Question

How many grams (to the nearest 0.01 g) of NH4Cl (Mm = 53.49 g/mol) must be added to 900. mL of 0.919-M solution of NH3 in order to prepare a pH = 9.95 buffer?

=_____g

Answer 1

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.7447

POH = 14 - pH

= 14 - 9.95

= 4.05

use formula for buffer

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

4.05 = 4.7447 + log ([NH4Cl]/[NH3])

log ([NH4Cl]/[NH3]) = -0.6947

[NH4Cl]/[NH3] = 0.202

[NH4Cl] = 0.1856

volume , V = 900 mL= 0.9 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.1856*0.9

= 0.167 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

= 0.167 mol * 53.49 g/mol

= 8.94 g

Answer: 8.94 g