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Give three circumstances in which an EDTA back titration might be necessary

Short Answer

Expert verified

The three circumstances where an EDTA back titration is needed.

  1. Precipitation of analyte in absence of EDTA
  2. If the reaction is too slow with EDTA
  3. If the indicator is blocked

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

Back Titration

In case of a direct titration an analyte is generally titrated with a standard solution of EDTA. Whereas in case of a back titration an excess amount (known amount) of EDTA is added with the analyte. The excess amount of EDTA takes part in a titration reaction with a solution of second metal ion.

03

Three circumstances in which an EDTA back titration is necessary

The three circumstances where an EDTA back titration is needed.

a) Precipitation of analyte in absence of EDTA

b) If the reaction is too slow with EDTA

c) If the indicator is blocked

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Most popular questions from this chapter

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

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

M+Lโ‡‹MLฮฒ1=[ML][M][L]M+2Lโ‡‹ML2ฮฒ2=[ML2][M][L]2

Let ฮฑM be the fraction of metal in the form M, ฮฑML be the fraction in the form ML, and ฮฑML2 be the fraction in the form ML2. Following the derivation in Section 12-5, you could show that these fractions are given by

ฮฑM=11+ฮฒ1[L]+ฮฒ2[L]2ฮฑML=(ฮฒ1[L])1+ฮฒ1[L]+ฮฒ2[L]2ฮฑML2=ฮฒ2[L]21+ฮฒ1[L]+ฮฒ2[L]2

The concentrations of ML and ML2 are

[ML]=ฮฑMLCMVMVM+VL[ML]=ฮฑML2CMVMVM+VL

becauseCMVMVM+VL is the total concentration of all metal in the solution. The mass balance for ligand is

[L]+[ML]+2[ML2]=CLVLVM+VL

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

ฯ•=CLVLCMVM=ฮฑML+2ฮฑML2+LCM1-LCL

Find pZn2+ after adding 30.0 and 51.0 mL of EDTA.

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

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

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