Question

Calculate the concentration (M) of sodium ions (Na ) in a solution made by diluting 47.8 mL of a 0.801 M solution of sodium sulfide to a total volume of 336.7 mL. Do not include units in your answer and significant figures matter.

Answer 1

The concentration (M) of sodium ions is 0.2274 M

Step-by-step explanation:

Step 1: Data given

Initial volume of the solution = 47.8 mL = 0.0478 L

Initial concentration of the solution = 0.801 M

The new volume = 336.7 mL = 0.3367 L

Step 2: The balanced equation

Na2S → 2Na + + S^2-

Step 3: Calculate the new concentration of Na2S

C1*V1 = C2*V2

⇒with C1 = the initial concentration of the solution = 0.801 M

⇒with v1 = the initial volume of the solution = 0.0478 L

⇒with V2 = the new volume = 0.3367 L

⇒with C2 = the new concentration of the solution = TO BE DETERMINED

0.801 M * 0.0478 L = 0.3367 L * C2

C2 = (0.801 M * 0.0478 L) / 0.3367 L

C2 = 0.1137 M

Step 4: Calculate concentration of sodium ions

For 1 mol Na2S we have 2 mol Na+

For 0.1137 M Na2S, the concentration of Na+ ions is 2*0.1137 = 0.2274 M

The concentration (M) of sodium ions is 0.2274 M

Answer 2

C Na+ = 0.228 M

Step-by-step explanation:

• Na2S → 2Na+ + S2-

M 2M M

∴ mol Na2S = (0.0478 L)*(0.801 mol/L) = 0.0383 mol

∴V sln = 336.7 mL = 0.3367 L

⇒ C Na2S sln = 0.0383 mol / 0.3367 L = 0.114 M

∴ mol Na+ = (0.0383 mol Na2S)*( 2 mol Na+/ mol Na2S) = 0.0766 mol

⇒ C Na+ = 0.0766 mol / 0.3367 L = 0.2275 M