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

Kinetics

Chemical Reactions

Making Measurements

Chemical Analysis

The Earths Atmosphere

Chemistry Branches

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