Answer:
pH = 3.36
Explanation:
Lidocaine is a weak base to be titrated with the strong acid HBr, therefore at equivalence point we wil have the protonated lidocaine weak conjugate acid of lidocaine which will drive the pH.
Thus to solve the question we will need to calculate the concentration of this weak acid at equivalence point.
Molarity = mol /V ∴ mol = V x M
mol lidocaine = (130 mL/1000 mL/L) x 0.4248 mol/L = 0.0552 mol
The volume of 0.4429 M HBr required to neutralize this 0.0552 mol is
0.0552 mol x (1L / 0.4429mol) = 0.125 L
Total volume at equivalence is initial volume lidocaine + volume HBr added
0 .130 L +0.125 L = 0.255L
and the concentration of protonated lidocaine at the end of the titration will be
0.0552 mol / 0.255 L = 0.22M
Now to calculate the pH we setup our customary ICE table for weak acids for the equilibria:
protonated lidocaine + H₂O ⇆ lidocaine + H₃O⁺
protonated lidocaine lidocaine H₃O⁺
Initial(M) 0.22 0 0
Change -x +x +x
Equilibrium 0.22 - x x x
We know for this equilibrium
Ka = [Lidocaine] [H₃O⁺] / [protonaded Lidocaine] = x² / ( 0.22 - x )
The Ka can be calculated from the given pKb for lidocaine
Kb = antilog( - 7.94 ) = 1.15 x 10⁻⁸
Ka = Kw / Kb = 10⁻¹⁴ / 1.15 x 10⁻⁸ = 8.71 x 10⁻⁷
Since Ka is very small we can make the approximation 0.22 - x ≈ 0.22
and solve for x. The pH will then be the negative log of this value.
8.71 x 10⁻⁷ = x² / 0.22 ⇒ x = √(/ 8.71 X 10⁻⁷ x 0.22) = 4.38 x 10⁻⁴
( Indeed our approximation checks since 4.38 x 10⁻⁴ is just 0.2 % of 0.22 )
pH = - log ( 4.4x 10⁻⁴) = 3.36
= 0.001 
. Therefore 200 
. Hope this helps.