A 1.12 g sample of soil containing Ni2+ and Zn2+ and other nonmetals was digeste

Question

A 1.12 g sample of soil containing Ni2+ and Zn2+ and other nonmetals was digested in acid and diluted to a final volume of 50.00 mL. This solution was treated with 25.0 mL of 0.0452 M EDTA to bind all the metal. The excess unreacted EDTA required 12.4 mL of 0.0123 M Mg2+ for complete reaction. An excess of the reagent 2, 3- dimercapto-1-propanol was then added to displace the EDTA from zinc only. Another 29.2 mL of Mg2+ were required for reaction with the liberated EDTA. Calculate the percent by weight of Ni2+ and percent by weight of Zn2+ in the original rock sample.

Explanation / Answer

EDTA : Mg2+=1:1

EDTA:Zn2+=1:1

EDTA:Ni2+=1:1

Initial EDTA moles=0.0452M*25.0/1000dm3 = 0.00113mol

The excess unreacted EDTA moles = 0.0123M * 12.4/1000dm3 =0.00015252 mol

above values got by EDTA;Mg2+ = 1;1

EDTA consumed for Ni2+ & Zn2+ = Initial EDTA - Unreacted EDTA =0.00097748 mol

Needed Mg2+ foe liberated EDTA from Zn = 0.0123M*29.3/1000dm3 = 0.00036039mol

Mg2+:EDTA:Zn2+ = 1:1:1

Zn2+ inthe sample = 0.00036039mol

EDTA:Zn2+ = 1:1

EDTA:Ni2+ = 1:1

Ni2+ moles in the sample = 0.00097748mol - 0.00036039mol = 0.00061709 mol

Zn2+ weight = 0.00036039mol * 65.38gmol- = 0.02356g

Ni2+ weight = 0.00061709mol * 58038gmol- = 0.03622g

Zn w/w % = 0.02356g * 100 /1.12g = 2.10%

Ni w/w % = 0.03622g * 100 / 1.12g = 3.23%

Firstly we can calculate initial EDTA. Then excess unreacted EDTA can be calculated. After that EDTA consume for Ni2+ & Zn2+ can get using initial EDTA substract by unreacted EDTA. 2,3-dimercapto-1-poropanol was displaced EDTA from zinc only. Them Mg2+ again reacted with displaced EDTA. After that using EDTA:Mg2+:Zn2+ ratio, we can get Zn2+ moles in the sample. Consumed full EDTA substracted by Zn2+ bonded EDTA , Wecan get Ni2+ moles

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