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 1.4 mol of HNO3 is added to 1.0 L of a 1.4 MNH3 solution. bases: other: 0.33 mol of KOH is added to 1.0 L of a solution that is 1.3 M in both NH3 and NH4CI acids: bases: other:

Explanation / Answer

NH3 is weak base ( pkb= 4.75)and HNO3 is strong acid. The reaction between NH3 and HNO3 is given by

NH3+ HNO3---------->NH4NO3. So NH3 and HNO3 react in 1:! moles ratio to give a salt which is due to strong acid HNO3 gives an acidic salt.

in the present case, moles of HNO3= 1.4 and moles of NH3= molarity* volume in liters= 1.4*1= 1.4 moles

so moles ratio of HNO3:NH3= 1:4 :1.4= 1:1

so all the HNO3 and all NH3 gets consumed giving an acidic salt. The salt will be acidic in nature.

moles of NH4Cl formed= 1.4, its concentration= 1.4/1= 1.4M

NH4Cl ionizes in water as NH4Cl+ H2O -------->NH3+H3O+

Kb of NH3= 10(-4.75)= 1.78*10-5

Ka= 10-14/Kb= 10-14/ (1.78*10-5)= 5.62*10-10 = [NH3][H3O+]/[NH4Cl]

let x= drop in concentration of NH4Cl to reach equilibrium

At equilibrium, [NH3]=[H3O+]= x and [NH4Cl]= 1.4-x

Ka= x2/(1.4-x)= 5.62*10-10, looking at the magnitude of RHS of the equation, it is reasonable to assume, 1.4-x approximatelt equal to 1.4

hence x2/1.4= 5.62*10-10, x= 2.8*10-5, pH= -log [H3O+]= 4.55, acidic range

2. NH3 is weak base whose pKb=5,

From Henderson Hasselbalch equation, pOH= pKb+ log [HB+]/[B]

where [HB+]is salt in this case NH4Cl and [B] is base in this NH3

[HB+]= moles/volume in liters= 1.3/1= 1.3 and [B]= 1.3/1=1.3

before addition of KOB, pOH= 4.75+ log [1.3/1/1.3/1] = 4.75

pH= 14-pOH= 14-4.75= 8.25

when KOH is added, it reacts with NH4Cl to form more NH3.

NH4Cl+ KOH -------->NH3+H2O+ KCl.

so more ammonia is formed. This increases the concentration of NH3 and decreases the concentration of NH4Cl.

moles of NH4Cl initially = molarity* volume in liters= 1.3*1= 1.3, NH3= 1.3*1=1.3

moles of KOH added= 0.33, this is the limiting reactant and moles of NH3 formed after reaction of NH4Cl to form additional NH3= 1.3+0.33= 1.63, moles of NH4Cl reamining =1.3-0.33= 0.97

concentrations : NH3= 1.63/1=1.63 and NH4Cl= 0.97/1=0.97

now from Henderson-Hasselbalch equation, pOH= 4.75+log (0.97/1.63)= 4.52

pH= 14-4.52= 8.48

so the pH is in basic range. ( pH lies between 0 to 7, it is acidic, if pH lies between 7 and 14, the solution is basic)

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