In a precipitation reaction between BaCl2(aq) and Na3PO4(aq),56.4mL of 0.100M BaCl2(aq) completely reacted with 24.3 mL of Na3PO4(aq).What was the molarity of Na3PO4(aq)?

Question

3BaCl2(aq)+2Na3PO4(aq)------>Ba3(PO4)2(s)+6NaCl(aq)

*Numerical value with three decimal places.(No units or chemical in answer)

Answer 1

3BaCl₂(aq) + 2Na₃PO₄(aq) ==> Ba₃(PO₄)₂(s) + 6NaCl(aq) ... balanced equation

First, we'll find moles BaCl₂ used in the titration. Then, using the mole ratio of the balanced equation, we'll find the moles of Na₃PO₄ that reacted. Finally, knowing moles and volume of the Na₃PO₄, we'll find the molarity.

moles BaCl₂ used = 56.4 ml x 1 L/1000 ml x 0.100 mol/L = 0.00564 moles BaCl₂

moles Na₃PO₄ present = 0.00564 mol BaCl₂ x 2 mol Na₃PO₄ / 3 mol BaCl₂ = 0.00376 moles Na₃PO₄

molarity Na₃PO₄ = moles/L = 0.00376 mol/0.0243 L = 0.155

Answer 2

Hello, and thank you for your great question! Thank you also for providingabalanced equation. Not only does it makes it easier for us to answer, but when it comes to molarity, it is of vital importance to correctly balance it, otherwise we would never get the right answer! Here we go.

Information given:

3BaCl₂(aq) + 2Na₃PO₄(aq) ------> Ba₃(PO₄)₂(s) + 6NaCl(aq)

56.4mL 24.3mL

0.100M XM?

Information to find out: Molarity Sodium phosphate

I have always found it helpful to convert milliliters into Liters, when it comes to problems involving molarity. Hence, 56.4mL/1000mL will give us Liters, and 24.3mL/1000mL as well. Thus, 0.0564L and 0.0243L. We also must know that 0.100M BaCl₂ can also be expressed as 0.100 mol BaCl₂ / 1 Liter.

I like to start doing conversion factors with the Liters we just got, from the side that has the most information given to us, which is Barium chloride, thus

0.0564L BaCl₂ x 0.100 Mol BaCl_{2 X} 2 mol Na₃PO₄ = 0.00376 mol Na₃PO_{4 (These are the moles of}

1 L BaCl₂ 3 mol BaCl_{2 Sodium phosphate that}

reacted with BaCl_{2 )}

We know that Molarity = moles / Liters, hence

We just found out the number of moles of sodium phosphate, and we were given the number of liters of it. All there is left to do is to divide these two numbers, and we would get our answer.

Thus,

0.00376 mol Na₃PO_{4 =} 0.15473 M Na₃PO_{4 (ANSWER!) Round it up to 0.155 to have 3 sig figs}

0.0243 L Na₃PO₄

Hope this helped, and that you were able to understand the procedure.

Take care!

Julio Torres