Question

Part A

What mass of ammonium chloride should be added to 2.45 L of a 0.155 M NH3 in order to obtain a buffer with a pH of 9.55?

Express your answer using two significant figures.

Answer 1

Kb = 1.8*10^-5

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

POH = 14 - pH
= 14 - 9.55
= 4.45

use formula for buffer
pOH = pKb + log ([NH4Cl]/[NH3])
4.45 = 4.7447 + log ([NH4Cl]/[NH3])
log ([NH4Cl]/[NH3]) = -0.2947
[NH4Cl]/[NH3] = 0.5073
[NH4Cl]/0.155 = 0.5073
[NH4Cl] = 0.0786
volume , V = 2.45 L


use:
number of mol,
n = Molarity * Volume
= 0.0786*2.45
= 0.1927 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
= 0.1927 mol * 53.492 g/mol
= 10.31 g
Answer: 10. g

Please let me know if you have any doubts by commenting on this answer. I reply very fast. Also please mark the answer as helpful, If it helped