determine the new molarity of calcium ions when 50.0 mL of a 1.50 M calcium chloride solution are combined with 750.0 mL of a 0.236 M calcium phosphate solution.
moles CaCl2 = M x L = ??
moles Ca^+2 = same
moles Ca3(PO4)2 = M x L = ??
moles Ca^+2 = 3x that.
Assuming the volumes are additive,
M = total moles/total volume.
To determine the new molarity of calcium ions when the two solutions are combined, we need to use the concept of moles and volume.
First, let's calculate the number of moles of calcium chloride and calcium phosphate solution:
Moles of calcium chloride = Molarity x Volume
Moles of calcium chloride = 1.50 M x 0.0500 L (since 50.0 mL is equivalent to 0.0500 L)
Moles of calcium chloride = 0.0750 moles
Moles of calcium phosphate = Molarity x Volume
Moles of calcium phosphate = 0.236 M x 0.750 L (since 750.0 mL is equivalent to 0.750 L)
Moles of calcium phosphate = 0.177 moles
Next, we need to determine the limiting reactant. Since we have more moles of calcium chloride than calcium phosphate, calcium phosphate is the limiting reactant. This means that calcium phosphate will completely react and determine the amount of calcium ions present in the solution.
Since calcium phosphate reacts in a 1:3 ratio with calcium ions, the number of moles of calcium ions formed will be three times the moles of calcium phosphate. Therefore:
Moles of calcium ions = 3 x Moles of calcium phosphate
Moles of calcium ions = 3 x 0.177 moles
Moles of calcium ions = 0.531 moles
Finally, we need to calculate the new molarity of calcium ions:
New Molarity of calcium ions = Moles of calcium ions / Total Volume
Total Volume = 50.0 mL + 750.0 mL = 800.0 mL = 0.800 L (since 800.0 mL is equivalent to 0.800 L)
New Molarity of calcium ions = 0.531 moles / 0.800 L
New Molarity of calcium ions = 0.664 M
Therefore, the new molarity of calcium ions when the two solutions are combined is 0.664 M.