How would you describe the preparation of 900 mL of 3.00 M #HNO_3# from the commercial reagent that is 70.5% #HNO_3# (w/w) and has a specific gravity of 1.42.?

Question

Noah Aldrich · Accepted Answer

Well first, we determine the molar concentration of the mother acid.

...........and finally we get a required volume of #170*mL# with respect to the starting acid.

#"Concentration of nitric acid"# #=# #"Moles of nitric acid"/"Volume of solution"# And if we specify a #1*mL# volume then.......... We have #((1.42*g*mL^-1xx1*mLxx70.5%)/(63.01*g*mol^-1))/(1xx10^-3*L)=15.9*mol*L^-1# So that is our mother acid. And as you know WE ALWAYS ADD ACID TO WATER AND NEVER THE REVERSE. Because if you spit in acid it will spit back............. And thus, we want a molar quantity of #900*mLxx10^-3*L*mL^-1xx3.00*mol*L^-1=2.70*mol#. And so we divide this molar quantity by the concentration of the conc. nitric.......... #=(2.70*mol)/(15.9*mol*L^-1)xx10^3*mL*L^-1=169.8*mL# And so we dilute (IN THIS ORDER), #170*mL# of the conc. acid TO a volume of #900*mL#.

Kai Bowman · Answer

undefined To prepare 900 mL of 3.00 M HNO3 solution from the commercial reagent that is 70.5% HNO3 (w/w) and has a specific gravity of 1.42, you would first calculate the mass of pure HNO3 present in the commercial reagent. Then, using the molarity equation, you would determine the volume of commercial reagent needed to make the desired concentration of HNO3 solution. Finally, you would dilute the commercial reagent to the desired volume using distilled water.