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Chapter 12: Q32P (page 285)

A 50.0-mL sample containing Ni2+ was treated with 25.0 mL of 0.050 0 M EDTA to complex all the Ni2+ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 5.00 mL of 0.050 0 M Zn2+. What was the concentration of Ni2+ in the original solution?

Short Answer

Expert verified

The concentration of Ni2+ in the original solution is 0.02 M

Step by step solution

01

Given Information

Amount of sample taken= 50.0 mL

Amount of EDTA required for titration =25 mL of 0.05M EDTA

Amount of Zn2+ required for back titration= 5 mL of 0.050 M Zn2+

02

Determine the number of moles of EDTA and Zn2+ required

Number of moles of EDTA required

=25mL0.050MEDTA=1.25mmol

Number of moles of Zn2+ required

=(5mL)(0.050EDTA)=0.25mmol

03

Determine the concentration of Ni2+

Let the number of mols of Ni2+ be x

mmolEDTA=mmolNi2++mmolZn2+1.250mmolEDTA=xmmolNi2++0.250mmolZn2+x=1mmolNi2+

50.0-mL of sample containing Ni2+ was taken

Therefore, concentration of Ni2+in the solution

=1mmol50mL=0.02M

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Most popular questions from this chapter

A 50.0-mL aliquot of solution containing 0.450 g of MgSO4 (FM 120.37) in 0.500 L required 37.6 mL of EDTA solution for titration. How many milligrams of CaCO3 (FM 100.09) will react with 1.00 mL of this EDTA solution?

A 1.000-mL sample of unknown containing Co2+ and Ni2+ was treated with 25.00 mL of 0.038 72 M EDTA. Back titration with 0.021 27 M Zn2+ at pH 5 required 23.54 mL to reach the xylenol orange end point. A 2.000-mL sample of unknown was passed through an ion-exchange column that retards Co2+ more than Ni2+. The Ni2+ that passed through the column was treated with 25.00 mL of 0.038 72 M EDTA and required 25.63 mL of 0.021 27 M Zn2+ for back titration. The Co2+ emerged from the column later. It, too, was treated with 25.00 mL of 0.038 72 M EDTA. How many milliliters of 0.021 27 M Zn2+ will be required for back 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

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+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
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