Q15P According to Appendix I, Cu2+ fo... [FREE SOLUTION]

91影视

Chapter 12: Q15P (page 283)

According to Appendix I, Cu²⁺ forms two complexes with acetate:
$\begin{array}{l} C u^{2 +} + C H_{3} C O_{2}^{鈭�} 鈬� C u {(C H_{3} C O_{2})}^{+} 鈥夆赌夆赌夆赌夆赌夆赌夆赌� 尾_{1} (= K_{1}) \\ C u^{2 +} + 2 C H_{3} C O_{2}^{鈭�} 鈬� C u {(C H_{3} C O_{2})}_{2} 鈥夆赌夆赌夆赌夆赌夆赌夆赌� 尾_{2} \end{array}$
(a) Referring to Box 6-2, find K₂ for the reaction
$C u {(C H_{3} C O_{2})}^{+} + C H_{3} C O_{2}^{鈭�} 鈬� C u {(C H_{3} C O_{2})}_{2} (a q) 鈥夆赌夆赌� K_{2}$
(b) Consider 1.00 L of solution prepared by mixing 1.00 脳 10^-4 mol Cu(ClO₄)₂ and 0.100 mol CH₃CO₂Na. Use Equation 12-16 to find the fraction of copper in the form Cu²⁺

Short Answer

Expert verified

(a) The value of K₂ for the reaction will be 25.12

(b) The fraction of copper in the form Cu²⁺ will be 0.017

Step by step solution

Information Given

Equations given

$\begin{matrix} C u^{2 +} + C H_{3} C O_{2}^{鈭�} 鈬� C u {(C H_{3} C O_{2})}^{+} 鈥夆赌夆赌夆赌夆赌夆赌夆赌� 尾_{1} (= K_{1}) \\ C u^{2 +} + 2 C H_{3} C O_{2}^{鈭�} 鈬� C u {(C H_{3} C O_{2})}_{2} 鈥夆赌夆赌夆赌夆赌夆赌夆赌� 尾_{2} \\ C u {(C H_{3} C O_{2})}^{+} + C H_{3} C O_{2}^{鈭�} 鈬� C u {(C H_{3} C O_{2})}_{2} (a q) 鈥夆赌夆赌� K_{2} \end{matrix}$

From Appendix I the following values were obtained for calculation

$\begin{array}{l} \log 鈥� 尾_{1} = 2.23 \\ \log 鈥� 尾_{2} = 3.63 \end{array}$

As per equation 12-16 we can write

Fraction of free metal ion(copper ion)

Determine the expression for K2

$\begin{matrix} 尾_{1} = \frac{[C u {(C H_{3} C O_{2})}^{+}]}{[C H_{3} C O_{2}^{鈭�}] [C u^{2 +}]} \\ 尾_{2} = \frac{[C u {(C H_{3} C O_{2})}_{2} 鈥夆赌�]}{{[C H_{3} C O_{2}^{鈭�}]}^{2} [C u^{2 +}]} \\ K_{2} = \frac{[C u {(C H_{3} C O_{2})}_{2} 鈥夆赌�]}{[C u {(C H_{3} C O_{2})}^{+}] [C H_{3} C O_{2}^{鈭�}]} \\ = \frac{[C u {(C H_{3} C O_{2})}_{2} 鈥夆赌�]}{{[C H_{3} C O_{2}^{鈭�}]}^{2} [C u^{2 +}]} 脳 \frac{[C H_{3} C O_{2}^{鈭�}] [C u^{2 +}]}{[C u {(C H_{3} C O_{2})}^{+}]} \\ = 尾_{2} 脳 \frac{1}{尾_{1}} \\ = \frac{尾_{2}}{尾_{1}} \end{matrix}$

a) Determine the value of K2

$\begin{matrix} 尾_{2} = K_{1} K_{2} \\ 尾_{2} = 尾_{1} K_{2} \\ K_{2} = \frac{尾_{2}}{尾_{1}} \\ K_{2} = \frac{10^{3.63}}{10^{2.23}} \\ = 10^{1.4} \\ = 25.12 \end{matrix}$

The value of K₂ for the reaction will be 25.12

b) Determine the fraction of copper ion

The fraction of copper in the form Cu²⁺ will be 0.017

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!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Short Answer

Step by step solution

Information Given

Determine the expression for K2

a) Determine the value of K2

b) Determine the fraction of copper ion

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Making Measurements

Kinetics

Chemical Reactions

Inorganic Chemistry

Organic Chemistry

Study anywhere. Anytime. Across all devices.