Question

How many grams (to the nearest 0.01 g) of NH4Cl (Mm = 53.49 g/mol) must be added to 550. mL of 0.991-M solution of NH3 in order to prepare a pH = 9.40 buffer?

Answer 1

Kb of NH3 = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.7447

POH = 14 - pH

= 14 - 9.4

= 4.6

use formula for buffer

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

4.6 = 4.7447 + log ([NH4Cl]/[NH3])

log ([NH4Cl]/[NH3]) = -0.1447

[NH4Cl]/[NH3] = 0.7166

[NH4Cl]/0.991 = 0.7166

[NH4Cl] = 0.7101

volume , V = 550.0 mL= 0.550 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.7101*0.55

= 0.3906 mol

Molar mass of NH4Cl = 53.49 g/mol

we have below equation to be used:

mass of NH4Cl,

m = number of mol * molar mass

= 0.3906 mol * 53.49 g/mol

= 20.89 g

Answer: 20.89 g