Cehmistry Question

Question

A 250.0-mL sample of 0.125 M Na₃PO₄ is mixed with 250.0 mL of 0.175 M Ca(NO₃)₂ to form Ca₃(PO₄)₂. The net ionic equation for the reaction is given below:

3 Ca²⁺_(aq) + 2 PO₄^3–_(aq) ➜ Ca₃(PO₄)_2(s)

What is the concentration of the excess reactant ion in the resulting solution assuming that the volumes are additive?

Select one:

a.

5.83 × 10^–2 M PO₄^3–

b.

2.5 × 10^–2 M Ca²⁺

c.

9.38 × 10^–3 M Ca²⁺

d.

4.17 × 10^–3 M PO₄^3–

Answer 1

moles Na₃PO₄ = 250.0 ml x 1 L/1000 ml x 0.125 mol/L = 0.03125 mols = 0.03125 mols PO₄^3-

moles Ca(NO₃)₂ = 250.0 ml x 1 L/1000 ml x 0.175 mol/L = 0.04375 mols = 0.04375 mol Ca²⁺

An easy way to determine limiting reactant is to divide each reactant by the corresponding coefficient in the balanced equation and whichever value is less is the limiting reactant. So, we have...

For PO₄^3-: 0.03125 / 2 = 0.015625

For Ca²⁺: 0.04375 / 3 = 0.01458

Ca²⁺ is limiting

How much PO₄^3- will be left over after the reaction? It will be the initial amount less the amount used up.

amount used = 0.04375 mols Ca²⁺ x 2 mols PO₄^3- / 3 mols Ca²⁺ = 0.02917 mols PO₄^3- used up

moles PO₄^3- left over = 0.03125 mols - 0.02917 mols = 0.00208 mols

Final volume of reaction = 250.0 ml + 250.0 mls = 500.0 mls = 0.500 L

Final concentration of PO₄^3- = 0.00208 mols / 0.500 L = 0.00416 M = 4.16x10^-3 M (ANSWER d)