Question

What mass of ammonium chloride should be added to 2.60 L of a 0.165 M NH3 to obtain a buffer with a pH of 9.50? (Kb for NH3 is 1.8×10−5.)

Answer 1

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

POH = 14 - pH

= 14 - 9.5

= 4.5

use formula for buffer

pOH = pKb + log ([NH4Cl]/[NH3])

4.5 = 4.7447 + log ([NH4Cl]/[NH3])

log ([NH4Cl]/[NH3]) = -0.2447

[NH4Cl]/[NH3] = 0.5692

[NH4Cl]/0.165 = 0.5692

[NH4Cl] = 0.0939

volume , V = 2.6 L

use:

number of mol,

n = Molarity * Volume

= 9.392*10^-2*2.6

= 0.2442 mol

Molar mass of NH4Cl,

MM = 1*MM(N) + 4*MM(H) + 1*MM(Cl)

= 1*14.01 + 4*1.008 + 1*35.45

= 53.492 g/mol

use:

mass of NH4Cl,

m = number of mol * molar mass

= 0.2442 mol * 53.49 g/mol

= 13.06 g

Answer: 13.1 g