ovský, 1958 (versu H,S0 2. A 1.12 g sample of soil containing Ni2, and Zn2 and o

Question

ovský, 1958 (versu H,S0 2. 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 Mg for complete reaction. An excess of the reagent 2, 3 dimercapto-1-propanol was then added to displace the ned s EDTA from zinc only. Another 29.2 ml of Mg were required for reaction with the liberated EDTA. Calculate the percent by weight of Ni?* and percent by weight of Zn2 in the original rock sample s ha tial n of

Explanation / Answer

Total EDTA added = 0.0452 M x 25 ml = 1.13 mmol

excess EDTA = 0.0123 M x 12.4 ml = 0.15252 mmol

Total (Ni2+ + Zn2+) in 50 ml = 1.13 - 0.15252 = 0.97748 mmol

amount of Zn2+ in 50 ml = 0.0123 M x 29.2 ml = 0.35916 mmol

mass Zn2+ in soil sample = 0.35916 x 10^-3 mol x 65.38 g/mol = 0.0235 g

mass% Zn2+ in soil sample = 0.0235 gx 100/1.12 g = 2.10%

amount of Ni2+ in 50 ml = 0.97748 - 0.35916 = 0.61832 mmol

mass Ni2+ in soil sample = 0.61832 x 10^-3 mol x 58.7 g/mol = 0.0363 g

mass% Ni2+ in soil sample = 0.0363 gx 100/1.12 g = 3.24%

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