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:

Bases:

Other:

Acids:

Bases:

Other:

.09 mol of HBr is added to 1.0L of a .5M NH3 solution.

Acids:

Bases:

Other:

.09 mol of HNO3 is added to 1.0L of a solution that is .5M in both NH3 and NH4Cl.

Acids:

Bases:

Other:

Explanation / Answer

HBr is a strong acid and NH3 is a weak base

So for the first part, the reaction happening will be

NH3 + HBr ------- NH4Br

Acid: HBr, Base: NH3, Salt: NH4Br

moles of HBr = 0.09 moles

Moles of NH3 = Molarity * Volume of solution in L = 0.5M * 1L = 0.5 moles

Moles of NH3 left after reaction = 0.5 - 0.09 = 0.41 moles

Hence NH3 will be the major species present in the equilibrium

2)

HNO3 is a strong acid and NH3 is a weak base

So for the second part, the reaction happening will be

NH3 + HNO3 ------- NH4NO3

Acid: HBr, Base: NH3, Salt: NH4NO3, NH4Cl

moles of HNO3= 0.09 moles

Moles of NH3 = Molarity * Volume of solution in L = 0.5M * 1L = 0.5 moles

Moles of NH3 left after reaction = 0.5 - 0.09 = 0.41 moles

Hence NH4Cl and NH3 will be the major species present in the equilibrium

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