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Chapter 12: Q24P (page 285)

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

Short Answer

Expert verified

The four methods for detecting the end point of an EDTA titration are

a) With metal ion indicators

b) With a glass electrode

c) With a mercury electrode

d) With an ion selective electrode.

Step by step solution

01

EDTA

EDTA, also known as ethylenediaminetetraacetic acid forms strong complexes with metal ions and plays a major role in quantitative analysis. In industries it is used as a metal binding agent and used in production of detergents, cleaning agents, food additives etc

02

EDTA titration

In most of the cases little amount of indicator is induced at the beginning of the titration experiment. EDTA is then added to the colorless metalcomplex solution and reacts until all the free metal ions are consumed.

03

List of methods

The four methods for detecting the end point of an EDTA titration are

a) With metal ion indicators

b) With a glass electrode

c) With a mercury electrode

d) With an ion selective electrode.

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Most popular questions from this chapter

Explain why the change from red to blue in Reaction 12-19 occurs suddenly at the equivalence point instead of gradually throughout the entire titration.

Consider the titration of 25.0 mL of 0.020 0 M MnSO4 with 0.010 0 M EDTA in a solution buffered to pH 8.00. Calculate pMn2+ at the following volumes of added EDTA and sketch the titration curve:

(a) 0 mL (b) 20.0 mL (c) 40.0 mL (d) 49.0 mL (e) 49.9 mL (f) 50.0 mL (g) 50.1 mL(h) 55.0 mL (i) 60.0 mL

A 50.0-mL solution containing Ni2+ and Zn2+ was treatedwith 25.0 mL of 0.045 2 M EDTA to bind all the metal. The excess unreacted EDTA required 12.4 mL of 0.012 3 M Mg2+ for complete reaction. An excess of the reagent 2,3-dimercapto-1-propanol was then added to displace the EDTA from zinc. Another 29.2 mL of Mg2+ were required for reaction with the liberated EDTA. Calculate the molarity of Ni2+ and Zn2+ in the original solution

What is the predominant form of arginine atpH11? What is the second major species?

According to Appendix I, Cu2+ forms two complexes with acetate:

Cu2++CH3CO2−⇌Cu(CH3CO2)+       β1(=K1)Cu2++2CH3CO2−⇌Cu(CH3CO2)2       β2

(a) Referring to Box 6-2, find K2 for the reaction

Cu(CH3CO2)++CH3CO2−⇌Cu(CH3CO2)2(aq)   K2

(b) Consider 1.00 L of solution prepared by mixing 1.00 × 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+
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