Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 68

Using the value of \(K_{s p}\) for \(\mathrm{Ag}_{2} \mathrm{~S}, K_{a 1}\) and \(\mathrm{K}_{a 2}\) for \(\mathrm{H}_{2} \mathrm{~S},\) and $K_{f}=1.1 \times 10^{5}\( for \)\mathrm{AgCl}_{2}^{-},$ calculate the equilibrium constant for the following reaction: $$ \mathrm{Ag}_{2} \mathrm{~S}(s)+4 \mathrm{Cl}^{-}(a q)+2 \mathrm{H}^{+}(a q) \rightleftharpoons 2 \mathrm{AgCl}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{~S}(a q) $$

Short Answer

Expert verified
The overall equilibrium constant for the given reaction is \(K = K_{sp}\cdot K_{a1}\cdot K_{a2}\cdot K_{f}^2\). Plugging the given values for \(K_{sp}\), \(K_{a1}\), \(K_{a2}\), and \(K_f\), we can calculate the value of \(K\).

Step by step solution

01

Write down the reactions with their equilibrium constants

The three separate reactions with their given equilibrium constants are: 1. Dissolution of silver sulfide (\(Ag_2S\)): \[Ag_2S(s) \rightleftharpoons 2Ag^+(aq) + S^{2-}(aq)\] Given: \(K_{sp} = [Ag^+]^2[S^{2-}]\) 2. The first dissociation of hydrogen sulfide (\(H_2S\)): \[H_2S(aq) \rightleftharpoons H^+(aq) + HS^-(aq)\] Given: \(K_{a1} = [H^+][HS^-]/[H_2S]\) 3. The second dissociation of hydrogen sulfide: \[HS^-(aq) \rightleftharpoons H^+(aq) + S^{2-}(aq)\] Given: \(K_{a2} = [H^+][S^{2-}]/[HS^-]\) 4. Formation of the silver chloride complex (\(AgCl_2^-\)): \[Ag^+(aq) + 2Cl^-(aq) \rightleftharpoons AgCl_2^-(aq)\] Given: \(K_f = [AgCl_2^-]/([Ag^+][Cl^-]^2)\)
02

Add the equations and find the overall equation

To obtain the overall balanced equation, we will sum up the individual reactions and cancel out the species that appear on both sides of the overall equation: \[Ag_2S(s) + 4Cl^-(aq) + 2H^+(aq) \rightleftharpoons 2AgCl_2^-(aq) + H_2S(aq)\]
03

Calculate the overall equilibrium constant

Since we have the individual equilibrium constants, we can now find the overall equilibrium constant for the reaction. When adding the individual reactions, we multiply their corresponding equilibrium constants. Overall equilibrium constant (\(K\)): \[K = K_{sp}\cdot K_{a1}\cdot K_{a2}\cdot K_{f}\cdot K_f\] (Note that we multiply \(K_f\) twice, as there are two moles of \(AgCl_2^-\) in the overall equation) Now, you can plug in the given values of \(K_{sp}\), \(K_{a1}\), \(K_{a2}\), and \(K_f\) into the equation above and solve for the overall equilibrium constant, \(K\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A 10.0-mL sample of \(0.250 \mathrm{MHNO}_{3}\) solution is titrated with $0.100 M$ KOH solution. Calculate the pH of the solution after the following volumes of base have been added: (a) \(20.0 \mathrm{~mL}\), (b) \(24.9 \mathrm{~mL}\) (c) \(25.0 \mathrm{~mL}\) (d) \(25.1 \mathrm{~mL}\), (e) \(30.0 \mathrm{~mL}\).

A buffer is prepared by adding \(15.0 \mathrm{~g}\) of sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) to \(500 \mathrm{~mL}\) of a \(0.100 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) solution. (a) Determine the pH of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of potassium hydroxide solution are added to the buffer.

A solution containing several metal ions is treated with dilute HCl; no precipitate forms. The pH is adjusted to about 1, and $\mathrm{H}_{2} \mathrm{~S}$ is bubbled through. Again, no precipitate forms. The \(\mathrm{pH}\) of the solution is then adjusted to about 8 . Again, \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled through. This time a precipitate forms. The filtrate from this solution is treated with \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\). No precipitate forms. Which of these metal cations are either possibly present or definitely absent: \(\mathrm{Al}^{3+}, \mathrm{Na}^{+}, \mathrm{Ag}^{+}, \mathrm{Mg}^{2+}\) ?

The acid-base indicator bromcresol green is a weak acid. The yellow acid and blue base forms of the indicator are present in equal concentrations in a solution when the pH is 4.68 . What is the \(\mathrm{p} K_{a}\) for bromcresol green?

What is the \(\mathrm{pH}\) of a solution made by mixing \(0.40 \mathrm{~mol}\) \(\mathrm{NaOH}, 0.25 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{HPO}_{4}\), and \(0.30 \mathrm{~mol} \mathrm{H}_{3} \mathrm{PO}_{4}\) with water and diluting to \(2.00 \mathrm{~L} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Organic 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.

Sign-up for free