Question

consider the reaction of 75.0 ml of 0.350 m c₅h₅n (kb = 1.7 x 10⁻⁹) with 100.0 ml of 0.425 m hcl.
after 0.0263 moles of c₅h₅n and 0.0425 moles of h⁺ have reacted, what species would be left in the beaker after the reaction goes to completion?
a only c₅h₅nh⁺
b c₅h₅nh⁺ and h⁺
c c₅h₅nh⁺ and c₅h₅n
d c₅h₅n and h⁺
e only c₅h₅n

Explanation:

Step1: Analyze the reaction

The reaction between \(C_5H_5N\) (a base) and \(HCl\) (a strong acid) is \(C_5H_5N + H^+
ightleftharpoons C_5H_5NH^+\).

Step2: Calculate the moles left

We are given that \(0.0263\) moles of \(C_5H_5N\) react with \(0.0425\) moles of \(H^+\).
From the reaction stoichiometry (\(1:1\) ratio between \(C_5H_5N\) and \(H^+\)), the moles of \(C_5H_5NH^+\) formed is \(0.0263\) moles (since \(C_5H_5N\) is the limiting reactant for this part of the reaction shown in the problem - like data).
The moles of \(H^+\) left \(=0.0425 - 0.0263= 0.0162\) moles.

Answer:

B. \(C_5H_5NH^+\) and \(H^+\)