Question

What mass of ammonium chloride should be added to 2.50 L of a 0.145 M NH3 in order to obtain a buffer with a pH of 9.65?

Answer 1

Total volume of buffer solution is 2.50 L,

[Base] = [NH3] = 0.145 M

[Acid] = [NH4⁺] = [NH4Cl] = ?

pH = 9.65

hence, pOH = 14 - pH................( pH + pOH = 14)

pOH = 14 - 9.65

pOH = 4.35.

For NH3, Kb = 1.8 x 10^-5.

pKb = -log(Kb) = -log(1.8 x 10^-5) = 4.744

pKb = 4.74

Now Handerson equation is,

pOH = pKb + log{[Salt]/[Base]}

4.35 = 4.74 + log([ NH4Cl] / 0.145)

log([ NH4Cl] / 0.145) = 4.35 - 4.74

log([ NH4Cl] / 0.145) = -0.39

[ NH4Cl] / 0.145 = Antilog(-0.39) ............(Taking antilogarthm to tha base 10)

[ NH4Cl] / 0.145 = 0.407

[ NH4Cl] = 0.407 x 0.145

[ NH4Cl] = 0.059 M

Molecular weight of NH4Cl (Ammonium chloride) = 53.5 g/mol

i.e. 1 M NH4Cl 53.5 g/mol in 1 L............( definition of molarity)

so, 0.059 M NH4Cl say 'A' g/mol .....in 1 L

A = (53.5 x 0.059) / 1

A = 3.157 g/mol........in 1L.

So,A = 3.157 x 2.5 g........... in 2.5 L

A = 7.893 g.

Hence, 7.893 g of NH4Cl need to be added to prepare 2.5 L buffer with 0.145 M NH3 and 9.65 pH.