Chapter 12: Q16P-d (page 284)

Calculate pCu²⁺ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu²⁺ with 0.00100 M EDTA at pH 11.00 in a solution with [NH₃] fixed at 1.00 M:
(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL

Short Answer

Expert verified

(d) For 50 mL the value ofpCu²⁺is 17.02.

Step by step solution

Introduction

Equations and data obtained in order to proceed for calculation are as follows

$\begin{matrix} C u^{2 +} + Y^{4 -} ⇌ C u Y^{2 -} \\ K_{f} = 10^{18.78} = 6.03 \times 10^{18} \\ A t p H 11 α_{Y^{4 -}} = 0.81 (T a b l e 12 - 1) \\ \log β_{1} = 3.99 \\ \log β_{2} = 7.33 \\ \log β_{3} = 10.06 \\ \log β_{4} = 12.03 \end{matrix}$

The beta(β) values were obtained from appendix-1 for Cu²⁺ and NH₃

Determine equilibrium constant

$\begin{matrix} α_{C u^{2 +}} = \frac{1}{1 + β_{1} (1.00) + β_{2} {(1.00)}^{2} + β_{3} {(1.00)}^{3} + β_{4} {(1.00)}^{4}} \\ = 9.23 \times 10^{- 13} \\ K_{f}^{'} = α_{Y^{4 -}} K_{f} \\ = 0.81 \times 6.03 \times 10^{18} \\ = 4.88 \times 10^{18} \\ K_{f}^{"} = α_{C u^{2 +}} \times K_{f}^{'} \\ = 9.23 \times 10^{- 13} \times 4.88 \times 10^{18} \\ = 4.51 \times 10^{6} \end{matrix}$

Equivalence point=50 mL

Determine the value of pCu2+

At equivalence point (50mL) the following can be written

$\begin{matrix} K_{f}^{"} = \frac{0.0005 - x}{x^{2}} \\ \frac{0.0005 - x}{x^{2}} = 4.51 \times 10^{6} \\ x = 1.04 \times 10^{- 5} M \\ C_{C u^{2 +}} = 1.04 \times 10^{- 5} M \end{matrix}$

$\begin{matrix} [C u^{2 +}] = α_{C u^{2 +}} \times C_{C u^{2 +}} \\ = (9.23 \times 10^{- 13} \times 1.04 \times 10^{- 5}) M \\ = 9.62 \times 10^{- 18} M \end{matrix}$

Therefore, the value of pCu²⁺

^{$\begin{matrix} p C u^{2 +} = - \log (C u^{2 +}) \\ = - \log (9.62 \times 10^{- 18}) \\ = 17.02 \end{matrix}$}

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Calculate pCu²⁺ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu²⁺ with 0.00100 M EDTA at pH 11.00 in a solution with [NH₃] fixed at 1.00 M:
(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL

Short Answer

Step by step solution

Introduction

Determine equilibrium constant

Determine the value of pCu2+

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Calculate pCu2+ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu2+ with 0.00100 M EDTA at pH 11.00 in a solution with [NH3] fixed at 1.00 M:(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL

Short Answer

Step by step solution

Introduction

Determine equilibrium constant

Determine the value of pCu2+

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemical Analysis

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Nuclear Chemistry

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Study anywhere. Anytime. Across all devices.

Calculate pCu²⁺ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu²⁺ with 0.00100 M EDTA at pH 11.00 in a solution with [NH₃] fixed at 1.00 M:
(a) 0 mL(b) 1.00 mL (c) 45.00 mL (d) 50.00 mL (e) 55.00 mL