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Pyrocatechol violet(Table 12-3) is to be used as a metal ion indicator in an EDTA titration. The procedure is as follows:

1. Add a known excess of EDTA to the unknown metal ion.

2. Adjust the pH with a suitable buffer.

3. Back-titrate the excess chelate with standard Al3+.

From the following available buffers, select the best buffer, and then state what color change will be observed at the end point. Explain your answer.

  1. pH 6–7 (ii) pH 7–8 (iii) pH 8–9 (iv) pH 9–10

Short Answer

Expert verified

From table 12-3 it can be identified that buffer solution of pH 6-7 is the most suitable one as in this range the color change occurs from yellow to blue while for the other pH ranges the color changes from violet to blue. Identifying change from yellow to blue is much easier than that of violet to blue. So (i) pH6-7 will be the best buffer to operate with.

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

Reaction procedure

A titration reaction between pyrocatechol violet (a metal ion indicator) and EDTA follows the steps below;

For an unknown concentration of metal ion, a known excess of EDTA must be added. Then the pH must be adjusted using a suitable buffer. After that we need to back titrate the excess amount of chelate with standard solution of Al3+

03

Selection of best buffer

From table 12-3 it can be identified that buffer solution of pH 6-7 is the most suitable one as in this range the color change occurs from yellow to blue while for the other pH ranges the color changes from violet to blue. Identifying change from yellow to blue is much easier than that of violet to blue. So (i) pH6-7 will be the best buffer to operate with.

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Most popular questions from this chapter

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+L𝆏MLβ1=[ML][M][L]M+2L𝆏ML2β2=[ML2][M][L]2

Let αM be the fraction of metal in the form M, αML be the fraction in the form ML, and αML2be the fraction in the form ML2. Following the derivation in Section 12-5, you could show that these fractions are given by

role="math" localid="1667801924683" αM1=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 ML2are

[ML]=αMLCMVMVM+VL[ML2]=αML2CMVMVM+VL

because CMVMVM+VLis the total concentration of all metal in the solution. The mass balance for ligand is

[L]+[ML]+2[ML2]=CMVMVM+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=CLVLVM+VM=αML+2αML2+LCM1-LCL

Spreadsheet equation for auxiliary complexing agent. Consider the titration of metal M (initial concentration = CM, initial volume = VM) with EDTA (concentration = CEDTA, volume added = VEDTA) in the presence of an auxiliary complexing ligand (such as ammonia). Follow the derivation in Section 12-4 to show that the master equation for the titration is

ϕ=CETDAVETDACMVM=1+Kf"[M]free-[M]free+Kf"[M]freeCMKf"[M]free+[M]free+Kf"[M]free2CETDA

where role="math"> is the conditional formation constant in the presence of auxiliary complexing agent at the fixed pH of the titration (Equation 12-18) and [M]free is the total concentration of metal not bound to EDTA. [M]free is the same as [M] in Equation 12-15. The result is equivalent to Equation 12-11, with [M] replaced by [M]free and Kf replaced by Kf".

A 25.00-mL sample containing Fe3+ and Cu2+ required 16.06 mL of 0.050 83 M EDTA for complete titration. A 50.00-mL sample of the unknown was treated with NH4F to protect the Fe3+. Then Cu2+ was reduced and masked by thiourea. Addition of 25.00 mL of 0.050 83 M EDTA liberated Fe3+ from its fluoride complex to form an EDTA complex. The excess EDTA required 19.77 mL of 0.018 83 M Pb2+ to reach a xylenol orange end point. Find [Cu2+] in the unknown.

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

Titration of M with L to form ML and ML2. Use theequation from Problem 12-21, where M is Cu2+ and L is acetate.Consider adding 0.500 M acetate to 10.00 mL of 0.050 0 M Cu2+ atpH 7.00 (so that all ligand is present as CH3CO2-, not CH3CO2H).Formation constants forCu(CH3CO2)+ and Cu(CH3CO2)2 aregiven in Appendix I. Construct a spreadsheet in which the inputis pL and the output is [L], VL, [M], [ML], and [ML2]. Prepare agraph showing concentrations of L, M, ML, and ML2as VL rangesfrom 0 to 3 mL

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