Question

1. complete the following equations and label the first reactant as either an acid or a base.

a) hcl₍ₐq₎ + h₂o₍ₗ₎ →
b) co₃²⁻₍ₐq₎ + h₂o₍ₗ₎ →
c) nh₃₍ₐq₎ + h₂o₍ₗ₎ →

Explanation:

Part (a)

Step 1: Identify the reaction type

HCl is a strong acid. In water, it donates a proton (\(H^+\)) to \(H_2O\). The reaction is an acid - base reaction (proton transfer).

Step 2: Write the products

HCl donates \(H^+\) to \(H_2O\) to form \(H_3O^+\) (hydronium ion) and \(Cl^-\) (chloride ion). The equation is \(HCl_{(aq)}+H_2O_{(l)}
ightarrow H_3O^+_{(aq)} + Cl^-_{(aq)}\). Since HCl donates a proton, it is an acid.

Part (b)

Step 1: Analyze the reactant

\(CO_3^{2 - }\) is the conjugate base of \(HCO_3^-\). In water, it will accept a proton from \(H_2O\) (acting as an acid here).

Step 2: Write the reaction

\(CO_3^{2 - }_{(aq)}+H_2O_{(l)}
ightleftharpoons HCO_3^-_{(aq)}+OH^-_{(aq)}\). Since \(CO_3^{2 - }\) accepts a proton, it is a base.

Part (c)

Step 1: Analyze the reactant

\(NH_3\) is a weak base. In water, it accepts a proton from \(H_2O\).

Step 2: Write the reaction

\(NH_{3(aq)}+H_2O_{(l)}
ightleftharpoons NH_4^+_{(aq)}+OH^-_{(aq)}\). Since \(NH_3\) accepts a proton, it is a base.

Answer:

a) \(HCl_{(aq)}+H_2O_{(l)}
ightarrow H_3O^+_{(aq)} + Cl^-_{(aq)}\), \(HCl\) is an acid.
b) \(CO_3^{2 - }_{(aq)}+H_2O_{(l)}
ightleftharpoons HCO_3^-_{(aq)}+OH^-_{(aq)}\), \(CO_3^{2 - }\) is a base.
c) \(NH_{3(aq)}+H_2O_{(l)}
ightleftharpoons NH_4^+_{(aq)}+OH^-_{(aq)}\), \(NH_3\) is a base.