Question

An aqueous solution contains 0.419 M ammonia (NH₃).

How many mL of 0.305 M perchloric acid would have to be added to 150 mL of this solution in order to prepare a buffer with a pH of 9.380.

mL

Answer 1

Pkb for NH3= 1.8*10-5,

The reaction between ammonia (NH3) and perchloric acid (HClO4)

NH3 + HClO4 ----> NH4+ + ClO4-

1 mole of ammonia requries 1 mole of HClO4 for neutralization.

mole of NH3 in the solution = molarity* volume in L =0.419*150/1000=0.063 moles

let x= volume of perchloric acid , moles of perchloric acid = x*0.305

since pH is in the basic, range, ammonia is excess and HClO4 will be limiting, Hence moles of NH3 remaining after the reaction = 0.063-x*0.305

moles of NH4+ formed= x*0.305, volume after mixing = 0.15+x, concentrations after mixing

[NH4+]= x*0.305/(0.15+x) and NH3= (0.063-x*0.305)(0.150+x)

since pOH= pKb+ log[BH+]/[B]

where [BH+]= [NH4+] and [B]= [NH3]

given pH= 9.38, pOH= 14-9.38= 4.62, pKb= -log (Kb)= 4.74

hence 4.62= 4.74+ log {x*0.305/(0.063-0.305x)}

x*0.305/(0.063-0.305x)= 0.76

x*0.305= 0.76*0.063-0.76*0.305x

x*0.305*1.76= 0.76*0.063

x= 0.089L =89 ml (1000ml= 1L)