Gravimetry in Analytical Chemistry

Question

What volume (mL) of 0.2000 M (NH4)2HPO4 is necessary to precipitate the calcium as Ca3(PO4)2 from 800.0 mg of a sample that contains 10.00% w/w Ca?

Answer 1

First, let's look at the equation for the reaction that is taking place (omitting spectator ions):

2HPO₄^2-(aq) + 3Ca2+(aq) ==> Ca₃(PO₄)₂(s) + 2H⁺(aq)

Next, let's find the moles of Ca²⁺ present:

800.0 mg sample x 10% Ca = 80 mg Ca

80 mg Ca x 1 g / 1000 mg x 1 mol Ca / 40.08 g = 0.001996 mols Ca

Next, we find the moles of HPO₄^2-- needed to react with 0.001996 mols Ca

0.001996 mols Ca x 2 mol HPO₄^2-- / 3 mols Ca²⁺ = 0.001331 mols HPO₄^2-- needed

Finally, we find what volume will contain this many moles

0.001331 mols HPO₄^2-- x 1 L / 0.200 mols = 0.006653 L = 6.653 mls needed