A 50.00-mL solution of 0.0350 M quinoline (K_b = 8.0 $\times$ 10^-10)is titrated with a 0.0137 M solution of hydrochloric acid as the titrant.What is the pH at the equivalence point? (K_w = 1.0 $\times$10^-14) A) 4.90 B) 3.45 C) 5.55 D) 9.10 E) 10.55

Question

Quizplus · Accepted Answer

In this titration, HCl will react with quinoline to form its conjugate acid.

The balanced chemical equation for the reaction is:

C9H7N + HCl → C9H8N+ + Cl-

At the start of the titration, before any HCl is added, the solution contains only quinoline and its conjugate base. Therefore, the pH of the solution is determined by the pKa of quinoline.

The pKa of quinoline is given by the equation:

pKa = -log(Ka) = -log(8.0×10-10) = 9.10

Therefore, at the start of the titration, the pH of the solution is:

pH = pKa + log([base]/[acid]) 
   = 9.10 + log(0.0350/0) 
   = 9.10

As the titration proceeds, HCl is added to the solution, which reacts with quinoline to form its conjugate acid. At the equivalence point, all of the quinoline has reacted with HCl, and the resulting solution contains only its conjugate acid.

The pH at the equivalence point can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([base]/[acid]) 
   = pKa + log(0/0.0500) 
   = pKa 
   = 9.10

Therefore, the correct answer is B) 3.45.

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