Question

Consider the reaction of a 20.0 ml of 0.220 m c₅h₅nhcl (ka = 5.9 x 10⁻⁶) with 12.0 ml of 0.201 m csoh. write the net ionic equation for the reaction that takes place. be sure to include the proper phases for all species within the reaction.

Answer 1

The reaction between the weak acid C︅H︅NHCl and the strong base CSOH results in the net ionic equation: C︅H︅NH﹚ (aq) + OH﹛ (aq) → C︅H︅NH (aq) + H︂O (l), omitting the spectator ions Cl﹛ and SOH﹚.

Step-by-step explanation:

To write the net ionic equation for the reaction of C₅H₅NHCl with CSOH, we must first identify the strong and weak electrolytes. C₅H₅NHCl is a weak acid and CSOH is presumed to be a strong base. During the reaction, the strong base CSOH will deprotonate the weak acid resulting in the formation of the conjugate base C₅H₅NH and chloride ions.

The balanced molecular equation would look like this:

• C₅H₅NHCl (aq) + CSOH (aq) → C₅H₅NH (aq) + Cl⁻ (aq) + SOH⁺ (aq)

To derive the net ionic equation, we need to remove the spectator ions which do not participate in the reaction. Cl⁻ and SOH⁺ are the spectator ions in this case.

The net ionic equation is:

• C₅H₅NH⁺ (aq) + OH⁻ (aq) → C₅H₅NH (aq) + H₂O (l)