To 20 mL of .2 m nitric acid are added successive amounts of a base of unknown concentration. Neutrality is reached after the addition of 40 mL of the base. What was the OH of the original base solution? assume that it was 100% ionized.

Question

To 20 mL of .2 m nitric acid are added successive amounts of a base of unknown concentration. Neutrality is reached after the addition of 40 mL of the base.  What was the OH of the original base solution? assume that it was 100% ionized.

DrBob222 · Answer

Assuming you meant 0.2 M and not 0.2 m, then HNO3 + OH^- ==> NO3^- + H2O mols HNO3 = M x L = 0.02 x 0.2 = 0.004 mols OH- = the same since 1 mol HNO3 reacts with 1 mol OH-. Then M OH- = mols OH-/L OH- = 0.004/0.04 = 0.1 M. Now let's be careful here. That 0.1M is the (OH^-); it is NOT the concentration of the original base. There is no way of knowing the concentration of the original base because nothing in the problem tells you that is is MOH, M(OH)2, or M(OH)3 or whatever.