How many mL of 0.0500 M EDTA are required to react with 54.0 mL of 0.0190 M Ca2+?
EDTA reacts with metals 1:1.
mols Ca^2+ = M x L = ?
mols EDTA = mols Ca^2+
Then M EDTA = mols EDTA/L EDTA. You know M and mols, solve for L and convert to mL.
To find out how many mL of 0.0500 M EDTA are required to react with 54.0 mL of 0.0190 M Ca2+, we need to use the balanced chemical equation and the concept of stoichiometry.
The balanced chemical equation for the reaction between EDTA (ethylenediaminetetraacetic acid) and Ca2+ (calcium ion) is:
Ca2+ + EDTA → Ca(EDTA)2-
From the balanced equation, we can see that the ratio of Ca2+ to EDTA is 1:1. This means that 1 mole of Ca2+ reacts with 1 mole of EDTA.
First, let's calculate the number of moles of Ca2+ in 54.0 mL of 0.0190 M Ca2+ solution:
moles of Ca2+ = volume of Ca2+ solution (in L) × molarity of Ca2+
moles of Ca2+ = 54.0 mL × 0.0190 M / 1000 mL/L
moles of Ca2+ = 0.00103 mol
Since the stoichiometry of the reaction is 1:1, this means that 0.00103 mol of Ca2+ will react with the same number of moles of EDTA.
Now, let's calculate the volume of 0.0500 M EDTA solution needed to react with 0.00103 mol of Ca2+:
volume of EDTA solution (in L) = moles of Ca2+ × (1 / Molarity of EDTA)
volume of EDTA solution = 0.00103 mol × (1 / 0.0500 M)
volume of EDTA solution = 0.0206 L
To convert this volume to mL, simply multiply it by 1000:
volume of EDTA solution = 0.0206 L × 1000 mL/L
volume of EDTA solution = 20.6 mL
Therefore, 20.6 mL of 0.0500 M EDTA solution is required to react with 54.0 mL of 0.0190 M Ca2+.