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Moles of EDTA added initially to the solution = (25.0 mL* 1 L / 1000 mL) * 0.0452 mol/L =0.00113 mol Excess EDTA present in the solution after reacting withNi+2 and Zn+2 = (12.4 mL * 1 L /1000mL) * 0.0123 mol/L 0.00015 mol Moles of EDTA reacting with Zn+2 andNi+2 = 0.00113 mol - 0.00015 mol = 0.00098 mol Zn+2 and Ni+2 react with EDTA in1:1 ratio Moles of Zn+2 and Ni+2 inthe solution = 0.00098 mol EDTA was displaced from Zn and the liberated EDTA wastitrated with Mg+2 Moles of Mg+2 reacting with EDTA liberatedfrom Zn+2 = (29.2 mL * 1 L / 1000 mL) * 0.0123mol/L = 0.00035916 mol Each mole of EDTA is bound to 1 mole ofZn+2 Moles of Zn+2 in the solution = 0.00035916mol Zn+2 Moles of Ni+2 in the solution = 0.00098 mol -0.00035916 mol =0.00062084 mol Ni+2 [Zn+2] = 0.00035916 molZn+2 / 0.050 L = 0.0072 M Zn+2 [Ni+2] = 0.00062084 mol Ni+2 /0.050 L = 0.0124 M Ni+2 Solution Moles of EDTA added initially to the solution = (25.0 mL* 1 L / 1000 mL) * 0.0452 mol/L =0.00113 mol Excess EDTA present in the solution after reacting withNi+2 and Zn+2 = (12.4 mL * 1 L /1000mL) * 0.0123 mol/L 0.00015 mol Moles of EDTA reacting with Zn+2 andNi+2 = 0.00113 mol - 0.00015 mol = 0.00098 mol Zn+2 and Ni+2 react with EDTA in1:1 ratio Moles of Zn+2 and Ni+2 inthe solution = 0.00098 mol EDTA was displaced from Zn and the liberated EDTA wastitrated with Mg+2 Moles of Mg+2 reacting with EDTA liberatedfrom Zn+2 = (29.2 mL * 1 L / 1000 mL) * 0.0123mol/L = 0.00035916 mol Each mole of EDTA is bound to 1 mole ofZn+2 Moles of Zn+2 in the solution = 0.00035916mol Zn+2 Moles of Ni+2 in the solution = 0.00098 mol -0.00035916 mol =0.00062084 mol Ni+2 [Zn+2] = 0.00035916 molZn+2 / 0.050 L = 0.0072 M Zn+2 [Ni+2] = 0.00062084 mol Ni+2 /0.050 L = 0.0124 M Ni+2.

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Moles of EDTA added initially to the solution = (25.0 mL* 1 L / 1000 mL) * 0.0452 mol/L =0.00113 mol Excess EDTA present in the solution after reacting withNi+2 and Zn+2 = (12.4 mL * 1 L /1000mL) * 0.0123 mol/L 0.00015 mol Moles of EDTA reacting with Zn+2 andNi+2 = 0.00113 mol - 0.00015 mol = 0.00098 mol Zn+2 and Ni+2 react with EDTA in1:1 ratio Moles of Zn+2 and Ni+2 inthe solution = 0.00098 mol EDTA was displaced from Zn and the liberated EDTA wastitrated with Mg+2 Moles of Mg+2 reacting with EDTA liberatedfrom Zn+2 = (29.2 mL * 1 L / 1000 mL) * 0.0123mol/L = 0.00035916 mol Each mole of EDTA is bound to 1 mole ofZn+2 Moles of Zn+2 in the solution = 0.00035916mol Zn+2 Moles of Ni+2 in the solution = 0.00098 mol -0.00035916 mol =0.00062084 mol Ni+2 [Zn+2] = 0.00035916 molZn+2 / 0.050 L = 0.0072 M Zn+2 [Ni+2] = 0.00062084 mol Ni+2 /0.050 L = 0.0124 M Ni+2 Solution Moles of EDTA added initially to the solution = (25.0 mL* 1 L / 1000 mL) * 0.0452 mol/L =0.00113 mol Excess EDTA present in the solution after reacting withNi+2 and Zn+2 = (12.4 mL * 1 L /1000mL) * 0.0123 mol/L 0.00015 mol Moles of EDTA reacting with Zn+2 andNi+2 = 0.00113 mol - 0.00015 mol = 0.00098 mol Zn+2 and Ni+2 react with EDTA in1:1 ratio Moles of Zn+2 and Ni+2 inthe solution = 0.00098 mol EDTA was displaced from Zn and the liberated EDTA wastitrated with Mg+2 Moles of Mg+2 reacting with EDTA liberatedfrom Zn+2 = (29.2 mL * 1 L / 1000 mL) * 0.0123mol/L = 0.00035916 mol Each mole of EDTA is bound to 1 mole ofZn+2 Moles of Zn+2 in the solution = 0.00035916mol Zn+2 Moles of Ni+2 in the solution = 0.00098 mol -0.00035916 mol =0.00062084 mol Ni+2 [Zn+2] = 0.00035916 molZn+2 / 0.050 L = 0.0072 M Zn+2 [Ni+2] = 0.00062084 mol Ni+2 /0.050 L = 0.0124 M Ni+2

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Moles of EDTA added initially to the solution = (.pdf

  • 1. Moles of EDTA added initially to the solution = (25.0 mL* 1 L / 1000 mL) * 0.0452 mol/L =0.00113 mol Excess EDTA present in the solution after reacting withNi+2 and Zn+2 = (12.4 mL * 1 L /1000mL) * 0.0123 mol/L 0.00015 mol Moles of EDTA reacting with Zn+2 andNi+2 = 0.00113 mol - 0.00015 mol = 0.00098 mol Zn+2 and Ni+2 react with EDTA in1:1 ratio Moles of Zn+2 and Ni+2 inthe solution = 0.00098 mol EDTA was displaced from Zn and the liberated EDTA wastitrated with Mg+2 Moles of Mg+2 reacting with EDTA liberatedfrom Zn+2 = (29.2 mL * 1 L / 1000 mL) * 0.0123mol/L = 0.00035916 mol Each mole of EDTA is bound to 1 mole ofZn+2 Moles of Zn+2 in the solution = 0.00035916mol Zn+2 Moles of Ni+2 in the solution = 0.00098 mol -0.00035916 mol =0.00062084 mol Ni+2 [Zn+2] = 0.00035916 molZn+2 / 0.050 L = 0.0072 M Zn+2 [Ni+2] = 0.00062084 mol Ni+2 /0.050 L = 0.0124 M Ni+2 Solution Moles of EDTA added initially to the solution = (25.0 mL* 1 L / 1000 mL) * 0.0452 mol/L =0.00113 mol Excess EDTA present in the solution after reacting withNi+2 and Zn+2 = (12.4 mL * 1 L /1000mL) * 0.0123 mol/L 0.00015 mol Moles of EDTA reacting with Zn+2 andNi+2 = 0.00113 mol - 0.00015 mol = 0.00098 mol Zn+2 and Ni+2 react with EDTA in1:1 ratio Moles of Zn+2 and Ni+2 inthe solution = 0.00098 mol EDTA was displaced from Zn and the liberated EDTA wastitrated with Mg+2 Moles of Mg+2 reacting with EDTA liberatedfrom Zn+2 = (29.2 mL * 1 L / 1000 mL) * 0.0123mol/L = 0.00035916 mol Each mole of EDTA is bound to 1 mole ofZn+2 Moles of Zn+2 in the solution = 0.00035916mol Zn+2 Moles of Ni+2 in the solution = 0.00098 mol -0.00035916 mol =0.00062084 mol Ni+2 [Zn+2] = 0.00035916 molZn+2 / 0.050 L = 0.0072 M Zn+2 [Ni+2] = 0.00062084 mol Ni+2 /0.050 L = 0.0124 M Ni+2