Question

If the [Na+] = 0.250 M in a Na3P solution, calculate the concentration of Phosphorous ions and the number of grams of Na3P required to make 1.50 L of the solution

Answer 1

[P³⁻] = 0.083 M

Mass of Na₃P required: 12.5 g

Step-by-step explanation:

We dissociate the salt, like this:

Na₃P → 3Na⁺ + P³⁻

It is a ionic salt that can be dissociated.

As [Na⁺] is 0.25M, ratio is 3:1.

For 3 moles of sodium cathion, I have 1 mol of salt

So, for 0.25 moles of Na⁺, we have (0.25 . 1)/3 = 0.083M

So ratio with the anion is 1:1. In conclussion, [P³⁻] = 0.083 M

Finally, we have to know that molarity is mol/L so:

M = mol/L → 0.083mol/L = moles /1.50L

To make 1.50 L of solution, we need 0.083 mol/L . 1.50L = 0.125 moles

Let's find out the mass, with the molecular weight:

0.125 mol . 99.97g / 1mol = 12.5 g