The preparations of two aqueous solutions are described in the table below. For

Question

The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that NH3 is a weak base. acids: 0.3 mol of HCl is added to 1.0 L of a 0.2 MNH3 solution. bases: other: ? 0.1 mol of NaOH is added to | 1.0 L of a solution that is 0.3 M in both NH, and acids: bases: NH4Br other:

Explanation / Answer

1) HCl(aq)+NH3(aq) <--->NH4Cl(aq) [neutralization of weak base]

Ionic equation: H+(aq) +Cl-(aq) +NH3(aq)< ---> NH4+(aq) +Cl-(aq)

net ionic eqn:H+(aq) +NH3(aq) <--->NH4+(aq)

acids: H+(aq) (or HCl)

bases:NH3(aq)

salts:NH4+(aq) (or NH4Cl)

2) NaOH+NH4Br --->NH3+H2O+NaBr

NaOH+NH3 -->no rxn

net rxn:NaOH+NH4Br +NH3 <--->2NH3+H2O+NaBr

Ionic equation:Na+ +OH- +NH4+ +Br- +NH3 <---->2NH3 +H2O(l) +Na+ +Br-

net ionic equation: NH4+(aq)+OH- <--->NH3+H2O(l)

acids:NH4+ (NH4Br)

bases:OH- ,NH3 (NaOH,NH3)

salts and others:H2O(l)

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