[rycharles]

How many milliliters of 0.246 M HNO3 should be added to 213 mL of 0.00666 M 2,2'-bipyridine to give a pH of 4.19?

[Grant N.]

Is there a reaction of water or H3O+ with 2,2'-bipyridine? Because if there is no reaction producing a buffer with the 2,2'-bipyridine, the pH will be solely based on amount of HNO3 you add.

If I recall correctly, HNO3 is a strong acid, therefore:

HA+ H20--> A- + H30+
meaning there is a full dissociation of strong acid.

pH=4.19

pH= -log[H3O+]
4.19=-log[H3O+]

[H3O+]= 6.45654229e-5 mol/L = [HNO3] because there is a 1:1 ratio of H30+ and HNO3

[Borek]

http://www.chemicalbook.com/ProductMSDSDetailCB5195697_EN.htm

pKb 9.67

[rycharles]

There is no reaction with water according to the problem, just a mix of the two solutions.

pKa (bipyridine) = 4.34

If I use the Henderson Hasselbach equation

pH = pKa + log[[A-]/[HA]]

4.19 = 4.34 + log[[A-]/[HA]]

-0.15 = log[[A-]/[HA]]

0.7079 = [[A-]/[HA]]

At this point I am stuck.  Does A- stand for the strong acid HNO3 and HA stands for the weak acid?  If so, is this the right way to set up the equation?

HA A^- + H⁺

[Borek]

At this point I am stuck.  Does A- stand for the strong acid HNO3 and HA stands for the weak acid?

No, HA and A^- in the Henderson-Hasselbalch equation are acid and its conjugate base. Do you know Brønsted-Lowry theory of acids and bases?

[rycharles]

bronsted-lowry acid: proton donor
bronsted-lowry base: proton acceptor

[rycharles]

H+ is the acid: HNO3
A- is the base: 2,2'-bipyridine

In order to decrease the pH of the solution (HNO3 + bipyridine) to 4.19 we need to add a strong acid (HNO3) resulting in a HA (weak acid) on the product side of the equation.

               H⁺       +       A^-          HA
Initial        x         0.001418               --
Final         --      0.001418 - x            x

H+ = Acid (HNO3)
A- = Base (bipyridine)
HA = weak acid

moles of A-    --> 0.213L*(0.00666mol/L) = 0.001418 mol

Hasselbach Henderson Equation

pH = pKa + log[[A-]/[HA]]

4.19 = 4.34 + log[[A-]/[HA]]

-0.15 = log[[A-]/[HA]]

0.7079 = [[A-]/[HA]] = (mol of A-/mol of HA) = (0.001418 - x/x)

0.7079x = 0.001418 - x

1.7079x = 0.001418
 
x = 0.0008306 mol

to get the milliliters of HNO3

0.0008306 mol*(L/0.246 mol)(10³ mL/L) = 3.38 mL

3.38 mL of HNO3 need to be added in order to change the pH of the solution to 4.19.

Does this logic make sense?  Can anyone give any details that might be helpful to this problem?

[Borek]

Looks OK. I got 3.43 mL with the pH calculator, close enough.

[rycharles]

Great! thanks very much