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Chapter 12: Q31P (page 285)

How many milliliters of 0.050 0 M EDTA are required to react with 50.0 mL of 0.010 0 M Ca2+? With 50.0 mL of 0.010 0 M Al3+?

Short Answer

Expert verified

10 milliliters(mL) of 0.050 0 M EDTA are required to react with 50.0 mL of 0.010 0 M Ca2+.

10 milliliters(mL) of 0.050 0 M EDTA are required to react with 50.0 mL of 0.010 0 M Al3+..

Step by step solution

01

Given Information

Amount of sample taken= 50.0 mL

Concentration of EDTA taken = 0.05M

Amount of Ca2+ required = 50 mL of 0.010 M Ca2+

Amount of Al3+ required = 50 mL of 0.010 M Al3+

02

Determine the amount of EDTA to react with Ca2+

Number of moles of Ca2+ required

=50mL0.010MCa2+

0.5 mmol of EDTA is required to titrate with 0.5 mmol Ca2+

Volume of EDTA required

=0.5mmol0.5mmol=10mL

03

Determine the amount of EDTA to react with Al3+

Number of moles of Al3+ required

=50mL0.010MAl3+=0.5mmolAl3+

0.5 mmol of EDTA is required to titrate with 0.5 mmol Al3+

Volume of EDTA required

=0.5mmol0.05M=10mL

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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

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 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.

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

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