Question

What is the Na+ concentration in each of the following solutions:

(A) 3.65 M Sodium Sulfate=
(B) 1.38 M Sodium Carbonate=
(C) 0.785 M Sodium Bicarbonate=

((B)) What is the concentration of a lithium carbonate solution that is 0.695 M in Li+?

Answer 1

The sodium ion concentration is double the molarity of sodium sulfate and sodium carbonate solutions, which are 7.30 M and 2.76 M respectively, and identical to the molarity of the sodium bicarbonate solution at 0.785 M. For the lithium carbonate solution, the molarity is half of the Li+ concentration for a total of 0.3475 M in Li2CO3.

Step-by-step explanation:

The concentration of Na+ in each solution can be determined by considering the stoichiometry of the dissolution process for each sodium compound.

• (A) Sodium sulfate (Na2SO4): Since each formula unit of sodium sulfate yields 2 Na+ ions upon dissociation in water, a 3.65 M solution of Na2SO4 will have a Na+ concentration of 7.30 M.
• (B) Sodium carbonate (Na2CO3): Similarly, a 1.38 M solution of Na2CO3 will have a Na+ concentration of 2.76 M because each formula unit yields 2 Na+ ions.
• (C) Sodium bicarbonate (NaHCO3): In this case, each formula unit yields only 1 Na+ ion. Thus, a 0.785 M solution of NaHCO3 will have an identical Na+ concentration of 0.785 M.

For part B of the question:

Lithium carbonate (Li2CO3) yields 2 Li+ ions for each formula unit dissolved in water. Therefore, to find the concentration of the lithium carbonate solution that is 0.695 M in Li+, we divide this value by 2, yielding a Li2CO3 solution concentration of 0.3475 M.

Answer 2

Sodium Sulfate = Na2(SO4) meaning there are two ions of Na+ in one mole of Sodium Sulfate the M stands for Molarity, defined as Molarity = (moles of solute)/(Liters of solution), So if the Na2SO4 solution is 3.65M that means one Liter of has 3.65 moles of Na2SO4, the stoichiometry of Na2SO4 shows that there would be two Na+ ions in solution for every one Na2SO4.

Therefore if 3.65 moles of Na2SO4 was to dissolve, it would produce 7.3 moles of Na+, and since this is still a theoretical solution, we can assume 1 L of solution.

Finally we find [Na+] = 2*3.65 = 7.3M

Use the same logic for parts b and c