1. You have a .246 M HNO3 solution. You also have 213mL of a 0.00666M 2,2' -bipyridine. You want a solution with pH 4.19. How man L of .246M HNO3 do you add to the 213mL other stuff? 2. pH=pKa+log[A-]/[HA] also, pKa of 2,2' -bipyridine is 4.34 I think. 3. pH=pKa+ log[A-]/[HA] 4.19=4.34+log[x-]/[0.00666M-x] -0.15=log[x-]/[0.00666M-x] 10^-.15=[x-]/[0.00666M-x] x=5.96x10^-5mol (5.96x10^-5mol) / (0.246M)= .0002423L or .2423 mL I know that is not the right answer, I'm just not sure what I am doing wrong. Is this the wrong equation for the problem or is there something else I need to do?