Question

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How many milliliters of 0.244 M HNO3 should be added to 214 mL of 0.00668 M 2,2'-bipyridine hydrochloride to give a pH of 4.16? (Assume pka = 4.34 and Kw = 1.01% 10-14) 49 3.9 x mL

Answer 1

Let's denote H+ as the acid - HNO3 and A- as the base - 2,2'-bipyridine hydrochloride

To reduce the pH of the solution to 4.16 we need to add a strong acid HNO3 resulting in a HA (weak acid) on the product side of the equation.

Total no. of moles of base = 0.214 L * 0.00668 mol/L = 0.0014295 mol


H⁺ + A^-    HA
Initial x 0.0014295 -
Final -    0.0014295 - x x

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


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

4.16 = 4.34 + log[[A-]/[HA]]

-0.18 = log[[A-]/[HA]]

0.660693 = [[A-]/[HA]] = (mol of A-/mol of HA) = ( 0.0014295 - x/x)

0.660693x = 0.0014295 - x

1.660693x = 0.0014295

x = 0.00086078 mol

To calculate volume of HNO3 from moles,

0.00086078 mol / (0.244 mol/L) = 0.0035 L or 3.5 mL

Therfore, 3.5 mL of HNO3 should be added in order to give pH of 4.16 of the solution.