Since this involves a monoprotic acid & a monobasic base, we can simply use the equation (MV)acid = (MV)base So, in rearranging this to solve for the Volume of Base, we have: Vbase = (MV)acid/Mbase which, when solved, gives you 125 mL of base, NaOH
In the second problem, we again have a monoprotic acid/monobasic base, so we can use the same equation, but this time solving for the Mbase. When putting in the given values you should get 0.156 M KOH