Question

Just as pH is the negative logarithm of [H3 O, pKa is the negative logarithm of Ka, The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions: base] [acid] pH — рКа + 1og Notice that the pH of a buffer has a value close to the pKa of the acid, differing only by the logarithm of the concentration ratio base/[acid]. The Henderson-Hasselbalch equation in terms of pOH and pKb is similar. acid base] РОН — рК, + 1оg =

Answer 1

Kb = 1.8*10^-5

pKb = - log (Kb)
= - log(1.8*10^-5)
= 4.745

POH = 14 - pH
= 14 - 8.86
= 5.14

use formula for buffer
pOH = pKb + log ([NH4Cl]/[NH3])
5.140000000000001 = 4.7447 + log ([NH4Cl]/[NH3])
log ([NH4Cl]/[NH3]) = 0.3953
[NH4Cl]/[NH3] = 2.4847
[NH4Cl]/0.1 = 2.4847
[NH4Cl] = 0.2485
volume , V = 1.9 L


use:
number of mol,
n = Molarity * Volume
= 0.2485*1.9
= 0.4721 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
= 0.4721 mol * 53.49 g/mol
= 25.25 g
Answer: 25.2 g