# Question #a0caa

Sep 30, 2014

The final molarity of the Na⁺ ion is 0.102 mol/L.

NaCl → Na⁺ + Cl⁻

To get the molarity, you divide the moles of Na⁺ by the litres of solution.

Moles of Na⁺ = $\text{1.19 g NaCl" ×"1 mol NaCl"/"58.44 g NaCl" × ("1 mol Na"^+)/"1 mol NaCl}$ =

0.020 36 mol Na⁺ (3 significant figures + 1 guard digit).

The ammonium sulfate contains no Na⁺ ions, and it does not form a precipitate with NaCl. Its only function is to dissolve the NaCl.

I assume that there is no volume change on adding the NaCl to the solution. Then

[Na⁺] = $\text{moles of NaCl"/"litres of solution" = "0.020 36 mol"/"0.200 L}$ = 0.102 mol/L