Chapter 12: Q16P-e (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

(e) For 55 mL the value ofpCu²⁺is 17.69.

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

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

Equivalence point=50 mL

Determine the value of pCu2+

Past equivalence point (55mL) we can calculate

$\begin{matrix} E D T A = (\frac{5}{105} \times 0.001 M) \\ = 4.76 \times 10^{- 5} M \\ C u Y^{2 -} = (\frac{50}{105} \times 0.001 M) \\ = 4.76 \times 10^{- 4} M \end{matrix}$

$\begin{matrix} K_{f}^{'} = \frac{[C u Y^{2 -}]}{[C u^{2 +}] [E D T A]} \\ 4.88 \times 10^{18} = \frac{[4.76 \times 10^{- 4}]}{[C u^{2 +}] [4.76 \times 10^{- 5}]} \\ C u^{2 +} = 2.05 \times 10^{- 18} M \end{matrix}$

Therefore, the value of pCu²⁺

^{$\begin{matrix} p C u^{2 +} = - \log (C u^{2 +}) \\ = - \log (2.05 \times 10^{- 18}) \\ = 17.69 \end{matrix}$}

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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

Making Measurements

Chemical Reactions

Ionic and Molecular Compounds

Organic Chemistry

Physical Chemistry

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