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The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of
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Answer #1

1. 2.3 mol of HBr is added to 1.0 L of 1.2 M NH3.

Mol(s) NH3 = (1.0 L)*(1.2 M) = 1.2 mol.

NH3 is a weak base and reacts with HBr (strong acid) as per the reaction below.

HBr (aq) + NH3 (aq) ------------> NH4Br (aq)

As per the stoichiometric equation above,

1 mol HBr = 1 mol NH3.

Therefore,

1.2 mol NH3 = 1.2 mol HBr.

Mol(s) of HBr left over = (2.3 – 1.2) mol = 1.1 mol.

Moreover, NH4Br is a weak acid and the solution contains 1.2 mol NH4Br as per the stoichiometric equation above.

Therefore, the major species present in the solution = NH4Br (weak acid).

Acids: HBr, NH4Br

Bases: NH3

Others: None

2. 0.2 mol KOH is added to 1.0 L of a solution containing 0.5 M each of NH3 and NH4Cl.

Mols NH3 in the solution = (1.0 L)*(0.5 M) = 0.5 mol.

Mols NH4Cl in the solution = (1.0 L)*(0.5 M) = 0.5 mol.

NH3 is a weak base while NH4Cl is a weak acid. KOH is a strong base.

The acidic and basic components in the solution react as

NH4Cl (aq) + KOH (aq) ---------> NH4OH (aq) + KCl (aq)

As per the stoichiometric equation,

1 mol NH4Cl = 1 mol KOH = 1 mol NH4OH.

NH4OH is a weak base and doesn’t react with the weak acid (NH4Cl).

Mol(s) NH4OH formed = mols KOH present = 0.2 mol.

Mol(s) NH4Cl left over = (0.5 – 0.2) mol = 0.3 mol.

Mol(s) KCl formed = 0.2 mol.

Since the solution contains excess NH3 (base), hence, the major species present in the solution at equilibrium = NH3.

Acids: NH4Cl.

Bases: NH3, NH4OH, KOH

Others: KCl

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