Question

question 1.a of 7
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.
write the net ionic equation for the reaction that takes place. be sure to include the proper phases for all species within the reaction.

Explanation:

Step1: Write the molecular equation

\(C_5H_5N(aq)+HCl(aq)
ightarrow C_5H_5NHCl(aq)\)

Step2: Dissociate ionic compounds

\(C_5H_5N(aq)+H^{+}(aq)+Cl^{-}(aq)
ightarrow C_5H_5NH^{+}(aq)+Cl^{-}(aq)\)

Step3: Cancel spectator ions

\(Cl^{-}\) is the spectator ion.

Answer:

\(C_5H_5N(aq)+H^{+}(aq)
ightarrow C_5H_5NH^{+}(aq)\)