Question

An experimental protocol calls for 25.4 mL of a solution in which the potassium ion concentration is 0.0872 M. How would you prepare this solution from pure solid potassium phosphate? (The solvent is water.)

(Please don’t forget to state how you would prepare the solution; don’t simply calculate the grams of potassium phosphate required.)

Answer 1

According to solution, then:

M = mol/V

mol = M x V = 0.0872 mol/L x 0.0254 L = 2.2149x10^-3 moles

Molar Mass: (3x39) + 31 + (4x16) = 212 g/mol

m = mol x MM = 2.2149x10^-3 mol x 212 g/mol = 0.4696 g of K₃PO₄

Now, fif K₃PO₄ is 100% pure, you can actually assume this mass to be pure too, so how can we prepare this step by step?

1. Weigh 0.4696 g of pure K₃PO₄ on a beaker (previously weighted).

2. Add some water first, (about 5 or 10 mL) to dissolve the salt into the water.

3. Transfer the solution to a volumetric flask of 25 mL.

4. Complete with water until 25 mL. This is an 0.0872 M solution.

But in this case, we can prepare the solution in 100 mL and from there, take the 25.4 mL. In this case, the moles would be:

moles = 0.0872 x 0.100 = 0.00872 moles

m = 0.00872 x 212 = 1.85 g.

And then use the same procedure of before, but instead of adding into a volumetric flask of 25 mL, use a volumetric flask of 100 mL. And then, it follows the next step:

5. Take a volumetric cilindric, and take 25.4 mL of solution.

If you need any other explanation or any doubt to prepare this, you may tell me and I'll gladly answer it for you.