If .0162 moles of sodium sulfide is dissolved in 100.0 mL of water, what is the concentration of sodium ions [Na⁺] in solution?

To determine the concentration of sodium ions ([Na⁺]) in the solution, we need to know the chemical formula for sodium sulfide (Na₂S) and its dissociation equation in water.

The dissociation equation for sodium sulfide in water is:
Na₂S → 2Na⁺ + S²⁻

From the equation, we can see that one formula unit of Na₂S produces two sodium ions (Na⁺).

Given that we have 0.0162 moles of sodium sulfide (Na₂S), we can use the stoichiometry from the balanced equation to calculate the number of moles of sodium ions (Na⁺).

1 mole of Na₂S produces 2 moles of Na⁺.
So, 0.0162 moles of Na₂S will produce (2 * 0.0162) moles of Na⁺.

Let's calculate the number of moles of sodium ions (Na⁺):
Number of moles of Na⁺ = (2 * 0.0162) moles = 0.0324 moles

Now, we need to determine the concentration of sodium ions ([Na⁺]) in the solution. Concentration is expressed in moles per liter (mol/L) or Molarity (M).

To find the concentration in moles per liter, we need to convert the volume of the solution (100.0 mL) to liters.

1 L = 1000 mL
So, 100.0 mL = 100.0 mL * (1 L / 1000 mL) = 0.100 L

Finally, we can calculate the concentration of sodium ions ([Na⁺]) using the formula:

Concentration of Na⁺ = Number of moles of Na⁺ / Volume of the solution in liters

Concentration of Na⁺ = 0.0324 moles / 0.100 L

Concentration of Na⁺ = 0.324 M

Therefore, the concentration of sodium ions ([Na⁺]) in the solution is 0.324 M.