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.

Short Answer

Expert verified

As very small of indicator is introduced into the sample, many of the Mg2+ ions is not bounded with the indicator. Free ions of Mg2+ are preferable for reaction with EDTA than MgIn. Therefore, concentration of MgIn remains constant until all the free Mg2+ ions have been consumed. Only after that MgIn gets chance to react with EDTA. As soon as the reaction starts with the indicator the color change from red to blue. This is the reason that the change from red to blue in Reaction 12-19 occurs suddenly at the equivalence point instead of gradually throughout the entire titration.

Step by step solution

01

Information given

Equation 12-19 is a typical illustration of a titration reaction between Mg2+ and EDTA at pH 10 in presence of calmagite indicator. The reaction is as follows

MgΙn+EDTAMgEDTA+ΙnRedColorlessColorlessBlue

02

How the reaction proceeds

Little amount of indicator is induced at the beginning of the experiment. It is added to the colorless Mg2+complex and forms a red complex. After that EDTA is added. This reacts with the Mg2+ solution first. After completion of reaction between Mg2+ and EDTA, the remained indicator sample reacts with EDTA and turns into blue unboundIn.

03

Reason

As very small of indicator is introduced into the sample, many of the Mg2+ ions is not bounded with the indicator. Free ions of Mg2+ are preferable for reaction with EDTA than MgIn . Therefore, concentration of MgIn remains constant until all the free Mg2+ ions have been consumed. Only after that MgIn gets chance to react with EDTA. As soon as the reaction starts with the indicator the color change from red to blue. This is the reason that the change from red to blue in Reaction 12-19 occurs suddenly at the equivalence point instead of gradually throughout the entire titration

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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

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

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+LMLβ1=[ML][M][L]M+2LML2β2=[ML2][M][L]2

Let αM be the fraction of metal in the form M, αML be the fraction in the form ML, and αML2 be the fraction in the form ML2. Following the derivation in Section 12-5, you could show that these fractions are given by

αM=11+β1[L]+β2[L]2αML=(β1[L])1+β1[L]+β2[L]2αML2=β2[L]21+β1[L]+β2[L]2

The concentrations of ML and ML2 are

[ML]=αMLCMVMVM+VL[ML]=αML2CMVMVM+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

ϕ=CLVLCMVM=αML+2αML2+LCM1-LCL

List four methods for detecting the end point of an EDTA 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

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