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

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Chapter 12: Q15P-b (page 283)

According to Appendix I, Cu²⁺ forms two complexes with acetate:
$\begin{matrix} C u^{2 +} + C H_{3} C O_{2}^{鈭�} 鈬� C u {(C H_{3} C O_{2})}^{+} 鈥夆赌夆赌夆赌夆赌夆赌夆赌� 尾_{1} (= K_{1}) \\ C u^{2 +} + 2 C H_{3} C O_{2}^{鈭�} 鈬� C u {(C H_{3} C O_{2})}_{2} 鈥夆赌夆赌夆赌夆赌夆赌夆赌� 尾_{2} \end{matrix}$
(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

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

Step by step solution

Reaction and their equilibrium constant

Equations given

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

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

Information Given

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)

$伪_{C u^{2 +}} = \frac{1}{1 + 尾_{1} [L] + 尾_{2} {[L]}^{2}}$

Determine the fraction of copper ion

$\begin{matrix} 伪_{C u^{2 +}} = \frac{1}{1 + 尾_{1} [L] + 尾_{2} {[L]}^{2}} \\ = \frac{1}{1 + 10^{2.23} 脳 [0.1] + 10^{3.63} 脳 {[0.1]}^{2}} \\ = 0.017 \end{matrix}$

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

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Short Answer

Step by step solution

Reaction and their equilibrium constant

Information Given

Determine the fraction of copper ion

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