.05 + [Ag^+]) [Cl^-] \approx 0.05[Cl^-] \]Solving for [Cl-]:\[ [Cl^-] = \frac{K_{sp}}{0.05} = \frac{1.8 \times 10^{-10}}{0.05} = 3.6 \times 10^{-9} \text{ M} \]Thus, \( S_4 = 3.6 \times 10^{-9} \text{ M} \).
07

Compare Solubilities

Now, compare the solubilities:- \( S_1 = 1.34 \times 10^{-5} \text{ M} \)- \( S_3 = 9 \times 10^{-9} \text{ M} \)- \( S_2 = 4.5 \times 10^{-9} \text{ M} \)- \( S_4 = 3.6 \times 10^{-9} \text{ M} \)Thus, we have: \( S_1 > S_3 > S_2 > S_4 \). Therefore, option (b) is correct.

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.

Solubility Product Constant (Ksp)
The Solubility Product Constant, often referred to as \(K_{sp}\), is fundamental in understanding the solubility of ionic compounds like silver chloride (AgCl) in water. \(K_{sp}\) represents the maximum product of the concentrations of the dissociated ions that can exist in a saturated solution at equilibrium. For AgCl, this is expressed as:\[ K_{sp} = [Ag^+][Cl^-] = 1.8 \times 10^{-10} \]This constant value is specific to each compound and temperature, helping us predict whether a precipitate will form under certain conditions. When the ion product exceeds the \(K_{sp}\), the solution becomes supersaturated, leading to precipitation.
Effect of Common Ion on Solubility
The presence of a common ion can significantly affect the solubility of a compound, an effect which is quite notable in AgCl when dissolved in solutions containing chloride ions. According to Le Chatelier鈥檚 principle, adding more of an ion that is a part of the equilibrium shifts the equilibrium to the left, favoring the formation of the solid AgCl and reducing its solubility.
For instance:
  • In a 0.02 M NaCl solution, the concentration of Cl- ions from NaCl reduces the solubility of AgCl. The equation \(K_{sp} = [Ag^+](0.02 + [Cl^-])\) considers this added chloride concentration.
  • Similarly, in a 0.02 M MgCl鈧� solution, since MgCl鈧� produces more chloride ions (0.04 M from MgCl鈧� alone), the solubility of AgCl is even further decreased.
Here, you see the impactful nature of the common ion effect in limiting the solubility of solutes in ionic equilibrium situations.
Ionic Equilibrium
Ionic equilibrium refers to the balanced status within a solution where the rates of the forward and reverse reactions of dissolution and precipitation are equal. This state of dynamic balance is crucial when calculating the solubility of ionic compounds like AgCl in various solutions.
Consider that:
  • In pure water, AgCl dissolves until equilibrium is reached, where the rate of dissolving equals the rate of precipitation, giving a solubility derived directly from the \(K_{sp}\).
  • In solutions containing common ions, this equilibrium is influenced by the additional ions. For example, in a solution with a higher Ag+ concentration like AgNO鈧�, less Cl- is needed to reach equilibrium, reducing AgCl solubility.
Understanding ionic equilibrium allows you to predict solubility changes under varying conditions.
Chemical Equilibrium Calculations
Chemical equilibrium calculations are essential for predicting how altering conditions can shift equilibria, especially in solubility-related scenarios. These calculations help one understand the concentrations of ions necessary to maintain equilibrium in different solvents.
Let's consider:
  • For pure water, the solubility of AgCl can be calculated by taking the square root of \(K_{sp}\), resulting in a straightforward determination of the solubility (S鈧�).
  • In solutions like NaCl or MgCl鈧�, equilibrium calculations account for additional chloride concentrations, derived from the salts themselves. This forms a necessity to adjust the calculated concentrations of Ag+ and Cl- to ensure the ion-product equals the \(K_{sp}\).
  • Similarly, when AgNO鈧� is present, it increases the Ag+ concentration, shifting the equilibrium and requiring less Cl- to satisfy the \(K_{sp}\) balance.
By leveraging these calculations, you can predict the influence of different ions and conditions on the solubility of compounds in equilibrium systems.

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

In the reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\), the equilibrium concentrations of \(\mathrm{PCl}_{5}\) and \(\mathrm{PCl}_{3}\) are \(0.4\) and \(0.2\) mole/litre respectively. If the value of \(\mathrm{K}_{\mathrm{c}}\) is \(0.5\), what is the concentration of \(\mathrm{Cl}_{2}\) in mole/litre: (a) \(2.0\) (b) \(1.5\) (c) \(1.0\) (d) \(0.5\)

At equilibrium total number of moles for the reaction \(2 \mathrm{HI} \rightleftharpoons \mathrm{H}_{2}+\mathrm{I}_{2}\), if \(\alpha\) is degree of dissociation, are: (a) 2 (b) \(2-\alpha\) (c) 1 (d) \(\mathrm{I}-\alpha\)

The law of chemical equilibrium was first given by: (a) Guldberg (b) Boyle (c) Waage (d) Both (a) and (c)

At \(100^{\circ} \mathrm{C}\) the vapour density of nitrogen peroxide \(\left(\mathrm{N}_{2} \mathrm{O}_{4}\right)\) is \(26.8\). The percentage dissociation into \(\mathrm{NO}_{2}\) molecules is: (a) \(71.64 \%\) (b) \(61.57 \%\) (c) \(83.56 \%\) (d) \(67.39 \%\)

In which of the following reactions, the concentration of reactant is equal to concentration of product at equilibrium \((\mathrm{K}=\) equilibrium constant \()\) : (a) \(\mathrm{A} \rightleftharpoons \mathrm{B} ; \mathrm{K}=0.01\) (b) \(\mathrm{R} \rightleftharpoons \mathrm{P} ; \mathrm{K}=1\) (c) \(\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}=10\) (d) \(\mathrm{L} \rightleftharpoons \mathrm{J} ;=0.025\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Chemistry Branches

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.