A 1.2485 g sample of CaCO3 is used to standardize an EDTA titrant. The CaCO3 was...
Homework Answers
Mass of CaCO3, W = 1.2485 g
Molar mass of CaCO3, MW = 100 g/mol
Moles of CaCO3 = W/MW = 1.2485/100=0.012485 moles = 12.485
millimoles
CaCO3 = Ca(+2) + CO3(-2)
[Ca+2] + [Y(-4)] = [CaY(-2)]
Moles of Ca(+2) = Moles of CaCO3 =12.485 millimoles
1 mole of CaCO3 complexes with 1 millimol of Y(-4)
12.485 millimoles of CaCO3 will be complexed by 12.485 millimoles
of Y(-4)
[Y(-4)] = 12.485 millimoles
pH = 9.0, alpha4 = [Y(-4)]/[EDTA] = 0.06
[EDTA] = [Y(-4)]/0.06 = 12.485/0.06 = 208 millimoles
Volume of EDTA = 39.53 ml
Molarity of EDTA = 208/39.53 = 5.26 M
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