Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: Problem 68

In the presence of \(\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\) forms the complex ion \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+} .\) If the equilibrium concentrations of \(\mathrm{Cu}^{2+}\) and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) are $1.8 \times 10^{-17} \mathrm{M}\( and \)1.0 \times 10^{-3} \mathrm{M},\( respectively, in a \)1.5-M \mathrm{NH}_{3}$ solution, calculate the value for the overall formation constant of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}.\) $$\mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \qquad K_{\mathrm{overall}}=?$$

Short Answer

Expert verified
To find the overall formation constant, \(K_{\mathrm{overall}}\), for the given reaction, use the equilibrium constant expression: \[K_{\mathrm{overall}} = \frac{\left[\mathrm{Cu}(\mathrm{NH}_{3})_{4}^{2+}\right]}{\left[\mathrm{Cu^{2+}}\right]\left[\mathrm{NH_{3}}\right]^4}\] Substitute the given concentrations into the equation: \[K_{\mathrm{overall}} = \frac{1.0 \times 10^{-3}}{(1.8 \times 10^{-17})\left(1.5\right)^4} \approx 2342812.5\] So, the overall formation constant, \(K_{\mathrm{overall}}\), for \(\mathrm{Cu}(\mathrm{NH}_{3})_{4}^{2+}\) is approximately \(2.34 \times 10^6\).

Step by step solution

01

Substitute the given concentrations into the equation

We have the following values: \(\begin{aligned} [\mathrm{Cu}(\mathrm{NH}_{3})_{4}^{2+}] &= 1.0 \times 10^{-3} \ \text{M}\\ [\mathrm{Cu^{2+}}] &= 1.8 \times 10^{-17} \ \text{M}\\ [\mathrm{NH_{3}}] &= 1.5 \ \text{M} \end{aligned}\) Now, we can substitute these values into the equilibrium constant expression for \(K_{\mathrm{overall}}\): \[K_{\mathrm{overall}} = \frac{\left[\mathrm{Cu}(\mathrm{NH}_{3})_{4}^{2+}\right]}{\left[\mathrm{Cu^{2+}}\right]\left[\mathrm{NH_{3}}\right]^4} = \frac{1.0 \times 10^{-3}}{(1.8 \times 10^{-17})\left(1.5\right)^4}\]
02

Calculate the value of \(K_{\mathrm{overall}}\)

To find the value of \(K_{\mathrm{overall}}\), we can now perform the calculation: \[K_{\mathrm{overall}} = \frac{1.0 \times 10^{-3}}{(1.8 \times 10^{-17})\left(1.5\right)^4} \approx 2342812.5\] So, the overall formation constant, \(K_{\mathrm{overall}}\), for \(\mathrm{Cu}(\mathrm{NH}_{3})_{4}^{2+}\) is approximately \(2.34 \times 10^6\).

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

Will a precipitate form when 100.0 \(\mathrm{mL}\) of \(4.0 \times 10^{-4} M\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is added to 100.0 \(\mathrm{mL}\) of \(2.0 \times 10^{-4} \mathrm{MNaOH}\)?

The salt MX has a solubility of $3.17 \times 10^{-8} \mathrm{mol} / \mathrm{L}\( in a solution with \)\mathrm{pH}=0.000 .\( If \)K_{\mathrm{a}}$ for \(\mathrm{HX}\) is \(1.00 \times 10^{-15}\) , calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{MX}\) .

A solution is prepared by adding 0.10 mole of \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6} \mathrm{Cl}_{2}\) to 0.50 \(\mathrm{L}\) of \(3.0 M\) \(\mathrm{NH}_{3}\) . Calculate \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]\) and \(\left[\mathrm{Ni}^{2+}\right]\) in this solution. \(K_{\text { overall }}\) for \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) is \(5.5 \times 10^{8} .\) That is, $$5.5 \times 10^{8}=\frac{\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\right]}{\left[\mathrm{Ni}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{6}}$$ for the overall reaction $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$

Silver cyanide \((\mathrm{AgCN})\) is an insoluble salt with \(K_{\mathrm{sp}}=2.2 \times 10^{-12}\) . Compare the effects on the solubility of silver cyanide by addition of \(\mathrm{HNO}_{3}(a q)\) or by addition of \(\mathrm{NH}_{3}(a q).\)

Sulfide precipitates are generally grouped as sulfides insoluble in acidic solution and sulfides insoluble in basic solution. Explain why there is a difference between the two groups of sulfide precipitates.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

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