Chemistry Hw

Question

 425.0 ml of 0.120 M Na2SO4 solution is added to 275.0 ml of 0.360 M NaNO3 solution. Calculate the final concentration of sodium ions in solution. You may assume that the volumes are additive. 

Answer 1

Hello Jana,

 

First find the concentration of sodium ions from each solution, and then the final Na+ concentration.

 

Na₂SO₄ solution:

mol of Na₂SO₄ = 0.4250 L x (0.120 mol/L)  = 0.0510 mol

 

mol of Na⁺ =  ( 2 mol Na⁺/1 mo Na₂SO₄ ) x 0.0510 mol Na₂SO₄  =  0.1020 mol

 

NaNO₃ solution:

 

mol of NaNO₃  = 0.2750 L x 0.360 mol/L  = 0.0990 mol 

 

mol of Na+ =   (1 mol Na+/1 mol NaNO3 ) x 0.0990 mol NaNO3 = 0.0990 mol

 

 

Final mol of Na⁺   =  0.1020 mol  + 0.0990 = 0.201 mol

 

Final volume of solution   =  425.0  mL + 275.0 mL = 700.0 mL =  0.7 L

 

Final Concentration of Na⁺  =  0.201 mol/ 0.7 L = 0. 3 M