Ion Concentration

Question

Part A

What is the concentration of K+ in 0.15M of K2S ?

Part B

If CaCl2 is dissolved in water, what can be said about the concentration of the Ca2+ ion?

Part C

A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of  Na3PO4 will be needed to produce 750. mL of a solution that has a concentration of Na+ ions of 0.900M?

Answer 1

K₂S(aq) ==> 2K⁺(aq) + S^2-(aq)

0.15 M ===>2 x 0.15M K⁺ + 0.15 M S^2-

[K⁺] = 0.30 M

CaCl₂(aq) ==> Ca²⁺ + 2Cl^-

The [Ca²⁺] will be 1/2 the concentration of the Cl^-

Na₃PO₄ molar mass = 164 g / mol

0.900 M Na+ = 0.900 mols Na⁺ / L, but since we are making only 750. mls we find mols of Na⁺ as follows:

0.900 mols Na⁺ / L x 0.750 L = 0.675 mols Na⁺ needed to make 750 mls (0.750 L)

Since each mol of Na₃PO₄ contains 3 mols of Na⁺, we can now find the mols Na₃PO₄ needed...

0.675 mols Na⁺ x 1 mol Na₃PO₄ / 3 mols Na⁺ = 0.225 mols Na₃PO₄ needed

Now using the molar mass of Na₃PO₄, we calculate the mass of Na₃PO₄ needed...

0.225 mols Na₃PO₄ x 164 g Na₃PO₄ / mol = 36.9 g Na₃PO₄ needed