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Chapter 12: Q14P (page 283)

State the purpose of an auxiliary complexing agent and give an example of its use.

Short Answer

Expert verified

The bond between metal cation and the auxiliary complexing agent must be strong enough so that precipitation of hydroxide can be prevented as well as without interference with EDTA titration.

NH₃ retains Zn²⁺ in solution at high pH , but easily displaced by EDTA

Step by step solution

01

Introduction

If an auxiliary complexing agent is induced while EDTA titration, formation of hydroxide became restricted. This agent is added if any of the titration solution is a metal cation.

02

Purpose

The bond between metal cation and the auxiliary complexing agent must be strong enough so that precipitation of hydroxide can be prevented as well as without interference with EDTA titration.

03

Example

NH₃ retains Zn²⁺ in solution at high pH , but easily displaced by EDTA

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Most popular questions from this chapter

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

List four methods for detecting the end point of an EDTA Titration

Spreadsheet equation for formation of the complexes ML and ML₂.Consider the titration of metal M (initial concentration = C_M, initial volume = V_M) with ligand L (concentration = C_L, volume added = V_L), which can form 1:1 and 2 : 1 complexes:

$M + L ⇋ ML β_{1} = \frac{[ML]}{[M] [L]} M + 2 L ⇋ {ML}_{2} β_{2} = \frac{[{ML}_{2}]}{[M] {[L]}^{2}}$

Let α_M be the fraction of metal in the form M, α_ML be the fraction in the form ML, and $α_{M L_{2}}$ be the fraction in the form ML₂. Following the derivation in Section 12-5, you could show that these fractions are given by

$α_{M} = \frac{1}{1 + β_{1} [L] + β_{2} [L]^{2}} α_{M} L = \frac{(β_{1} [L])}{1 + β_{1} [L] + β_{2} [L]^{2}} α_{M L_{2}} = \frac{β_{2} [L]^{2}}{1 + β_{1} [L] + β_{2} [L]^{2}}$

The concentrations of ML and ML₂ are

$[ML] = α_{ML} \frac{C_{M} V_{M}}{V_{M} + V_{L}} [ML] = {α_{ML}}_{2} \frac{C_{M} V_{M}}{V_{M} + V_{L}}$

because $\frac{C_{M} V_{M}}{V_{M} + V_{L}}$ is the total concentration of all metal in the solution. The mass balance for ligand is

$[L] + [ML] + 2 [{ML}_{2}] = \frac{C_{L} V_{L}}{V_{M} + V_{L}}$

By substituting expressions for [ML] and [ML₂] into the mass balance, show that the master equation for a titration of metal by ligand is

$ϕ = \frac{C_{L} V_{L}}{C_{M} V_{M}} = \frac{α_{ML} + 2 α_{{ML}_{2}} + \frac{L}{C_{M}}}{1 - \frac{L}{C_{L}}}$

Cyanide solution (12.73 mL) was treated with 25.00 mL of Ni²⁺ solution (containing excess Ni²⁺) to convert the cyanide into tetracyano nickelate (II):

$4 {CN}^{-} + {Ni}^{2 +} \to Ni (CN)_{4}^{2 -}$

Excess Ni²⁺ was then titrated with 10.15 mL of 0.01307 M EDTA. $Ni {(CN)}_{4}^{2 -}$ does not react with EDTA. If 39.35 mL of EDTA were required to react with 30.10 mL of the original Ni²⁺ solution, calculate the molarity of CN^- in the 12.73-mL sample.

Calculate pCu²⁺ (to the 0.01 decimal place) at each of the following points in the titration of 50.0 mL of 0.040 0 M EDTA with 0.080 0 M Cu (NO₃)₂ at pH 5.00: 0.1, 5.0, 10.0, 15.0, 20.0, 24.0, 25.0, 26.0, and 30.0 mL. Make a graph of pCu²⁺ versus volume of titrant.

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