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. 0.45 mol of HCl is added to 1.0L of a 1.0MNH solution acids: ? bases: other: 0.2 mol of KOH is added toacids: 1.0L of a solution that is 0.7 Min both NH, and NH Br 07 Min both NH, and bases: other:

Explanation / Answer

1) Write the balanced chemical equation for the reaction between HCl and NH3 as below.

HCl (aq) + NH3 (aq) ---------> NH4+Cl- (aq)

As per the stoichiometric equation,

1 mole HCl = 1 mole NH3.

Mole(s) of NH3 present initially (1.0 L)*(1.0 M) = 1.0 mole.

Moles NH4+Cl- present at equilibrium = moles HCl added = moles NH3 neutralized = (1.0 – 0.45) mole = 0.55 mole.

NH3 is a weak base while NH4+Cl- is a weak acid. Since NH3 is present in higher amount, hence the major species is NH3.

2) The solution contains both NH3 and NH4Br. KOH reacts with NH4Br (itself a weak acid) to form NH3 (a weak base). KOH (a strong base) reacts with NH4Br to form NH3 as below.

NH4Br (aq) + KOH (aq) --------> NH3 (aq) + KBr (aq) + H2O (l)

As per the stoichiometric equation,

1 mole NH4Br = 1 mole KOH = 1 mole NH3.

Moles of NH3 = moles of NH4Br in 1.0 L solution = (1.0 L)*(0.7 M) = 0.7 mole.

Moles NH4Br reacted with 0.2 mole KOH = moles NH3 formed = moles NH4Br reacted = 0.2 mole.

Moles NH3 at equilibrium = (0.7 + 0.2) mole = 0.9 mole.

Moles NH4Br at equilibrium = (0.7 – 0.2) mole = 0.5 mole.

KBr is neutral.

Since the solution contains excess NH3, the major species is the weak base NH3.

Solution

Species at equilibrium

1

Acids – NH4Cl

Bases – NH3

Others - None

2

Acids – NH4Br

Bases – NH3

Others - KBr

Solution

Species at equilibrium

1

Acids – NH4Cl

Bases – NH3

Others - None

2

Acids – NH4Br

Bases – NH3

Others - KBr

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