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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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