Question

What would be the final concentration of nitrate ion if you mixed 50.0 mL of 2.5 M KNO3 with 100.0 mL of .750 M NaNO3? assume that the volumes are additive.

Answer 1

number of moles of nitrate ions from KNO3, n1=2.5*0.05=0.125 mol

number of moles of nitrate ions from NaNO3, n2=0.75*0.1=0.075 mol

Total number of moles of nitrate ions = 0.125+0.075=0.2 mol

Total volume = 0.05+0.1= 0.15 L

Final concentration = total moles/ total volume
                         =0.2/0.15
                          =1.33 M