How many grams of dry NaOH must be added to 200 mL of 0.4 M H2SO4 to neutralize it completely? Molar mass of NaOH is 40 g/mol.
I'm going through a practice test (my professor doesn't post answers for it) and I wanted to see if I did this correctly:
I set up the equation NaOH + H2SO4 <-> HSO4- + H2O (l)
Then I set up an sRf table:
NaOH + H2SO4 <-> HSO4-
x .08 mol 0
-x .08 - x +x
0 0 x
From this I figured the moles of NaOH = .08. Is this correct or did I totally miss the point of the question?