An excess of EDTA (ethylenediaminetetraacetic acid) must be added to Cr^3+ solut

Question

An excess of EDTA (ethylenediaminetetraacetic acid) must be added to Cr^3+ solutions for quantitative reaction. A 123.0 mg sample containing chromium was dissolved and the chromium was then converted to Cr^3+ 25.00 mL of 0.01000 MEDTA solution was added and allowed to react. The excess EDTA was titrated with 0.00950 M Zn^2+ solution, requiring 10.00 mL. (AW for Cr = 52.00, FW for Cr_2 O_3 = 152.0 g/mol) The sample contained a. 38.31% C_r2 0_3 b. 19.2% Cr_2 0_3 c. 9.58% Cr_2 0_3 d. 6.55% Cr_2 0_3

Explanation / Answer

Moles of Cr2O3 = mass/FW = x*0.123/152 = x*0.000809 moles

Here x denotes the fraction of Cr2O3 in the sample

1 molecule Cr2O3 gives 2 Cr3+ ions, so total Cr3+ ions formed = 2*x*0.000809 = x*0.001618 moles

Moles of EDTA added = 0.025*0.01 = 0.00025

Moles of Zn2+ added = 0.01*0.0095 = 0.000095 moles

So excess moles of EDTA = moles that reacted with Zn2+ = 0.000095

So, moles reacting with Cr3+ = 0.00025-0.000095 = 0.000155 moles

So moles of Cr3+ present = x*0.001618 = 0.000155

Solving we get :
x = 0.0958

Thus % Cr2O3 present = 0.0958*100 = 9.58%

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