Question

What mass of ammonium chloride should be added to 2.45 L of a 0.145 M NH3 to obtain a buffer with a pH of 9.65? (Kb for NH3 is 1.8×10−5.)

Answer 1

According to Henderson-Hasselbalch equation,

pH = pKa + log([NH3]/[NH4Cl]) ......(1)

pKb= -log( 1.8×10^-5) = 4.74

pKa = 14.00 - pkb

pKa = 14.00-4.74 = 9.26

Number of moles of NH3,

2.45 L × (0.145 mol NH3 / 1L) = 0.355 mol NH3

Substitute all the values in equation (1)

9.65 = 9.26 + log ( 0.355 mol / NH4Cl)

log( 0.355 mol/NH4Cl) = 0.39

0.355 mol / NH4Cl = 10^0.39

0.355 mol / NH4Cl = 2.45

NH4Cl = 0.355 mol / 2.45

NH4Cl = 0.145 mol

Mass of ammonium chloride is,

0.145 mol NH4Cl × ( 53.49 g NH4Cl / 1 mol NH4Cl)

= 7.76 g NH4Cl

Hence amount of NH4Cl added is 7.76 g