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A solution was prepared by dissolving 3.5 grams of EDTA (Na2H2Y·2H2O) in exactly 1L of water. This solution was then standardized against 40.00 mL aliquots of 0.006000 M Mg2+. An average titration of 27.00 mL was required. Calculate the molar concentration of the EDTA. The formula weight of EDTA is 372.2 grams.
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Answered by
DrBob222
M = moles/L
moles = grams/molar mass = 3.5/372.2 = ?/1 L = 0.00940 M. NOTE: You don't dissolve 3.5 g EDTA in exactly 1 L of water (or if you do it does NOT prepare any known concentration of anything unless you know the density of the solution). INSTEAD, you dissolve 3.5 g EDTA is exactly 1 L OF SOLUTION. What's the difference. Well 1,000 mL H2O + 3.5 g EDTA produces a solution which is greater than1 L volume and that's not the definition of molarity.)
Mg ion firms a 1:1 complex with EDTA so
millimoles EDTA = 27.00 mL x M = ?
millimoles Mg^2+ = 40.00 mL x 0.00600 M = ? solve for M in
27.00 mL x M EDTA = 40.00 x 0.0600.
moles = grams/molar mass = 3.5/372.2 = ?/1 L = 0.00940 M. NOTE: You don't dissolve 3.5 g EDTA in exactly 1 L of water (or if you do it does NOT prepare any known concentration of anything unless you know the density of the solution). INSTEAD, you dissolve 3.5 g EDTA is exactly 1 L OF SOLUTION. What's the difference. Well 1,000 mL H2O + 3.5 g EDTA produces a solution which is greater than1 L volume and that's not the definition of molarity.)
Mg ion firms a 1:1 complex with EDTA so
millimoles EDTA = 27.00 mL x M = ?
millimoles Mg^2+ = 40.00 mL x 0.00600 M = ? solve for M in
27.00 mL x M EDTA = 40.00 x 0.0600.
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