How many milliliters of 0.0500 M EDTA are required to react with 50.0 mL of 0.0100 M Ca2+? With 50.0 mL of 0.0100 M Al3+?
Added by Xavier H.
Step 1
For Ca2+: moles of Ca2+ = Molarity × Volume moles of Ca2+ = 0.0100 M × 0.0500 L = 0.000500 mol For Al3+: moles of Al3+ = Molarity × Volume moles of Al3+ = 0.0100 M × 0.0500 L = 0.000500 mol Now, we need to find the volume of 0.0500 M EDTA required to react with Show more…
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