If 30.0 mL of 0.2 M nitric acid and 30.0 mL of 0.25 M solution of sodium hydroxide are mixed, what is the molarity of sodium nitrate? Assume that total volume of the solution after mixing is equal 60.0 mL.
HNO3 + NaOH---> H2O + NaNO3
Nitric acid is the limiting reagent.
Molarity=moles of sodium nitrate/60 x 10^-3L=(30 x 10^-3L *0.2M)/60 x 10^-3L