How Much HNO3 to Achieve pH 4.19 in 2,2'-Bipyridine Solution?

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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?
 
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First things first - what is your HA and what is your A-?

Note you have not used volume of the pyridine solution. You are right about using Henderson-Hasselbalch equation, but you need to pay attention to the way you calculate concentrations.