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 quantity in moles of h⁺ would be left in the beaker after the reaction goes to completion?

Explanation:

Step1: Analyze the reaction

The reaction between \(C_5H_5N\) (a base) and \(HCl\) (which provides \(H^+\)) is \(C_5H_5N + H^+
ightarrow C_5H_5NH^+\). The moles of \(H^+\) left is calculated by subtracting the moles of \(H^+\) that reacted with \(C_5H_5N\) from the initial moles of \(H^+\).

Step2: Calculate the moles of \(H^+\) left

We know the moles of \(H^+\) that reacted (\(n_{reacted}= 0.0263\) mol) and the initial moles of \(H^+\) (\(n_{initial}=0.0425\) mol). Using the formula \(n_{left}=n_{initial}-n_{reacted}\)

$$n_{left}=0.0425 - 0.0263$$

$$n_{left}=0.0162$$

Answer:

\(0.0162\) moles