Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 80

Calculate the concentrations of \(\mathrm{Cd}^{2+}, \mathrm{Cd}(\mathrm{CN})_{4}^{2-}\) and \(\mathrm{CN}^{-}\) at equilibrium when \(0.50 \mathrm{~g}\) of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) dissolves in \(5.0 \times 10^{2} \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{NaCN}\).

Short Answer

Expert verified
The equilibrium concentrations of \(\mathrm{Cd^{2+}}\), \(\mathrm{CN^-}\) and \(\mathrm{Cd(CN)_4^{2-}}\) are \(4.22 \times 10^{-3} - x M\), \(0.5 - 4x M\), and \(x M\) respectively, where x is obtained by solving the formation constant equation

Step by step solution

01

Calculate initial concentration of Cd2+

First, we compute the molar mass of \(\mathrm{Cd(NO_3)_2}\) which is 236.42 g/mol. Now we can find the moles of \(\mathrm{Cd(NO_3)_2}\) using moles = mass/molar mass = 0.50 g / 236.42 g/mol = \(2.11 \times 10^{-3}\) mol. The volume of the solution is 500.0 mL which equals 0.500 L. Thus we can find the initial concentration of \(\mathrm{Cd^{2+}}\) as moles/volume which is \(2.11 \times 10^{-3}\) mol / 0.500 L = \(4.22 \times 10^{-3} M\)
02

Stoichiometric Relationships

Next, we write the net ionic equation that describes this equilibrium process: \(\mathrm{Cd^{2+}} + 4\mathrm{CN^-} \leftrightarrow \mathrm{Cd(CN)_4^{2-}}\). From the stoichiometry of this equilibrium, we know that, for each \(\mathrm{Cd^{2+}}) ion that reacts, four \(\mathrm{CN^-}\) ions are consumed and one \(\mathrm{Cd(CN)_4^{2-}}\) ion is produced. As such, the change in \(\mathrm{Cd^{2+}}\) concentration can be taken as -x, for \(\mathrm{CN^-}\) is -4x and for \(\mathrm{Cd(CN)_4^{2-}}\) is +x.
03

Finding equilibrium concentrations

Thus, the equilibrium concentration of \(\mathrm{Cd^{2+}}) is \(4.22 \times 10^{-3} - x M\), for \(\mathrm{CN^-}\) is \(0.5 - 4x M\) and for \(\mathrm{Cd(CN)_4^{2-}}\) is x M. Lastly, we use the formation constant, Kf, for \(\mathrm{Cd(CN)_4^{2-}}\) which is given as \(1.0 \times 10^{19}\), and write out an expression for Kf using these concentrations. Solving this equation gives us the value of x, giving us the equilibrium concentrations of all ions.

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

When lemon juice is squirted into tea, the color becomes lighter. In part, the color change is due to dilution, but the main reason for the change is an acid-base reaction. What is the reaction? (Hint: Tea contains "polyphenols" which are weak acids and lemon juice contains citric acid.)

A student carried out an acid-base titration by adding \(\mathrm{NaOH}\) solution from a buret to an Erlenmeyer flask containing HCl solution and using phenolphthalein as indicator. At the equivalence point, she observed a faint reddish-pink color. However, after a few minutes, the solution gradually turned colorless. What do you suppose happened?

What is the pH of the buffer \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{HPO}_{4} /\) \(0.15 M \mathrm{KH}_{2} \mathrm{PO}_{4} ?\)

A volume of \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) is titrated against a \(0.100 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) solution added to it from a buret. Calculate the \(\mathrm{pH}\) values of the solution (a) after \(10.0 \mathrm{~mL}\) of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) solution have been added, (b) after \(25.0 \mathrm{~mL}\) of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) solution have been added, (c) after \(35.0 \mathrm{~mL}\) of \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) solution have been added.

Amino acids are building blocks of proteins. These compounds contain at least one amino group \(\left(-\mathrm{NH}_{2}\right)\) and one carboxyl group \((-\mathrm{COOH})\) Consider glycine \(\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\right) .\) Depending on the pH of the solution, glycine can exist in one of three possible forms: $$ \begin{array}{l} \text { Fully protonated: } \mathrm{NH}_{3}-\mathrm{CH}_{2}-\mathrm{COOH} \\ \text { Dipolar ion: } \mathrm{NH}_{3}-\mathrm{CH}_{2}-\mathrm{COO}^{-} \\ \text {Fully ionized: } \mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{COO}^{-} \end{array} $$ Predict the predominant form of glycine at \(\mathrm{pH} 1.0\), 7.0, and 12.0. The \(\mathrm{p} K_{\mathrm{a}}\) of the carboxyl group is 2.3 and that of the ammonium group \(\left(-\mathrm{NH}_{3}^{+}\right)\) is 9.6
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Physical 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