Question

How many moles of NH3 are required in a 1.00L solution to create a buffer with a pH of 9.13 if the solution contains 0.61 moles of NH4Cl? Kb NH3 = 1.8 x 10-5?

Answer 1

For NH4Cl/NH3 (i.e. NH4Cl/NH4OH) buffer,

NH3 + H-OH <------> NH4⁺ + OH^- . For NH3, Kb = 1.8 x 10^-5 .

pH = 9.13 hence, pOH = 14 - 9.13 = 4.87,.........( pH+ pOH = 14)

Kb = 1.8 x 10^-5,

pKb = -log(Kb)

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

pKb = 4.74

[NH4Cl] = 0.61

[NH3] = ?

WE have Handerson's equation as,

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

pOH = pKb + log{[NH₄Cl]/[NH₃]}

Let us put all the known values,

4.87 = 4.74 + log(0.61 / [NH₃])

log(0.61 / [NH₃] ) = 4.87-4.74

log(0.61 / [NH₃] ) = 0.13

0.61 / [NH₃] = Antilog(0.13)..........(say 10^0.13 i.e. antilog to the base 10)

0.61 / [NH₃] = 1.349

[NH₃] = 0.61 / 1.349

[NH₃] = 0.45 mole.

Hence 0.45 moles of NH3 are required in a 1 L solution to create a buffer of pH 9.13 if solution contains 0.61 moles of NH4Cl.